Author Archives: Priyanka

Exercise Ex. 7

Question 1

Without using trigonometric tables, evaluate:

Solution 1

(i)

(ii)              

(iii)

(iv)

(v)

(vi)

Question 2

Without using trigonometric tables, prove that:

(i) 

(ii) 

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)Solution 2

(i)                LHS = cos81° – sin9°

= cos(90° -9°)- sin9° = sin9° – sin9°

= 0 = RHS

(ii)              LHS = tan71° – cot19°

=tan(90° – 19°) – cot19° =cot19° – cot19°

=0 = RHS

(iii)               LHS = cosec80° – sec10°

= cosec(90° – 10°) – sec(10°)

= sec10° – sec10° = 0

= RHS

(iv)               LHS= 

(v)            

(vi)             

(vii)                 

LHS = RHS

(viii) 

(ix)              LHS = (sin65° + cos25°) (sin65° – cos25°)

Question 3(i)

Without using trigonometric tables, prove that:

sin 53^ᵒ cos37^ᵒ + cos53^ᵒ sin37^ᵒ = 1Solution 3(i)

Question 3(ii)

cos 54^ᵒ cos 36^ᵒ – sin 54^ᵒ sin36^ᵒ = 0Solution 3(ii)

Question 3(iii)

sec 70^ᵒ sin 20^ᵒ + cos 20^ᵒ cosec 70^ᵒ = 2Solution 3(iii)

Question 3(iv)

sin 35^ᵒ sin 55^ᵒ – cos 35^ᵒ cos 55^ᵒ = 0Solution 3(iv)

Question 3(v)

(sin 72^ᵒ + cos 18^ᵒ)(sin72^ᵒ – cos18^ᵒ) = 0Solution 3(v)

Question 3(vi)

tan 48^ᵒ tan 23^ᵒ tan 42^ᵒ tan 67^ᵒ = 1Solution 3(vi)

Question 4(i)

Prove that:

Solution 4(i)

Question 4(ii)

Solution 4(ii)

Question 4(iii)

Solution 4(iii)

Question 4(iv)

Solution 4(iv)

Question 4(v)

Solution 4(v)

Question 5(i)

Prove that:

sin θ cos (90^ᵒ – θ) + sin (90^ᵒ – θ) cos θ = 1Solution 5(i)

Question 5(ii)

Solution 5(ii)

Question 5(iii)

Solution 5(iii)

Question 5(iv)

Solution 5(iv)

Question 5(v)

Solution 5(v)

Question 5(vi)

Solution 5(vi)

Question 5(vii)

Solution 5(vii)

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 6(iii)

Solution 6(iii)

Question 6(iv)

cos 1^ᵒ cos2^ᵒ cos3^ᵒ … cos 180^ᵒ= 0Solution 6(iv)

Question 6(v)

Solution 6(v)

Question 7

Prove that:

(i)

(ii) 

(iii)

(iv)

(v)Solution 7

(i)                  LHS =

= RHS

(ii)                LHS =

= RHS

(iii)                  LHS =

= RHS

(iv)             LHS = cosec(65° + ) – sec(25°-  ) – tan(55° –  ) + cot(35° +  )

= RHS

(v)        LHS =

             = 0 + 1 = 1

             = RHS
Question 8(i)

Express each of the following in terms of T-ratios of angles lying between 0^ᵒ and 45^ᵒ:

sin 67^ᵒ + cos 75^ᵒSolution 8(i)

Question 8(ii)

cot 65^ᵒ + tan 49^ᵒSolution 8(ii)

Question 8(iii)

sec 78^ᵒ + cosec 56^ᵒSolution 8(iii)

Question 8(iv)

cosec 54^ᵒ + sin 72^ᵒSolution 8(iv)

Question 9

If A, B, C are the angles of a triangle ABC, prove that:Solution 9

A + B + C = 180°

So, B + C= 180° – A

Question 10

If cos 2θ = sin 4θ, where 2θ and 4θ are acute angles, find the value of θ.Solution 10

Question 11

If sec 2A= cosec (A – 42ᵒ), where 2A is an acute angle, find the value of A.Solution 11

Question 12

If sin3A = cos(A – 26^o), where 3A is an acute angle, find the value of A.Solution 12

Question 13

If tan 2A = cot(A – 12^o), where 2A is an acute angle, find the value of A.Solution 13

Question 14

If sec 4A = cosec(A – 15^o), where 4A is an acute angle, find the value of A.Solution 14

Question 15

Prove that:

Solution 15

Exercise 9A

Question 1:

Steps of construction:
Step 1 : Draw a line segment AB = 6.5 cm
Step 2: Draw a ray AX making ∠ BAX.
Step 3: Along AX mark (4+7) = 11 points
A₁, A₂, A₃, A₄, A₅, A₆, A₇, A₈, A₉, A₁₀, A₁₁, such that
AA₁ = A₁A₂
Step 4: Join A₁₁ and B.
Step 5: Through A₄ draw a line parallel to A₁₁ B meeting AB at C.
Therefore, C is the point on AB, which divides AB in the ratio 4 : 7
On measuring,
AC = 2.4 cm
CB = 4.1 cm

Question 2:

Steps of Construction:
Step 1 : Draw a line segment PQ = 5.8 cm
Step 2: Draw a ray PX making an acute angle QPX.
Step 3: Along PX mark (5 + 3) = 8 points
A₁, A₂, A₃, A₄, A₅, A₆, A₇ and A₈ such that
PA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A6 = A₆A7 = A₇A₈
Step 4: Join A₈Q.
Step 5: From A₅ draw A₅C || A₈Q meeting PQ at C.
C is the point on PQ, which divides PQ in the ratio 5 : 3
On measurement,
PC = 3.6 cm, CQ = 2.2 cm

Question 3:

Steps of construction:
Step 1: Draw a line segment BC = 6 cm
Step 2: With B as centre and radius equal to 5 cm draw an arc.
Step 3: With C as centre and radius equal to 7 cm draw another arc cutting the previous arc at A.
Step 4: Join AB and AC. Thus, ∆ABC is obtained.
Step 5: Below BC draw another line BX.
Step 6: Mark 7 points B₁B₂B₃B₄B₅B₆B₇ such that
BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇
Step 7: Join  B₇C.
Step 8: from B₅, draw B₅D || B₇C.
Step 9: Draw a line DE through D parallel to CA.
Hence ∆ BDE is the required triangle.

Question 4:

Steps of construction:
Step 1: Draw a line segment QR = 6 cm
Step 2: At Q, draw an angle RQA of 60◦.
Step 3: From QA cut off a segment QP = 5 cm.
Join PR. ∆PQR is the given triangle.
Step 4: Below QR draw another line QX.
Step 5: Along QX cut – off equal distances Q₁Q₂Q₃Q₄Q₅
QQ₁ = Q₁Q2 = Q₂Q₃ = Q3Q₄ = Q₄Q₅
Step 6: Join Q₅R.
Step 7: Through Q₃ draw Q₃S || Q₅R.
Step 8: Through S, draw ST || PR.
∆ TQS is the required triangle.

Question 5:

Steps of construction:
Step 1: Draw a line segment BC = 6 cm
Step 2: Draw a right bisector PQ of BC meeting it at M.
Step 3: From QP cut – off a distance MA = 4 cm
Step 4: Join AB, AC.
∆ ABC is the given triangle.
Step 5: Below BC, draw a line BX.
Step 6: Along BX, cut – off 3 equal distances such that
BR₁ = R₁R₂ = R₂R₃ 
Step 7: Join R₂C.
Step 8: Through R₃ draw a line R₃C₁ || R₂C.
Step 9 : Through C₁ draw line C₁A₁ || CA _.
∆ A₁BC₁ is the required triangle.

Question 6:

Steps of Construction:
Step 1: Draw a line segment BC = 5.4 cm
Step 2. At B, draw ∠ CBM = 45°
Step 3: Now ∠ A = 105°, ∠ B = 45°, ∠ C = 180° – (105°+ 45°) = 30°
At C draw ∠ BCA = 30°.
∆ ABC is the given triangle.
Step 4: Draw a line BX below BC.
Step 5: Cut-off equal distances such that  BR₁ = R₁R₂ = R₂R₃ = R₃R₄
Step 6: Join R₃C.
Step 7: Through R₄, draw a line R₄C₁ || R₃C.
Step 8: Through C₁ draw a line C₁A₁ parallel to CA.
∆ A₁BC₁ is the required triangle.

Question 7:

Steps of Construction:
Step 1: Draw a line segment BC = 4 cm
Step 2: Draw a right- angle CBM at B.
Step 3: Cut-off BA = 3cm from BM.
Step 4: Join AC.
ΔABC is the given triangle.
Step 5: Below BC draw a line BX.
Step 6: Along BX, cut-off 7 equal distances such that
BR₁ = R₁R₂ = R₂R₃ = R₃R₄ = R₄R₅ = R₅R₆ = R₆R₇
Step 7: Join R₅C.
Step 8: Through R₇ draw a line parallel to R₅C cutting BC produced at C₁
Step 9: Through C₁ draw a line parallel to CA cutting BA at A₁
∆ A₁BC₁ is the required triangle.

Question 8:

Steps of Construction:
Step 1: draw a line segment BC = 5 cm
Step 2: With B as centre and radius 7cm an arc is drawn.
Step 3: With C as centre and radius 6 cm another arc is drawn intersecting the previous arc at A.
Step 4: Join AB and AC.
Step 5: ∆ ABC is the given triangle.
Step 6: Draw a line BX below BC.
Step 7: Cut- off equal distances from DX such that
BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇
Step 8: join B₅C.
Step 9: Draw a line through B₇ parallel to B₅C cutting BC produced at C’.
Step 10: Through C’ draw a line parallel to CA, cutting BA produced at A’.
Step 11: ∆ A’BC’ is the required triangle.

Question 9:

Steps of construction:
Step 1: Draw a line segment AB = 6.5 cm
Step 2: With B as centre and some radius draw an arc cutting AB at D.
Step 3: With centre D and same radius draw another arc cutting previous arc at E. ∠ ABE = 60°
Step 4: Join BE and produce it to a point X.
Step 5: With centre B and radius 5.5 cm draw an arc intersecting BX at C.
Step 6: Join AC.
∆ ABC is the required triangle.
Step 7: Draw a line AP below AB.
Step 8: Cut- off 3 equal distances such that
AA₁ = A₁A₂ = A₂A₃
Step 9: Join BA₂
Step 10: Draw A₃B’ through A₃ parallel to A₃B.
Step 11: Draw a line parallel to BC through B’ intersecting AY at C’.
∆ AB’C’ is the required triangle.

Question 10:

Steps of construction:
Step 1: Draw a line segment BC = 6.5 cm
Step 2: Draw an angle of 60° at B so that ∠ XBC = 60°.
Step 3: With centre B and radius 4.5cm, draw an arc intersecting XB at A.
Step 4: Join AC.
∆ ABC is the required triangle.
Step 5: Draw a line BY below BC.
Step 6: Cut- off 4 equal distances from BY.
Such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄
Step 7: Join CB₄
Step 8: draw B₃C’ parallel to CB₄
Step 9: Draw C’A’ parallel to CA through C’ intersecting BA produced at A’.
∆ A’BC’ is the required similar triangle.

Question 11:

Steps of Construction:
Step 1: Draw a line segment BC = 9 cm
Step 2: with centre B and radius more than 1/2 BC, draw arcs on both sides of BC.
Step 3: With centre C and same radius draw other arcs on both sides of BC intersecting previous arcs at P and Q.
Step 4: join PQ and produce it to a point X. PQ meets BC at M.
Step 5: With centre M and radius 5 cm, draw an arc intersecting MX at A.
Step 6: Join AB and AC.
∆ ABC is the required triangle.
Step 7: Draw a line BY below BC.
Step 8: Cut off 4 equal distances from BY so that
BB₁ = B₁B₂ = B₂B₃ = B₃B₄
Step 9: Join CB₄
Step 10: Draw C’B₃ parallel to CB₄
Step 11: Draw C’A’ parallel to CA, through C’ intersecting BA at A’.
∆ A’BC’ is the required similar triangle.

Exercise 9B

https://youtube.com/watch?v=HEdUc-q9_es%3Ffeature%3Doembed

Question 1:

Steps of construction:
Step 1: Draw a circle of radius 5 cm with centre O.
Step 2: A point P at a distance of 8cm from O is taken.
Step 3: A right bisector of OP meeting OP at M is drawn.
Step 4: With centre M radius OM a circle is drawn intersecting the previous circle at T₁ and T₂
Step 5: Join PT₁ and PT₂
PT₁ and PT₂ and the required tangents. Measuring PT₁ and PT₂
We find, PT₁= PT₂ =6.2 cm

Question 2:

Steps of construction:
Step 1: Two concentric circles with centre O and radii 4 cm and 6 cm are drawn.
Step 2: A point P is taken on outer circle and O, P are joined.
Step 3: A right bisector of OP is drawn bisecting OP at M.
Step 4: With centre M and radius OM a circle is drawn cutting the inner circle at T₁ and T₂
Step 5: Join PT₁ and PT₂
PT₁ and PT₂ are the required tangents. Further PT₁= PT₂ =4.8 cm

Question 3:

Steps of construction:
Step 1: Draw a circle with centre O and radius 3.5 cm
Step 2: the diameter P₁P₂ is extended to the points A and B such that AO = OB = 7 cm
Step 3: With centre P₁ and radius 3.5 cm draw a circle cutting the first circle at T₁ and T₂
Step 4: join AT₁ and AT₂
Step 5: With centre P₂ and radius 3.5 cm draw another circle cutting the first circle at T₃ and T₄
Step 6: Join BT₃ and BT₄ . Thus AT₁, AT₂ and BT₃, BT₄ are the required tangents to the given circle from A and B.

Question 4:

Steps of construction:
(i) A circle of radius 4.2 cm at centre O is drawn.
(ii) A diameter AB is drawn.
(iii) With OB as base, an angle BOC of 45° is drawn.
(iv) At A, a line perpendicular to OA is drawn.
(v) At C, a line perpendicular to OC is drawn.
(vi) These lines intersect each other at P.
PA and PC are the required tangents.

Question 5:

Steps of construction:
(i) A line segment AB = 8,5 cm is drawn.
(ii) Draw a right bisector of AB which meets AB at M.
(iii) With M as centre AM as radius a circle is drawn intersecting the given circles at T₁, T₂, T₃ and T₄
(iv) Join AT₃, AT₄ and BT₁, BT₂.
Thus AT₃, AT₄, BT₁, BT₂ are the required tangents.

Question 6:

Steps of construction:
(i) Draw a line segment AB = 7cm
(ii) Taking A as centre and radius 3 cm, a circle is drawn.
(iii) With centre B and radius 2.5 cm, another circle is drawn.
(iv) With centre A and radius more than 1/2 AB, arcs are drawn on both sides of AB.
(v) With centre B and the same radius, [as in step (iv)] arcs are drawn on both sides of AB intersecting previous arcs at P and Q.
(vi) Join PQ which meets AB at M.
(vii) With centre M and radius AM, a circle is drawn which intersects circle with centre A at T₁ and T₂ and the circle with centre B at T₃ and T₄
(viii) Join AT₃, AT₄, BT₁ and BT₂
Thus, AT₃, AT₄, BT₁, BT₂ are the required tangents.

Question 7:

Steps of construction:
(i) A circle of radius 3 cm with centre O is drawn.
(ii) A radius OC is drawn making an angle of 60° with the diameter AB.
(iii) At C, ∠OCP = 90° is drawn.
CP is required tangent.

Question 8:

Steps of construction :
(i) Draw a circle of radius 4 cm with centre O.
(ii) With diameter AB, a line OC is drawn making an angle of 30° i.e., ∠BOC = 30°
(iii) At C a perpendicular to OC is drawn meeting OB at P.
PC is the required tangent.

Practice Test – MCQs test series for Term 2 ExamsENROLL NOW×Contact UsOrRequest a callAsk a DoubtLog InSign Up

Exercise Ex. 8A

Question 1

Find the length of tangent drawn to a circle with radius 8 cm from a point 17 cm away from the centre of the circle.Solution 1

PA is the tangent to the circle with center O and radius AO = 8 cm. The point P is at a distance of 17 cm from O.

In PAO, A = 90

By Pythagoras theorem:

Hence, the length of the tangent = 15 cm.Question 2

A point P is 25 cm away from the centre of a circle and the length of tangent drawn from P to the circle is 24 cm. Findthe radius of the circle.Solution 2

PA is the tangent to the circle with centre O and radius, such that PO = 25 cm, PA = 24 cm

In PAO, A = 90,

By Pythagoras theorem:

Hence, the radius of the circle is 7 cm.Question 3

Two concentric circles are of radii 6.5 cm and 2.5 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Solution 3

Question 4

In the given figure, a circle inscribed in a triangle ABC, touches the sides AB, BC and AC at points D, E and F respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, find the lengths of AD, BE and CF.

Solution 4

Question 5

In the given figure, PA and PB are the tangent segments to a circle with centre O. Show that the points A, O, B and P are concyclic.

Solution 5

Given AP is a tangent at A and OA is radius through A and PA and PB are the tangent segments to circle with centre O.

Therefore, OA is perpendicular to AP, similarly, OB is perpendicular to BP.

OAP = 90

And OBP = 90

So, OAP = OBP = 90

OBP + OAP = (90 + 90) = 180

Thus, the sum of opposite angles of quad. AOBP is 180

AOBP is a cyclic quadrilateralQuestion 6

In the given figure, the chord AB of the larger of the two concentric circles, with centre O, touches the smaller circle at C. Prove that AC = CB.

Solution 6

Question 7

From an external point P, tangents PA and PB are drawn to a circle with centre O. if CD is the tangent to the circle at a point E and PA = 14 cm, find the perimeter of PCD.

Solution 7

Given: From an external point P, tangent PA and PB are drawn to a circle with centre O. CD is the tangent to the circle at a point E and PA = 14cm.

Since the tangents from an external point are equal, we have

PA = PB,

Also, CA = CE and DB = DE

Perimeter of PCD = PC + CD + PD

=(PA – CA) + (CE + DE) +(PB – DB)

= (PA – CE) + (CE + DE) + (PB – DE)

= (PA + PB) = 2PA = (2 14) cm

= 28 cm

Hence, Perimeter of PCD = 28 cmQuestion 8

A circle is inscribed in a ABC, touching AB, BC and AC at P, Q and R respectively. If AB = 10 cm, AR = 7cm and CR = 5 cm, find the length of BC.

Solution 8

A circle is inscribed in a triangle ABC touching AB, BC and CA at P, Q and R respectively.

Also, AB = 10 cm, AR = 7cm, CR = 5cm

AR, AP are the tangents to the circle

AP = AR = 7cm

AB = 10 cm

BP = AB – AP = (10 – 7)= 3 cm

Also, BP and BQ are tangents to the circle

BP = BQ = 3 cm

Further, CQ and CR are tangents to the circle

CQ = CR = 5cm

BC = BQ + CQ = (3 + 5) cm = 8 cm

Hence, BC = 8 cm
Question 9

In the given figure, a circle touches all the four sides of a quadrilateral ABCD whose three sides are AB = 6cm, BC = 7cm and CD = 4 cm. Find AD.

Solution 9

Let the circle touches the sides AB, BC, CD and DA at P, Q, R, S respectively

We know that the length of tangents drawn from an exterior point to a circle are equal

AP = AS —-(1)    {tangents from A}

BP = BQ —(2)     {tangents from B}

CR = CQ —(3)    {tangents from C}

DR = DS—-(4)    {tangents from D}

Adding (1), (2) and (3) we get

AP + BP + CR + DR = AS + BQ + CQ + DS

(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)

AB + CD = AD + BC

AD = (AB + CD) – BC = {(6 + 4) – 7} cm = 3 cm

Hence, AD = 3 cm
Question 10

In the given figure, an isosceles triangle ABC with AB = AC, circumscribes a circle. Prove that the point of contact P bisects the base BC.

Solution 10

Question 11

In the given figure, O is the centre of two concentric circles of radii 4 cm and 6 cm respectively. PA and PB are tangents to the outer and inner circle respectively. If PA = 10 cm, find the length of PB up to one place of decimal

Solution 11

Given O is the centre of two concentric circles of radii 4 cm and 6 cm respectively. PA and PB are tangents to the outer and inner circle respectively. PA = 10cm. Join OA, OB and OP.

Then, OB = 4 cm, OA= 6 cm and PA = 10 cm

In triangle OAP,

Hence, BP = 10.9 cm
Question 12

In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 3 cm such that the segments BD and DC into which BC is divided by the point of contact D, are of lengths 6 cm and 9 cm respectively. If the area of ∆ABC = 54 cm² then find the lengths of sides AB and AC.

Solution 12

Question 13

PQ is a chord of length 4.8 cm of a circle of radius 3 cm. The tangents at P and Q intersect at a point T as shown in the figure. Find the length of TP.

Solution 13

Question 14

Prove that the line segment joining the points of contact of two parallel tangents of a circle, passes through its centre.Solution 14

Question 15

In the given figure, a circle with centre O, is inscribed in a quadrilateral ABCD such that it touches the side BC, AB, AD and CD at points P, Q, R and S respectively. If AB = 29 cm, AD = 23 cm, ∠B = 90° and DS = 5 cm then find the radius of the circle.

Solution 15

Question 16

In the given figure, O is the centre of the circle and TP is the tangent to the circle from an external point T. If ∠PBT = 30°, prove that BA: AT = 2:1.

Solution 16

Exercise Ex. 8B

Question 1

In the adjoining figure, a circle touches all the four sides of a quadrilateral ABCD whose sides are AB = 6 cm, BC = 9 cm and CD = 8 cm. Find the length of side AD.

Solution 1

Question 2

In the given figure, PA and PB are two tangents to the circle with centre O. If ∠APB = 50° then what is the measure of ∠OAB.

Solution 2

Question 3

In the given figure, O is the centre of a circle. PT and PQ are tangents to the circle from an external point P. If ∠TPQ = 70°, find ∠TRQ.

Solution 3

Question 4

In the given figure, common tangents AB and CD to the two circles with centres O₁ and O₂ intersect at E. Prove that AB = CD.

Solution 4

Question 5

If PT is a tangent to a circle with centre O and PQ is a chord of the circle such that ∠QPT = 70°, then find the measure of ∠POQ.

Solution 5

Question 6

In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 2 cm such that the segments BD and DC into which BC is divided by the point of contact D, are of lengths 4 cm and 3 cm respectively. If the area of ∆ABC = 21 cm² then find the lengths of sides AB and AC.

Solution 6

Question 7

Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle (in cm) which touches the smaller circle.Solution 7

Question 8

Prove that the perpendicular at the point of contact of the tangent to a circle passes through the centre.Solution 8

Question 9

In the given figure, two tangents RQ and RP are drawn from an external point R to the circle with centre O. If ∠PRQ =120°, then prove that OR = PR + RQ.

Solution 9

Question 10

In the given figure, a circle inscribed in a triangle ABC touches the sides AB, BC and CA at points D, E and F respectively. If AB = 14 cm, BC = 8 cm and CA = 12 cm. Find the lengths AD, BE and CF.

Solution 10

Question 11

In the given figure, O is the centre of the circle. PA and PB are tangents. Show that AOBP is a cyclic quadrilateral.

Solution 11

Question 12

In two concentric circles, a chord of length 8 cm of the larger circle touches the smaller circle. If the radius of the larger circle is 5 cm then find the radius of the smaller circle.Solution 12

Question 13

In the given figure, PQ is a chord of a circle with centre O and PT is a tangent. If ∠QPT = 60°, find ∠PRQ.

Solution 13

Question 14

In the given figure, PA and PB are two tangents to the circle with centre O. If ∠APB = 60° then find the measure of ∠OAB.

Solution 14

Exercise MCQ

Question 1

The number of tangents that can be drawn from an external circle is

(a) 1

(b) 2

(c) 3

(d) 4Solution 1

Correct option : (b)

We can draw only 2 tangents from an external point to a circle.Question 2

In the given figure, RQ is a tangent to the circle with centre O. If SQ = 6 cm and QR = 4 cm, then OR is equal to

(a) 2.5 cm

(b) 3 cm

(c) 5 cm

(d) 8 cmSolution 2

Question 3

In a circle of radius 7 cm, tangent PT is drawn from a point P such that PT = 24 cm. If 0 is the centre of the circle, then length OP = ?

(a) 30 cm

(b) 28 cm

(c) 25 cm

(d) 18 cmSolution 3

Question 4

Which of the following pairs of lines in a circle cannot be parallel?

(a) two chords

(b) a chord and a tangent

(c) two tangents

(d) two diametersSolution 4

Correct option: (d)

The diameter of the circle always passes through the centre. This means all the diameters of a given circle will intersect at the centre, and hence they cannot be parallel.Question 5

The chord of a circle of radius 10 cm subtends a right angle at its centre. The length of the chord (in cm) is

Solution 5

Question 6

In the given figure, PT is a tangent to the circle with centre O. If OT = 6 cm and OP =10 cm, then the length of tangent PT is

(a) 8 cm

(b) 10 cm

(c) 12 cm

(d) 16 cmSolution 6

Question 7

In the given figure, point P is 26 cm away from the centre O of a circle and the length PT of the tangent drawn from P to the circle is 24 cm. Then, the radius of the circle is

(a) 10 cm

(b) 12 cm

(c) 13 cm

(d) 15 cmSolution 7

Question 8

PQ is a tangent to a circle with centre O at the point P. If ∆OPQ is an isosceles triangle, then ∠OQP is equal to

(a) 30°

(b) 45°

(c) 60°

(d) 90°Solution 8

Question 9

In the given figure, AB and AC are tangents to the circle with centre O such that ∠BAC = 40°.

Then, ∠BOC is equal to

(a) 80°

(b) 100°

(c) 120°

(d) 140°

Solution 9

Question 10

If a chord AB subtends an angle of 60° at the centre of a circle, then the angle between the tangents to the circle drawn from A and B is

(a) 30° 

(b) 60° 

(c) 90° 

(d) 120° Solution 10

Question 11

In the given figure, O is the centre of two concentric circles of radii 6 cm and 10 cm. AB is a chord of outer circle which touches the inner circle. The length of chord AB is

(a) 8 cm

(b) 14 cm

(c) 16 cm

(d) 136 cm

Solution 11

Question 12

In the given figure, AB and AC are tangents to a circle with centre O and radius 8 cm. If OA =17 cm, then the length of AC (in cm) is

(a) 9

(b) 15

(c)  

(d) 25

Solution 12

Question 13

In the given figure, O is the centre of a circle, AOC is its diameter such that ∠ACB = 50°. If AT is the tangent to the circle at the point A then ∠BAT=?

(a) 40°

(b) 50°

(c) 60°

(d) 65° 

Solution 13

Question 14

In the given figure, O is the centre of a circle, PQ is a chord and PT is the tangent at P. If ∠POQ = 70°, then ∠TPQ is equal to

(a) 35°

(b) 45°

(c) 55°

(d) 70°

Solution 14

Question 15

In the given figure, AT is a tangent to the circle with centre O such that OT = 4 cm and ∠OTA = 30°. Then, AT= ?

(a) 4 cm

(b) 2 cm

(c)  

(d) 

Solution 15

Question 16

If PA and PB are two tangents to a circle with centre O such that ∠AOB =110° then ∠APB is equal to

(a) 55°

(b) 60°

(c) 70°

(d) 90°

Solution 16

Question 17

In the given figure, the length of BC is

(a) 7 cm

(b) 10 cm

(c) 14 cm

(d) 15 cm

Solution 17

Question 18

In the given figure, if ∠AOD = 135° then ∠BOC is equal to

(a) 25°

(b) 45°

(c) 52.5°

(d) 62.5°

Solution 18

Question 19

In the given figure, O is the centre of a circle and PT is the tangent to the circle. If PQ is a chord such that ∠QPT = 50° then ∠POQ = ?

(a) 100°

(b) 90°

(c) 80°

(d) 75° 

Solution 19

Question 20

In the given figure, PA and PB are two tangents to the circle with centre O. If ∠APB = 60° then ∠OAB is

(a) 15°

(b) 30°

(c) 60°

(d) 90° 

Solution 20

Question 21

If two tangents inclined at an angle of 60° are drawn to a circle of radius 3 cm then the length of each tangent is

Solution 21

Question 22

In the given figure, PQ and PR are tangents to a circle with centre A. If ∠QPA = 27° then ∠QAR equals

(a) 63°

(b) 117°

(c) 126°

(d) 153° 

Solution 22

Question 23

In the given figure, PA and PB are two tangents drawn from an external point P to a circle with centre C and radius 4 cm. If PA ⊥ PB, then the length of each tangent is

(a) 3 cm

(b) 4 cm

(c) 5 cm

(d) 6 cm

Solution 23

Question 24

If PA and PB are two tangents to a circle with centre O such that ∠APB = 80°. Then, ∠AOP=?

(a) 40°

(b) 50°

(c) 60°

(d) 70° 

Solution 24

Question 25

In the given figure, Q is the centre of the circle. AB is the tangent to the circle at the point P. If ∠APQ = 58° then the measure of ∠PQB is

(a) 32°

(b) 58°

(c) 122°

(d) 132° 

Solution 25

Question 26

In the given figure, O is the centre of the circle. AB is the tangent to the circle at the point P. If ∠PAO = 30° then ∠CPB + ∠ACP is equal to

(a) 60°

(b) 90°

(c) 120°

(d) 150° 

Solution 26

Question 27

In the given figure, PQ is a tangent to a circle with centre O. A is the point of contact. If ∠PAB = 67°, then the measure of ∠AQB is

(a) 73°

(b) 64°

(c) 53°

(d) 44° 

Solution 27

Question 28

In the given figure, two circles touch each other at C and AB is a tangent to both the circles. The measure of ∠ACB is

(a) 45°

(b) 60°

(c) 90°

(d) 120° 

Solution 28

Question 29

O is the centre of a circle of radius 5 cm. At a distance of 13 cm from O, a point P is taken. From this point, two tangents PQ and PR are drawn to the circle. Then, the area of quad. PQOR is

(a) 60 cm²

(b) 32.5 cm²

(c) 65 cm²

(d) 30 cm²

Solution 29

Question 30

In the given figure, PQR is a tangent to the circle at Q, whose centre is O and AB is a chord parallel to PR such that ∠BQR = 70°. Then, ∠AQB = ?

(a) 20°

(b) 35°

(c) 40°

(d) 45°

Solution 30

Question 31

The length of the tangent from an external point P to a circle of radius 5 cm is 10 cm. The distance of the point from the centre of the circle is

Solution 31

Question 32

In the given figure, O is the centre of a circle, BOA is its diameter and the tangent at the point P meets BA extended at T. If ∠PBO = 30° then ∠PTA =?

(a) 60°

(b) 30°

(c) 15°

(d) 45°

Solution 32

Question 33

In the given figure, a circle touches the side DF of ∆EDF at H and touches ED and EF produced at K and M respectively. If EK = 9 cm then the perimeter of ∆EDF is

(a) 9 cm

(b) 12 cm

(c) 13.5 cm

(d) 18 cm

Solution 33

Question 34

To draw a pair of tangents to a circle, which are inclined to each other at an angle of 45°, we have to draw tangents at the end points of those two radii, the angle between which is

(a) 105°

(b) 135°

(c) 140°

(d) 145°Solution 34

Question 35

In the given figure, O is the centre of a circle; PQL and PRM are the tangents at the points Q and R respectively and S is a point on the circle such that ∠SQL = 50° and ∠SRM = 60°. Then, ∠QSR =?

(a) 40°

(b) 50°

(c) 60°

(d) 70°

Solution 35

Question 36

In the given figure, a triangle PQR is drawn to circumscribe a circle of radius 6 cm such that the segments QT and TR into which QR is divided by the point of contact T, are of lengths 12 cm and 9 cm respectively. If the area of ∆PQR = 189 cm² then the length of side PQ is

(a) 17.5 cm

(b) 20 cm

(c) 22.5 cm

(d) 25 cm

Solution 36

Question 37

In the given figure, QR is a common tangent to the given circles, touching externally at the point T. The tangent at T meets QR at P. If PT = 3.8 cm then the length of QR is (a) 1.9 cm

(b) 3.8 cm

(c) 5.7 cm

(d) 7.6 cm

Solution 37

Question 38

In the given figure, quad. ABCD is circumscribed, touching the circle at P, Q, R and S. If AP = 5 cm, BC =7 cm and CS = 3 cm. Then, the length AB =?

(a) 9 cm

(b) 10 cm

(c) 12 cm

(d) 8 cm

Solution 38

Question 39

In the given figure, quad. ABCD is circumscribed, touching the circle at P, Q, R and S. If AP = 6cm, BP = 5 cm, CQ = 3 cm and DR = 4 cm then perimeter of quad. ABCD is

(a) 18 cm

(b) 27 cm

(c) 36 cm

(d) 32 cm

Solution 39

Question 40

In the given figure, O is the centre of a circle, AB is a chord and AT is the tangent at A. If ∠AOB =100° then ∠BAT is equal to

(a) 40°

(b) 50°

(c) 90°

(d) 100°

Solution 40

Question 41

In a right triangle ABC, right-angled at B, BC = 12 cm and AB =5 cm. The radius of the circle inscribed in the triangle is

(a) 1 cm

(b) 2 cm

(c) 3 cm

(d) 4 cmSolution 41

Question 42

In the given figure, a circle is inscribed in a quadrilateral ABCD touching its sides AB, BC, CD and AD at P, Q, R and S respectively. If the radius of the circle is 10 cm, BC = 38 cm, PB = 27 cm and AD ⊥ CD then the length of CD is

(a) 11 cm

(b) 15 cm

(c) 20 cm

(d) 21 cm

Solution 42

Question 43

In the given figure, ∆ABC is right-angled at B such that BC= 6 cm and AB = 8 cm. A circle with centre O has been inscribed inside the triangle. OP ⊥ AB, OQ ⊥ BC and OR ⊥ AC. If OP = OQ =OR= x cm then x = ?

(a) 2 cm

(b) 2.5 cm

(c) 3 cm

(d) 3.5 cm

Solution 43

Question 44

Quadrilateral ABCD is circumscribed to a circle. If AB = 6 cm, BC =7 cm, and CD = 4 cm then the length of AD is

(a) 3 cm

(b) 4 cm

(c) 6 cm

(d) 7 cmSolution 44

Question 45

In the given figure, PA and PB are tangents to the given circle such that PA = 5 cm and ∠APB = 60°. The length of chord AB is

Solution 45

Question 46

In the given figure, DE and DF are tangents from an external point D to a circle with centre A. If DE = 5 cm and DE ⊥ DF then the radius of the circle is

(a) 3 cm

(b) 4 cm

(c) 5 cm

(d) 6 cm

Solution 46

Question 47

In the given figure, three circles with centres A, B, C respectively touch each other externally. If AB = 5 cm, BC = 7 cm and CA = 6 cm then the radius of the circle with centre A is

(a) 1.5 cm

(b) 2 cm

(c) 2.5 cm

(d) 3 cm

Solution 47

Question 48

In the given figure, AP, AQ and BC are tangents to the circle. If AB = 5 cm, AC = 6 cm and BC = 4 cm then the length of AP is

(a) 15 cm

(b) 10 cm

(c) 9 cm

(d) 7.5 cm

Solution 48

Question 49

In the given figure, O is the centre of two concentric circles of radii 5 cm and 3 cm. From an external point P tangents PA and PB are drawn to these circles. If PA =12 cm then PB is equal to

Solution 49

Question 50

Which of the following statements is not true?

a. If a point P lies inside a circle, no tangent can be drawn to the circle, passing through P.

b. If a point P lies on the circle, then one and only one tangent can be drawn to the circle at P.

c. If a point P lies outside the circle, then only two tangents can be drawn to the circle from P.

d. A circle can have more than two parallel tangents, parallel to a given line.Solution 50

Correct option: (d)

Options (a), (b) and (c) are all true.

However, option (d) is false since we can draw only parallel tangents on either side of the diameter, which would be parallel to a given line.Question 51

Which of the following statements is not true?

1. A tangent to a circle intersects the circle exactly at one point.
2. The point common to the circle and its tangent is called the point of contact.
3. The tangent at any point of a circle is perpendicular to the radius of the circle through the point of contact.
4. A straight line can meet a circle at one point only.

Solution 51

Correct option: (d)

Options (a), (b) and (c) are all true.

However, option (d) is false since a straight line can meet a circle at two points even as shown below.

Question 52

Which of the following statements is not true?

1. A line which intersects a circle in two points, is called a secant of the circle.
2. A line intersecting a circle at one point only, is called a tangent to the circle.
3. The point at which a line touches the circle, is called the point of contact.
4. A tangent to the circle can be drawn from a point inside the circle.

Solution 52

Correct option: (d)

Options (a), (b) and (c) are true.

However, option (d) is false since it is not possible to draw a tangent from a point inside a circle.Question 53

Assertion-and-Reason

Type Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:

1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
3. Assertion (A) is true and Reason (R) is false.
4. Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
At a point P of a circle with centre O and radius 12 cm, a tangent  PQ of length 16 cm is drawn. Then, OQ = 20 cm.	The tangent at any point of a circle is perpendicular to the radius through the point of contact.

The correct answer is (a)/(b)/(c)/(d).Solution 53

Question 54

Assertion (A)	Reason (R)
If two tangents are drawn to a circle from an external point then they subtend equal angles at the centre.	A parallelogram circumscribing a circle is a rhombus.

The correct answer is (a)/(b)/(c)/(d).Solution 54

Question 55

Assertion (A)	Reason (R)
In the given figure, a quad. ABCD is drawn to circumscribe a given circle, as shown.Then, AB + BC = AD + DC	In two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.

The correct answer is (a) / (b) / (c) / (d).Solution 55

Exercise FA

Question 1

In the given figure, O is the centre of a circle, PQ is a chord and the tangent PT at P makes an angle of 50° with PQ. Then, ∠POQ = ?

(a) 130°

(b) 100°

(c) 90°

(d) 75°

Solution 1

Question 2

If the angle between two radii of a circle is 130° then the angle between the tangents at the ends of the radii is

(a) 65°

(b) 40°

(c) 50°

(d) 90°Solution 2

Question 3

If tangents PA and PB from a point P to a circle with centre O are drawn so that ∠APS = 80° then ∠POA = ?

(a) 40°

(b) 50°

(c) 80°

(d) 60°

Solution 3

Question 4

In the given figure, AD and AE are the tangents to a circle with centre O and BC touches the circle at F. If AE = 5 cm then perimeter of  ∆ABC is

(a) 15 cm

(b) 10 cm

(c) 22.5 cm

(d) 20 cm

Solution 4

Question 5

In the given figure, a quadrilateral ABCD is drawn to circumscribe a circle such that its sides AB, BC, CD and AD touch the circle at P, Q, R and S respectively. If AB = x cm, BC = 7 cm, CR = 3 cm and AS = 5 cm, find x.

Solution 5

Question 6

In the given figure, PA and PB are the tangents to a circle with centre O. Show that the points A, O, B, P are concyclic.

Solution 6

Question 7

In the given figure, PA and PB are two tangents from an external point P to a circle with centre O. If ∠PBA = 65°, find ∠OAB and ∠APB.

Solution 7

Question 8

Two tangent segments BC and BD are drawn to a circle with centre O such that ∠CBD = 120°. Prove that OB = 2BC.

Solution 8

Question 9

Fill in the blanks.

i. A line intersecting a circle in two distinct points is called a ______ .

ii. A circle can have ________  parallel tangents at the most.

iii. The common point of a tangent to a circle and the circle is called the ________ .

iv. A circle can have _________ tangents.Solution 9

1. A line intersecting a circle in two distinct points is called a secant.
2. A circle can have two parallel tangents at the most.
3. This is since we can draw only parallel tangents on either side of a diameter.
4. The common point of a tangent to a circle and the circle is called the point of contact.
5. A circle can have infinitely many tangents.

Question 10

Prove that the lengths of two tangents drawn from an external point to a circle are equal.Solution 10

Question 11

Prove that the tangents drawn at the ends of the diameter of a circle are parallel.Solution 11

Question 12

In the given figure, if AB = AC, prove that BE = CE.

Solution 12

Question 13

If two tangents are drawn to a circle from an external point, show that they subtend equal angles at the centre.Solution 13

Question 14

Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord.Solution 14

Question 15

Prove that the parallelogram circumscribing a circle, is a rhombus.Solution 15

Question 16

Two concentric circles are of radii 5 cm and 3 cm respectively. Find the length of the chord of the larger circle which touches the smaller circle.Solution 16

Question 17

A quadrilateral is drawn to circumscribe a circle. Prove that the sums of opposite sides are equal.Solution 17

Question 18

Prove that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.Solution 18

Question 19

Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segments joining the points of contact at the centre.Solution 19

Question 20

PQ is a chord of length 16 cm of a circle of radius 10 cm. The tangents at P and Q intersect at a point T as shown in the figure. Find the length of TP.Solution 20

Exercise MCQ

Question 1

Solution 1

Question 2

In a ΔABC, if ∠A – ∠B = 42° and ∠B – ∠C = 21° then ∠B = ?

(a) 32° 

(b) 63° 

(c) 53° 

(d) 95° Solution 2

Correct option: (c)

∠A – ∠B = 42° 

⇒ ∠A = ∠B + 42° 

∠B – ∠C = 21° 

⇒ ∠C = ∠B – 21° 

In ΔABC,

∠A + ∠B + ∠C = 180° 

⇒ ∠B + 42° + ∠B + ∠B – 21° = 180° 

⇒ 3∠B = 159

⇒ ∠B = 53° Question 3

In a ΔABC, side BC is produced to D. If ∠ABC = 50° and ∠ACD = 110° then ∠A = ?

(a) 160° 

(b) 60° 

(c) 80° 

(d) 30° Solution 3

Correct option: (b)

∠ACD = ∠B + ∠A (Exterior angle property)

⇒ 110° = 50° + ∠A

⇒ ∠A = 60° Question 4

Side BC of ΔABC has been produced to D on left hand side and to E on right hand side such that ∠ABD = 125° and ∠ACE = 130°. Then ∠A = ?

1. 50°
2. 55°
3. 65°
4. 75°

Solution 4

Correct option: (d)

Question 5

In the given figure, sides CB and BA of ΔABC have been produced to D and E respectively such that ∠ABD = 110° and ∠CAE = 135°. Than ∠ACB =?

1. 65°
2. 45°
3. 55°
4. 35°

Solution 5

Correct option: (a)

Question 6

The sides BC, CA and AB of ΔABC have been produced to D,E and F respectively.  ∠BAE + ∠CBF + ∠ACD =?

1. 240°
2. 300°
3. 320°
4. 360°

Solution 6

Question 7

In the given figure, EAD ⊥ BCD. Ray FAC cuts ray EAD at a point A such that ∠EAF = 30°. Also, in ΔBAC, ∠BAC = x° and ∠ABC = (x + 10)°. Then, the value of x is

(a) 20

(b) 25

(c) 30

(d) 35Solution 7

Correct option: (b)

∠EAF = ∠CAD (vertically opposite angles)

⇒ ∠CAD = 30° 

In ΔABD, by angle sum property

∠A + ∠B + ∠D = 180° 

⇒ (x + 30)° + (x + 10)° + 90° = 180° 

⇒ 2x + 130° = 180° 

⇒ 2x = 50° 

⇒ x = 25° Question 8

In the given figure, two rays BD and CE intersect at a point A. The side BC of ΔABC have been produced on both sides to points F and G respectively. If ∠ABF = x°, ∠ACG = y° and ∠DAE = z° then z = ?

(a) x + y – 180

(b) x + y + 180

(c) 180 – (x + y)

(d) x + y + 360° Solution 8

Correct option: (a)

∠ABF + ∠ABC = 180° (linear pair)

⇒ x + ∠ABC = 180° 

⇒ ∠ABC = 180° – x

∠ACG + ∠ACB = 180° (linear pair)

⇒ y + ∠ACB = 180° 

⇒ ∠ACB = 180° – y

In ΔABC, by angle sum property

∠ABC + ∠ACB + ∠BAC = 180° 

⇒ (180° – x) + (180° – y) + ∠BAC = 180° 

⇒ ∠BAC – x – y + 180° = 0

⇒ ∠BAC = x + y – 180° 

Now, ∠EAD = ∠BAC (vertically opposite angles)

⇒ z = x + y – 180°  Question 9

In the given figure, lines AB and CD intersect at a point O. The sides CA and OB have been produced to E and F respectively. such that ∠OAE = x° and ∠ DBF = y°.

If ∠OCA = 80°, ∠COA = 40° and ∠BDO = 70° then x° + y° = ?

(a) 190° 

(b) 230° 

(c) 210° 

(d) 270° Solution 9

Correct option: (b)

In ΔOAC, by angle sum property

∠OCA + ∠COA + ∠CAO = 180° 

⇒ 80° + 40° + ∠CAO = 180° 

⇒ ∠CAO = 60° 

∠CAO + ∠OAE = 180° (linear pair)

⇒ 60° + x = 180° 

⇒ x = 120° 

∠COA = ∠BOD (vertically opposite angles)

⇒ ∠BOD = 40° 

In ΔOBD, by angle sum property

∠OBD + ∠BOD + ∠ODB = 180° 

⇒ ∠OBD + 40° + 70° = 180° 

⇒ ∠OBD = 70° 

∠OBD + ∠DBF = 180° (linear pair)

⇒ 70° + y = 180° 

⇒ y = 110° 

∴ x + y = 120° + 110° = 230° Question 10

In a ΔABC it is given that ∠A:∠B:∠C = 3:2:1 and ∠ACD = 90^o. If it is produced to E, Then ∠ECD =?

1. 60°
2. 50°
3. 40°
4. 25°

Solution 10

Question 11

In the given figure , BO and CO are the bisectors of ∠B and ∠C respectively. If ∠A = 50°, then ∠BOC= ?

1. 130°
2. 100°
3. 115°
4. 120°

Solution 11

Question 12

In the given figure, side BC of ΔABC has been produced to a point D. If ∠A = 3y°, ∠B = x°, ∠C = 5y° and ∠ACD = 7y°. Then, the value of x is

(a) 60

(b) 50

(c) 45

(d) 35Solution 12

Correct option: (a)

∠ACB + ∠ACD = 180° (linear pair)

⇒ 5y + 7y = 180° 

⇒ 12y = 180° 

⇒ y = 15° 

Now, ∠ACD = ∠ABC + ∠BAC (Exterior angle property)

⇒ 7y = x + 3y

⇒ 7(15°) = x + 3(15°)

⇒ 105° = x + 45° 

⇒ x = 60° 

Exercise Ex. 8

Question 1

In ABC, if B = 76^o and C = 48^o, find A.Solution 1

Since, sum of the angles of a triangle is 180^o

A + B + C = 180^o

 A + 76^o + 48^o = 180^o

 A = 180^o – 124^o = 56^o

 A = 56^oQuestion 2

The angles of a triangle are in the ratio 2:3:4. Find the angles.Solution 2

Let the measures of the angles of a triangle are (2x)^o, (3x)^o and (4x)^o.

Then, 2x + 3x + 4x = 180         [sum of the angles of a triangle is 180^o ]     

 9x = 180

 The measures of the required angles are:

2x = (2  20)^o = 40^o

3x = (3  20)^o = 60^o

4x = (4  20)^o = 80^oQuestion 3

In ABC, if 3A = 4B = 6C, calculate A, B and C.Solution 3

Let 3A = 4B = 6C = x (say)

Then, 3A = x

 A = 

4B = x

and 6C = x

 C = 

As A + B + C = 180^o

 A = 

B = 

C = Question 4

In ABC, if A + B = 108^o and B + C = 130^o, find A, B and C.Solution 4

A + B = 108^o [Given]

But as A, B and C are the angles of a triangle,

A + B + C = 180^o

 108^o + C = 180^o

 C = 180^o – 108^o = 72^o

Also, B + C = 130^o [Given]

 B + 72^o = 130^o

 B = 130^o – 72^o = 58^o

Now as, A + B = 108^o

 A + 58^o = 108^o

 A = 108^o – 58^o = 50^o

 A = 50^o, B = 58^o and C = 72^o.Question 5

In ABC, A + B = 125^o and A + C = 113^o. Find A, B and C.Solution 5

Since. A , B and C are the angles of a triangle .

So, A + B + C = 180^o

Now, A + B = 125^o [Given]

 125^o + C = 180^o

 C = 180^o – 125^o = 55^o

Also, A + C = 113^o [Given]

 A + 55^o = 113^o

 A = 113^o – 55^o = 58^o

Now as A + B = 125^o

 58^o + B = 125^o

 B = 125^o – 58^o = 67^o

 A = 58^o, B = 67^o and C = 55^o.Question 6

In PQR, if P – Q = 42^o and Q – R = 21^o, find P, Q and R.Solution 6

Since, P, Q and R are the angles of a triangle.

So,P + Q + R = 180^o(i)

Now,P – Q = 42^o[Given]

P = 42^o + Q(ii)

andQ – R = 21^o[Given]

R = Q – 21^o(iii)

Substituting the value of P and R from (ii) and (iii) in (i), we get,

42^o + Q + Q + Q – 21^o = 180^o

3Q + 21^o = 180^o

3Q = 180^o – 21^o = 159^o

Q = 

P = 42^o + Q

= 42^o + 53^o = 95^o

R = Q – 21^o

= 53^o – 21^o = 32^o

P = 95^o, Q = 53^o and R = 32^o.Question 7

The sum of two angles of a triangle is 116^o and their difference is 24^o. Find the measure of each angle of the triangle.Solution 7

Given that the sum of the angles A and B of a ABC is 116^o, i.e., A + B = 116^o.

Since, A + B + C = 180^o

So, 116^o + C = 180^o

 C = 180^o – 116^o = 64^o

Also, it is given that:

A – B = 24^o

 A = 24^o + B

Putting, A = 24^o + B in A + B = 116^o, we get,

24^o + B + B = 116^o

 2B + 24^o = 116^o

 2B = 116^o – 24^o = 92^o

 B = 

Therefore, A = 24^o + 46^o = 70^o

 A = 70^o, B = 46^o and C = 64^o.Question 8

Two angles of a triangle are equal and the third angle is greater than each one of them by 18^o. Find the angles.Solution 8

Let the two equal angles, A and B, of the triangle be x^o each.

We know,

A + B + C = 180^o

x^o + x^o + C = 180^o

2x^o + C = 180^o(i)

Also, it is given that,

C = x^o + 18^o(ii)

Substituting C from (ii) in (i), we get,

2x^o + x^o + 18^o = 180^o

3x^o = 180^o – 18^o = 162^o

x = 

Thus, the required angles of the triangle are 54^o, 54^o and x^o + 18^o = 54^o + 18^o = 72^o.Question 9

Of the three angles of triangle, one is twice the smallest and another one is thrice the smallest. Find the angles.Solution 9

Let C be the smallest angle of ABC.

Then, A = 2C and B = 3C

Also, A + B + C = 180^o

 2C + 3C + C = 180^o

 6C = 180^o

 C = 30^o

So, A = 2C = 2  30^o = 60^o

B = 3C = 3  30^o = 90^o

 The required angles of the triangle are 60^o, 90^o, 30^o.Question 10

In a right-angled triangle, one of the acute angles measures 53^o. Find the measure of each angle of the triangle.Solution 10

Let ABC be a right angled triangle and C = 90^o

Since, A + B + C = 180^o

 A + B = 180^o – C = 180^o – 90^o = 90^o

Suppose A = 53^o

Then, 53^o + B = 90^o

 B = 90^o – 53^o = 37^o

 The required angles are 53^o, 37^o and 90^o.Question 11

In a right-angled triangle, one of the acute angles measures 53^o. Find the measure of each angle of the triangle.Solution 11

Let ABC be a right angled triangle and C = 90^o

Since, A + B + C = 180^o

 A + B = 180^o – C = 180^o – 90^o = 90^o

Suppose A = 53^o

Then, 53^o + B = 90^o

 B = 90^o – 53^o = 37^o

 The required angles are 53^o, 37^o and 90^o.Question 12

If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.Solution 12

Let ABC be a triangle.

So, A < B + C

Adding A to both sides of the inequality,

2 A < A + B + C

2 A < 180^o [Since A + B + C = 180^o]

Similarly, B <A + C

B < 90^o

and C < A + B

C < 90^o

ABC is an acute angled triangle.Question 13

If one angle of a triangle is greater than the sum of the other two, show that the triangle is obtuse angled.Solution 13

Let ABC be a triangle and B > A + C

Since, A + B + C = 180^o

 A + C = 180^o – B

_{Therefore, we get,}

B > 180^o – B

Adding B on both sides of the inequality, we get,

B + B > 180^o – B + B

 2B > 180^o

 B > 

i.e., B > 90^o which means B is an obtuse angle.

 ABC is an obtuse angled triangle.Question 14

In the given figure, side BC of ABC is produced to D. If ACD = 128^o and ABC = 43^o, find BAC and ACB.

Solution 14

Since ACB and ACD form a linear pair.

So, ACB + ACD = 180^o

 ACB + 128^o = 180^o

 ACB = 180^o – 128 = 52^o

Also, ABC + ACB + BAC = 180^o

 43^o + 52^o + BAC = 180^o

 95^o + BAC = 180^o

 BAC = 180^o – 95^o = 85^o

 ACB = 52^o and BAC = 85^o.Question 15

In the given figure, the side BC of ABC has been produced on the right-hand side from B to D and  on the right-hand side from C and E. If ABD = 106^o and ACE= 118^o, find the measure of each angle of the triangle.

Solution 15

As DBA and ABC form a linear pair.

So,DBA + ABC = 180^o

106^o + ABC = 180^o

ABC = 180^o – 106^o = 74^o

Also, ACB and ACE form a linear pair.

So,ACB + ACE = 180^o

ACB + 118^o = 180^o

ACB = 180^o – 118^o = 62^o

In ABC, we have,

ABC + ACB + BAC = 180^o

74^o + 62^o + BAC = 180^o

136^o + BAC = 180^o

BAC = 180^o – 136^o = 44^o

In triangle ABC, A = 44^o, B = 74^o and C = 62^oQuestion 16

Calculate the value of x in each of the following figures.

(i) 

(ii) 

(iii) 

Given: AB || CD

(vi) Solution 16

(i) EAB + BAC = 180^o [Linear pair angles]

110^o + BAC = 180^o

 BAC = 180^o – 110^o = 70^o

Again, BCA + ACD = 180^o [Linear pair angles]

 BCA + 120^o = 180^o

 BCA = 180^o – 120^o = 60^o

Now, in ABC,

ABC + BAC + ACB = 180^o

x^o + 70^o + 60^o = 180^o

 x + 130^o = 180^o

 x = 180^o – 130^o = 50^o

 x = 50

(ii)

In ABC,

A + B + C = 180^o

 30^o + 40^o + C = 180^o

 70^o + C = 180^o

 C = 180^o – 70^o = 110^o

Now BCA + ACD = 180^o [Linear pair]

 110^o + ACD = 180^o

 ACD = 180^o – 110^o = 70^o

In ECD,

ECD + CDE + CED = 180^o

 70^o + 50^o + CED = 180^o

 120^o + CED = 180^o

 CED = 180^o – 120^o = 60^o

Since AED and CED from a linear pair

So, AED + CED = 180^o

 x^o + 60^o = 180^o

 x^o = 180^o – 60^o = 120^o

 x = 120

(iii)

EAF = BAC [Vertically opposite angles]

 BAC = 60^o

In ABC, exterior ACD is equal to the sum of two opposite interior angles.

So, ACD = BAC + ABC

 115^o = 60^o + x^o

 x^o = 115^o – 60^o = 55^o

 x = 55

(iv)

Since AB || CD and AD is a transversal.

So, BAD = ADC

 ADC = 60^o

In ECD, we have,

E + C + D = 180^o

 x^o + 45^o + 60^o = 180^o

 x^o + 105^o = 180^o

 x^o = 180^o – 105^o = 75^o

 x = 75

(v)

In AEF,

Exterior BED = EAF + EFA

 100^o = 40^o + EFA

 EFA = 100^o – 40^o = 60^o

Also, CFD = EFA [Vertically Opposite angles]

 CFD = 60^o

Now in FCD,

Exterior BCF = CFD + CDF

 90^o = 60^o + x^o

 x^o = 90^o – 60^o = 30^o

 x = 30

(vi)

In ABE, we have,

A + B + E = 180^o

 75^o + 65^o + E = 180^o

 140^o + E = 180^o

 E = 180^o – 140^o = 40^o

Now, CED = AEB [Vertically opposite angles]

 CED = 40^o

Now, in CED, we have,

C + E + D = 180^o

 110^o + 40^o + x^o = 180^o

 150^o + x^o = 180^o

 x^o = 180^o – 150^o = 30^o

 x = 30Question 17

In the figure given alongside, AB ∥ CD, EF ∥ BC, ∠BAC = 60° and ∠DHF = 50°. Find ∠GCH and ∠AGH.

Solution 17

AB ∥ CD and AC is the transversal.

⇒ ∠BAC = ∠ACD = 60° (alternate angles)

i.e. ∠BAC = ∠GCH = 60° 

Now, ∠DHF = ∠CHG = 50° (vertically opposite angles)

In ΔGCH, by angle sum property,

∠GCH + ∠CHG + ∠CGH = 180° 

⇒ 60° + 50° + ∠CGH = 180° 

⇒ ∠CGH = 70° 

Now, ∠CGH + ∠AGH = 180° (linear pair)

⇒ 70° + ∠AGH = 180° 

⇒ ∠AGH = 110° Question 18

Calculate the value of x in the given figure.

Solution 18

Produce CD to cut AB at E.

Now, in BDE, we have,

Exterior CDB = CEB + DBE

 x^o = CEB + 45^o     …..(i)

In  AEC, we have,

Exterior CEB = CAB + ACE

= 55^o + 30^o = 85^o

Putting CEB = 85^o in (i), we get,

x^o = 85^o + 45^o = 130^o

 x = 130Question 19

In the given figure, AD divides BAC in the ratio 1: 3 and AD = DB. Determine the value of x.

Solution 19

The angle BAC is divided by AD in the ratio 1 : 3.

Let BAD and DAC be y and 3y, respectively.

As BAE is a straight line,

BAC + CAE = 180^o        [linear pair]

 BAD + DAC +  CAE = 180^o

 y + 3y + 108^o = 180^o

 4y = 180^o – 108^o = 72^o

Now, in ABC,

ABC + BCA + BAC = 180^o

y + x + 4y = 180^o

[Since, ABC = BAD (given AD = DB) and BAC = y + 3y = 4y]

 5y + x = 180

 5  18 + x = 180

 90 + x = 180

 x = 180 – 90 = 90Question 20

If the sides of a triangle are produced in order, prove that the sum of the exterior angles so fomed is equal to four right angles.

Solution 20

Given : A ABC in which BC, CA and AB are produced to D, E and F respectively.

To prove : Exterior DCA + Exterior BAE + Exterior FBD = 360^o

Proof : Exterior DCA = A + B(i)

Exterior FAE = B + C(ii)

Exterior FBD = A + C(iii)

Adding (i), (ii) and (iii), we get,

Ext. DCA + Ext. FAE + Ext. FBD

= A + B + B + C + A + C

= 2A +2B + 2C

= 2 (A + B + C)

= 2 180^o

[Since, in triangle the sum of all three angle is 180^o]

= 360^o

Hence, proved.Question 21

In the given figure, show that

A + B + C + D + E + F = 360^o.

Solution 21

In ACE, we have,

A + C + E = 180^o (i)

In BDF, we have,

B + D + F = 180^o (ii)

Adding both sides of (i) and (ii), we get,

A + C+E + B + D + F = 180^o + 180^o

A + B + C + D + E + F = 360^o.Question 22

In the given figure, AM ⊥ BC and AN is the bisector of ∠A. If ∠ABC = 70° and ∠ACB = 20°, find ∠MAN.

Solution 22

In ΔABC, by angle sum property,

∠A + ∠B + ∠C = 180° 

⇒ ∠A + 70° + 20° = 180° 

⇒ ∠A = 90° 

In ΔABM, by angle sum property,

∠BAM + ∠ABM + ∠AMB = 180° 

⇒ ∠BAM + 70° + 90° = 180° 

⇒ ∠BAM = 20° 

Since AN is the bisector of ∠A,

Now, ∠MAN + ∠BAM = ∠BAN

⇒ ∠MAN + 20° = 45° 

⇒ ∠MAN = 25° Question 23

In the given figure, BAD ∥ EF, ∠AEF = 55° and ∠ACB = 25°, find ∠ABC.

Solution 23

BAD ∥ EF and EC is the transversal.

⇒ ∠AEF = ∠CAD (corresponding angles)

⇒ ∠CAD = 55° 

Now, ∠CAD + ∠CAB = 180° (linear pair)

⇒ 55° + ∠CAB = 180° 

⇒ ∠CAB = 125° 

In ΔABC, by angle sum property,

∠ABC + ∠CAB + ∠ACB = 180° 

⇒ ∠ABC + 125° + 25° = 180° 

⇒ ∠ABC = 30° Question 24

In the given figure, ABC is a triangle in which A : B : C = 3 : 2 : 1 and AC  CD. Find the measure of 

Solution 24

In the given ABC, we have,

A : B : C = 3 : 2 : 1

Let A = 3x, B = 2x, C = x. Then,

A + B + C = 180^o

 3x + 2x + x = 180^o

 6x = 180^o

 x = 30^o

 A = 3x = 3  30^o = 90^o

B = 2x = 2  30^o = 60^o

and, C = x = 30^o

Now, in ABC, we have,

Ext ACE = A + B = 90^o + 60^o = 150^o

 ACD + ECD = 150^o

 ECD = 150^o – ACD 

 ECD = 150^o – 90^o    [since , ]

  ECD= 60^o

Question 25

In the given figure, AB ∥ DE and BD ∥ FG such that ∠ABC = 50° and ∠FGH = 120°. Find the values of x and y.

Solution 25

∠FGH + ∠FGE = 180° (linear pair)

⇒ 120° + y = 180° 

⇒ y = 60° 

AB ∥ DF and BD is the transversal.

⇒ ∠ABC = ∠CDE (alternate angles)

⇒ ∠CDE = 50° 

BD ∥ FG and DF is the transversal.

⇒ ∠EFG = ∠CDE (alternate angles)

⇒ ∠EFG = 50° 

In ΔEFG, by angle sum property,

∠FEG + ∠FGE + ∠EFG = 180° 

⇒ x + y + 50° = 180° 

⇒ x + 60° + 50° = 180° 

⇒ x = 70° Question 26

In the given figure, AB ∥ CD and EF is a transversal. If ∠AEF = 65°, ∠DFG = 30°, ∠EGF = 90° and ∠GEF = x°, Find the value of x.

Solution 26

AB ∥ CD and EF is the transversal.

⇒ ∠AEF = ∠EFD (alternate angles)

⇒ ∠AEF = ∠EFG + ∠DFG

⇒ 65° = ∠EFG + 30° 

⇒ ∠EFG = 35° 

In ΔGEF, by angle sum property,

∠GEF + ∠EGF + ∠EFG = 180° 

⇒ x + 90° + 35° = 180° 

⇒ x = 55° Question 27

In the given figure, AB ∥ CD, ∠BAE = 65° and ∠OEC = 20°. Find ∠ECO.

Solution 27

AB ∥ CD and AE is the transversal.

⇒ ∠BAE = ∠DOE (corresponding angles)

⇒ ∠DOE = 65° 

Now, ∠DOE + ∠COE = 180° (linear pair)

⇒ 65° + ∠COE = 180° 

⇒ ∠COE = 115° 

In ΔOCE, by angle sum property,

∠OEC + ∠ECO + ∠COE = 180° 

⇒ 20° + ∠ECO + 115° = 180° 

⇒ ∠ECO = 45° Question 28

In the given figure, AB ∥ CD and EF is a transversal, cutting them at G and H respectively. If ∠EGB = 35° and QP ⊥ EF, find the measure of ∠PQH.

Solution 28

AB ∥ CD and EF is the transversal.

⇒ ∠EGB = ∠GHD (corresponding angles)

⇒ ∠GHD = 35° 

Now, ∠GHD = ∠QHP (vertically opposite angles)

⇒ ∠QHP = 35° 

In DQHP, by angle sum property,

∠PQH + ∠QHP + ∠QPH = 180° 

⇒ ∠PQH + 35° + 90° = 180° 

⇒ ∠PQH = 55° Question 29

In the given figure, AB ∥ CD and EF ⊥ AB. If EG is the transversal such that ∠GED = 130°, find ∠EGF.

Solution 29

AB ∥ CD and GE is the transversal.

⇒ ∠EGF + ∠GED = 180° (interior angles are supplementary)

⇒ ∠EGF + 130° = 180° 

⇒ ∠EGF = 50°

Exercise Ex. 6D

Question 1

Points A(-1, y) and B(5, 7) lie on a circle with centre O(2, -3y). Find the values of y.Solution 1

Question 2

If the point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), find p.Solution 2

Question 3

ABCD is a rectangle whose three vertices are B(4, 0), C(4, 3) and D(0, 3). Find the length of one of its diagonal.Solution 3

Question 4

If the point P(k – 1, 2) is equidistant from the points A(3, k) and B(k, 5), find the values of k.Solution 4

Question 5

Find the ratio in which the point P(x, 2) divides the join of A(12, 5) and B(4, -3).Solution 5

Question 6

Prove that the diagonals of a rectangle ABCD with vertices A(2, -1), B(5, -1), C(5, 6) and D(2, 6) are equal and bisect each other.Solution 6

Question 7

Find the lengths of the medians AD and BE of ∆ABC whose vertices are A(7, -3), B(5, 3) and C(3, -1).Solution 7

Question 8

If the point C(k, 4) divides the join of A(2, 6) and B(5, 1) in the ratio 2 : 3 then find the value of k.Solution 8

Question 9

Find the point on x-axis which is equidistant from points A(-1, 0) and B(5, 0).Solution 9

Question 10

Find the distance between the points .Solution 10

Distance between the points 

Question 11

Find the value of a, so that the point (3, a) lies on the line represented by 2x – 3y = 5.Solution 11

Question 12

If the points A(4, 3) and B(x, 5) lie on the circle with centre O(2, 3), find the value of x.Solution 12

The points A(4,3) and B(x, 5) lie on the circle with center O(2,3)

OA and OB are radius of the circle.

Question 13

If P(x, y) is equidistant from the points A(7, 1) and B(3, 5), find the relation between x and y.Solution 13

The point P(x, y) is equidistant from the point A(7, 1) and B(3, 5)

Question 14

If the centroid of ABC having vertices A(a, b), B(b, c) and C(c, a) is the origin, then find the value of (a + b + c).Solution 14

The vertices of ABC are (a, b), (b, c) and (c, a)

Centroid is 

But centroid is (0, 0)

 a + b + c = 0Question 15

Find the centroid of ABC whose vertices are A(2, 2), B(-4, -4) and C(5, -8).Solution 15

The vertices of ABC are A(2, 2), B(-4, -4) and C(5, -8)

Centroid of ABC is given by

Question 16

In what ratio does the point C(4, 5) divide the join of A(2, 3) and B(7 , 8)?Solution 16

Let the point C(4, 5) divides the join of A(2, 3) and B(7, 8) in the ratio k : 1

The point C is 

But C is (4, 5)

Thus, C divides AB in the ratio 2 : 3Question 17

If the points A(2, 3), B(4, k)and C(6, -3) are collinear, find the value of k.Solution 17

The points A(2, 3), B(4, k) and C(6, -3) are collinear if area of ABC is zero

But area of ABC = 0,

k = 0

Exercise Ex. 6C

Question 9

A(6, 1), B(8, 2) and C(9, 4) are the vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of ∆ADE.Solution 9

Question 1

Find the area of ABC, whose vertices are:

(i)A(1, 2), B(-2, 3) and C(-3, -4)

(ii)A(-5, 7), B(-4, -5) and C(4, 5)

(iii)A(3, 8), B(-4, 2) and C(5, -1)

(iv)A(10, -6), B(2, 5) and C(-1, 3)Solution 1

(i)Let A(1, 2), B(-2, 3) and C(-3, -4) be the vertices ofthe given ABC, then

(ii)The coordinates of vertices of ABC are A(-5, 7), B(-4, -5) and C(4, 5)

Here, 

(iii)The coordinates of ABC are A(3, 8), B(-4, 2) and C(5, -1)

(iv)Let P(10, -6), Q(2, 5) and R(-1, 3) be the vertices of the given PQR. Then,

Question 2

Find the area of quadrilateral ABCD whose vertices are A(3, -1), B(9, -5), C(14, 0) and D(9, 19).Solution 2

Question 3

Find the area of quadrilateral PQRS whose vertices are P(-5, -3), Q(-4, -6), R(2, -3) and S(1, 2).Solution 3

Question 4

Find the area of quadrilateral ABCD whose vertices are A(-3, -1), B(-2, -4), C(4, -1) and D(3, 4).Solution 4

Question 5

Find the area of quadrilateral ABCD whose vertices are A(-5, 7), B(-4, -5), C(-1, -6) and D(4, 5).Solution 5

Question 6

Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are A(2, 1), B(4, 3) and C(2, 5).Solution 6

Question 7

A(7, -3), B(5, 3) and C(3, -1) are the vertices of a ∆ABC and AD is its median. Prove that the median AD divides ∆ABC into two triangles of equal areas.Solution 7

Question 8

Find the area of ∆ABC with A(1, -4) and midpoints of sides through A being (2, -1) and (0, -1).Solution 8

Question 10(i)

If the vertices of ∆ABC be A(1, -3), B(4, p) and C(-9, 7) 15 square units, find the values of p.Solution 10(i)

Question 11

Find the value of k so that the area of the triangle with vertices A(k + 1, 1), B(4, -3) and C(7, -k) is 6 square units.Solution 11

Question 12

For what value of k(k > 0) is the area of the triangle with vertices (-2, 5), (k, -4) and (2k + 1, 10) equal to 53 square units?Solution 12

Question 13(i)

Show that the following points are collinear.

A(2, -2), B(-3, 8) and C(-1, 4)Solution 13(i)

Question 13(ii)

Show that the following points are collinear.

A(-5, 1), B(5, 5) and C(10, 7)Solution 13(ii)

Question 13(iii)

Show that the following points are collinear.

A(5, 1), B(1, -1) and C(11, 4)Solution 13(iii)

Question 13(iv)

Show that the following points are collinear.

A(8, 1), B(3, -4) and C(2, -5)Solution 13(iv)

Question 14

Find the value of x for which the points A(x, 2), B(-3, -4) and C(7, -5) are collinear.Solution 14

Question 15

For what value of x are the points A(-3, 12), B(7, 6) and C(x, 9) collinear?Solution 15

The given points are A(-3, 12), B(7, 6) and C(x, 9)

Question 16

For what value of y are the points P(1, 4), Q(3, y) and R(-3, 16) collinear?Solution 16

Let P(1, 4), Q(3, y) and R(-3, 16)

Question 17

Find the value of y for which the points A(-3, 9), B(2, y) and C(4, -5) are collinear.Solution 17

Question 18

For what values of k are the points A(8, 1), B(3, -2k) and C(k, -5) collinear.Solution 18

Question 19

Find a relation between x and y, if the points A(2, 1), B(x, y) and C(7, 5) are collinear.Solution 19

Vertices of ABC are A(2, 1), B(x, y) and C(7, 5)

The points A, B and C are collinear

area of ABC =0

Or 4x – 5y – 3 = 0Question 20

Find a relation between x and y, if the points A(x, y), B(-5,7) and C(-4, 5) are collinear.Solution 20

Question 21

Prove that the points A(a, 0), B(0, b) and C(1, 1) are collinear, if .Solution 21

The vertices of ABC are (a, 0), (0, b), C(1, 1)

The points A, B, C are collinear

Area of ABC = 0

ab – a – b = 0 a + b = ab

Dividing by ab

Question 22

If the points P(-3, 9), Q(a, b) and R(4, -5) are collinear and a + b = 1. find the values of a and b.Solution 22

Question 23

Find the area of ∆ABC with vertices A(0, -1), B(2, 1) and C(0, 3). Also, find the area of the triangle formed by joining the midpoints of its sides. Show that the ratio of the areas of two triangles is 4 :1.Solution 23

Exercise Ex. 6A

Question 12

If P(x, y) is a point equidistant from the points A(6, -1) and B(2, 3). Show that x – y = 3Solution 12

Let A(6, -1) and B(2,3) be the given point and P(x,y) be the required point, we get

Question 1

Find the distance between the points:

(i)A(9,3) and B(15, 11)

(ii)A(7, -4) and B(-5, 1)

(iii)A(-6, -4) and B(9, -12)

(iv)A(1, -3) and B(4, -6)

(v)P(a + b, a – b) and Q(a – b, a + b)

(vi)P(a sin a, a cos a) and Q(a cos a, -a sin a)Solution 1

(i)The given points are A(9,3) and B(15,11)

(ii)The given points are A(7,4) and B(-5,1)

(iii)The given points are A(-6, -4) and B(9,-12)

(iv)The given points are A(1, -3) and B(4, -6)

(v)The given points are P(a + b, a – b) and Q(a – b, a + b)

(vi)The given points are P(a sin a, a cos a) and Q(a cos a, – a sina)

Question 2

Find the distance of each of the following points from the origin:

(i)A(5, -12)

(ii)B(-5, 5)

(iii)C(-4, -6)Solution 2

(i)The given point is A(5, -12) and let O(0,0) be the origin

(ii)The given point is B(-5, 5) and let O(0,0) be the origin

(iii)The given point is C(-4, -6) and let O(0,0) be the origin

Question 3

Find all possible values of a for which the distance between the points A(a, -1) and B(5, 3) is 5 units.Solution 3

The given points are A(a, -1) and B(5,3)

Question 4

Find all possible values of y for which the distance between the points A(2, -3) and B(10, y) is 10 units.Solution 4

Question 5

Find the values of x for which the distance between the points P(x, 4) and Q(9, 10) is 10 units.Solution 5

Question 6

If the point A(x, 2) is equidistant from the points B(8, -2) and C(2, -2), find the value of x. Also, find the length of AB.Solution 6

Question 7

If the point A(0, 2) is equidistant from the points B(3, p) and C(p,5), find the value of p. Also, find the length of AB. Solution 7

Question 8

Find the point on the x-axis which is equidistant from the points (-2, 5) and (-2, 9).Solution 8

Let any point P on x – axis is (x,0) which is equidistant from A(-2, 5) and B(-2, 9)

This is not admissible

Hence, there is no point on x – axis which is equidistant from A(-2, 5) and B(-2, 9)Question 9

Find points on the x – axis, each o f which is at a distance of 10 units from the point A(11, -8).Solution 9

Let A(11, -8) be the given point and let P(x,0) be the required point on x – axis

Then,

Hence, the required points are (17,0) and (5,0)Question 10

Find the point on the y-axis which is equidistant from the points A(6, 5) and B(-4, 3).Solution 10

Question 11

If the point P(x, y) is equidistant from the points A(5, 1) and B(-1, 5), prove that 3x = 2y.Solution 11

Question 13

Find the coordinates of the point equidistant from three given points A(5, 3), B(5, -5) and C(1, -5).Solution 13

Let the required points be P(x,y), then

PA = PB = PC. The points A, B, C are (5,3), (5, -5) and (1, -5) respectively

Hence, the point P is (3, -1)Question 14

If the points A(4, 3) and B(x, 5) lie on a circle with the centre O(2, 3), find the value of x.Solution 14

Question 15

If the point C(-2, 3) is equidistant from the points A(3, -1) and B(x, 8), find the values of x. Also, find the distance BC.Solution 15

Question 16

If the point P(2, 2) is equidistant from the points A(-2, k) and B(-2k, – 3), find k. Also, find the length of AP.Solution 16

6A Question 17

If the point (x, y) is equidistant from the points (a + b, b – a) and (a – b, a + b), prove that bx = ay.Solution 17

Question 18(i)

Using the distance formula, show that the given points are collinear.

(1, -1), (5, 2) and (9, 5)Solution 18(i)

Question 18(ii)

Using the distance formula, show that the given points are collinear.

(6, 9), (0, 1) and (-6, -7)Solution 18(ii)

Question 18(iii)

Using the distance formula, show that the given points are collinear

(-1, -1), (2, 3) and (8, 11)Solution 18(iii)

Question 18(iv)

Using the distance formula, show that the given points are collinear.

(-2, 5), (0, 1) and (2, -3).Solution 18(iv)

Question 19

Show that the points A(7, 10), B(-2, 5) and C(3, -4) are the vertices of an isosceles right triangle.Solution 19

Question 20

Show that the points A(3, 0), B(6, 4) and C(-1, 3) are the vertices of an isosceles right triangle.Solution 20

Question 21

If A(5, 2), B(2, -2) and C(-2, t) are the vertices of a right triangle with ∠B = 90°, then find the value of t.Solution 21

Question 22

Prove that the points A(2, 4), B(2, 6) and C(2 +, 5) are the vertices of an equilateral triangle.Solution 22

Question 23

Show that the points (-3, -3), (3, 3) and (-3, 3) are the vertices of an equilateral triangle.Solution 23

Question 24

Show that the points A(-5, 6), B(3, 0) and C(9, 8) are the vertices of isosceles triangle. Calculate its area.Solution 24

Let A(-5,6), B(3,0) and C(9,8) be the given points. Then

Question 25

Show that the points O(0, 0), A(3, ) and B(3, –) are the vertices of an equilateral triangle. Find the area of this triangle.Solution 25

are the given points

Hence, DABC is equilateral and each of its sides being 

Question 26

Show that the following points are the vertices of a square:

(i)A(3, 2), B(0, 5), C(-3, 2) and D(0, -1)

(ii)A(6, 2), B(2, 1), C(1, 5) and D(5, 6)

(iii)P(0, -2), Q(3, 1), R(0, 4) and S(-3, 1)Solution 26

(i)The angular points of quadrilateral ABCD are A(3,2), B(0,5), C(-3,2) and D(0,-1)

     Thus, all sides of quad. ABCD are equal and diagonals are also equal

      Quad. ABCD is a square

(ii)Let A(6,2), B(2,1), C(1,5) and D(5,6) be the angular points of quad. ABCD. Join AC and BD

Thus, ABCD is a quadrilateral in which all sides are equal and the diagonals are equal.

Hence, quad ABCD is a square.

(iii)Let P(0, -2), Q(3,1), R(0,4) and S(-3,1) be the angular points of quad. ABCD

Join PR and QSD

Thus, PQRS is a quadrilateral in which all sides are equal and the diagonals are equal

Hence, quad. PQRS is a squareQuestion 27

Show that the points A(-3, 2), B(-5, -5), C(2, -3) and D(4, 4) are the vertices of a rhombus. Find the area of this rhombus.Solution 27

Let A(-3,2), B(-5, -5), C(2, -3) and D(4,4) be the angular point of quad ABCD. Join AC and BD.

Thus, ABCD is a quadrilateral having all sides equal but diagonals are unequal.

Hence, ABCD is a rhombus

Question 28

Show that the points A(3, 0), B(4, 5), C(-1, 4) and D(-2, -1) are the vertices of a rhombus. Find its area.Solution 28

Question 29

Show that the points A(6, 1), B(8, 2), C(9, 4) and D(7, 3) are the vertices of a rhombus. Find its area.Solution 29

Question 30

Show that the points A(2, 1), B(5, 2), C(6, 4) and D(3, 3) are the angular points of a parallelogram. Is this figure a rectangle?Solution 30

Let A(2,1), B(5,2), C(6,4) and D(3,3) are the angular points of a parallelogram ABCD. Then

Diagonal AC  Diagonal BD

Thus ABCD is not a rectangle but it is a parallelogram because its opposite sides are equal and diagonals are not equalQuestion 31

Show that A(1, 2), B(4, 3), C(6, 6) and D(3, 5) are the vertices of a parallelogram. Show that ABCD is not a rectangle.Solution 31

Question 32

Show that the following points are the vertices of a rectangle:

(i)A(-4, -1), B(-2, -4), C(4, 0) and D(2, 3)

(ii)A(2, -2), B(14, 10), C(11, 13) and D(-1, 1)

      (i)A(0, -4), B(6, 2), C(3, 5) and D(-3, -1)Solution 32

(i) Let A(-4, -1), B(-2, -4), C(4, 0) and D(2, 3) are the vertices of quad. ABCD. Then

Thus, ABCD is a quadrilateral whose opposite sides are equal and the diagonals are equal

Hence, quad. ABCD is a rectangle.

(ii)Let A(2, -2), B(14, 10), C(11, 13) and D(-1, 1) be the angular points of quad. ABCD, then

Thus, ABCD is a quadrilateral whose opposite sides are equal and diagonals are equal.

Hence, quad. ABCD is rectangle.

(iii)Let A(0, -4), B(6,2), C(3,5) and D(-3,-1) are the vertices of quad. ABCD. Then

  Thus, ABCD is a quadrilateral whose opposite sides are equal and the diagonals are equal

Hence, quad. ABCD is a rectangle

Exercise MCQ

Question 1

The distance of the point P(-6,8) from the origin is

Solution 1

Question 2

The distance of the point (-3, 4) from x-axis is

(a) 3

(b) -3

(c) 4

(d) 5Solution 2

Question 3

The point on x-axis which is equidistant from points

A(-1, 0) and B(5, 0) is

(a) (0, 2)

(b) (2, 0)

(c) (3, 0)

(d) (0, 3)Solution 3

Question 4

If R(5, 6) is the midpoint of the line segment AB joining the points A(6, 5) and B(4, y) they y equals

(a) 5

(b) 7

(c) 12

(d) 6Solution 4

Question 5

If the point C(k, 4) divides the join of the points A(2, 6) and B(5,1) in the ratio 2:3 then the value of k is

Solution 5

Question 6

The perimeter of the triangle with vertices (0, 4), (0, 0) and (3, 0) is

Solution 6

Correct option: (d)

Question 7

If A(1, 3), B(-1, 2), C(2, 5) and D(x, 4) are the vertices of a ‖gm ABCD then the value of x is

Solution 7

Correct option: (b)

Question 8

If the points A(x, 2), B(-3, -4) and C(7, -5) are collinear then the value of x is

(a) -63

(b) 63

(c) 60

(d) -60Solution 8

Question 9

The area of a triangle with vertices A(5, 0),  B(8, 0) and C(8,4) in square units is

(a) 20

(b) 12

(c) 6

(d) 16Solution 9

Question 10

The area of ABC with vertices A(a, 0), O(0, 0) and

B(0, b) in square units is

Solution 10

Question 11

(a) -8

(b) 3

(c) -4

(d) 4Solution 11

Question 12

ABCD is a rectangle whose three vertices are B(4, 0), C(4, 3) and D(0, 3). The length of one of its diagonals is

(a) 5

(b) 4

(c) 3

(d) 25Solution 12

Correct option: (a)

Question 13

The coordinates of the point P dividing the line segment joining the points A(1, 3) and B(4, 6) in the ratio 2:1 is

(a) (2, 4)

(b) (3, 5)

(c) (4, 2)

(d) (5, 3)Solution 13

Question 14

If the coordinates of one end of a diameter of a circle are (2, 3) and the coordinates of its centre are (-2, 5), then the coordinates of the other end of the diameter are

(a) (-6, 7)

(b) (6, -7)

(c) (4, 2)

(d) (5, 3)Solution 14

Question 15

In the given figure P(5, -3) and Q(3, y) are the points of trisection of the line segment joining A(7, -2) and

B(1, -5). Then y equals

Solution 15

Question 16

The midpoint of segment AB is P(0, 4). If the coordinates of B are (-2, 3), then the coordinates of A are

(a) (2, 5)

(b) (-2, -5)

(c) (2, 9)

(d) (-2, 11)Solution 16

Question 17

The point P which divides the line segment joining the points A(2, -5) and B(5, 2) in the ratio 2:3 lies in the quadrant

(a) I

(b) II

(c) III

(d) IV Solution 17

Question 18

If A(6, -7) and B(-1, -5) are two given points then the distance 2AB is

(a) 13

(b) 26

(c) 169

(d) 238Solution 18

Question 19

Which point on x-axis is equidistant from the points

A(7, 6) and B(-3, 4)?

(a) (0, 4)

(b) (-4, 0)

(c) (3, 0)

(d) (0, 3)Solution 19

Question 20

The distance of P(3, 4) from the x-axis is

(a) 3 units

(b) 4 units

(c) 5 units

(d) 1 unitsSolution 20

Question 21

In what ratio does the x-axis divide the join of A(2, -3) and B(5, 6)?

(a) 2:3

(b) 3:5

(c) 1:2

(d) 2:1Solution 21

Question 22

In what ratio does the y-axis divide the join of P(-4, 2) and Q(8, 3)?

(a) 3:1

(b) 1:3

(c) 2:1

(d) 1:2Solution 22

Question 23

If P(-1, 1) is the midpoint of the line segment joining

A(-3, b) and B(1, b + 4) then b =?

(a) 1

(b) -1

(c) 2

(d) 0Solution 23

Question 24

The line 2x + y – 4 = 0 divides the line segment joining A(2, -2) and (3, 7) in the ratio

(a) 2:5

(b) 2:9

(c) 2:7

(d) 2:3Solution 24

Question 25

If A(4, 2), B(6, 5) and C(1,4) be the vertices of ∆ABC and AD is a median, then the coordinates of D are

Solution 25

Question 26

If A(-1, 0), B(5, -2) and C(8,2) are the vertices of a ∆ABC then its centroid is

(a) (12, 0)

(b) (6, 0)

(c) (0, 6)

(d) (4, 0)Solution 26

Question 27

Two vertices of ∆ABC are A (-1, 4) and B(5, 2) and its centroid is G(0, -3). Then, the coordinates of are

(a) (4, 3)

(b) (4, 15)

(c) (-4, -15)

(d) (-15, -4)Solution 27

Question 28

The points A(-4, 0), B(4, 0) and C(0,3) are the vertices of a triangle, which is

(a) isosceles

(b) equilateral

(c) scalene

(d) right-angledSolution 28

Question 29

The point P(0, 6), Q(-5, 3) and R(3, 1)are the vertices of a triangle, which is

(a) equilateral

(b) isosceles 

(c) scalene

(d) right-angledSolution 29

Question 30

If the points A(2, 3), B(5, k) and C(6, 7) are collinear then

Solution 30

Question 31

If the point A (1, 2), O(0, 0) and C(a, b) are collinear then

(a) a = b

(b) a = 2b

(c) 2a = b

(d) a + b = 0Solution 31

Question 32

The area of ∆ABC with vertices A(3, 0), B(7, 0) and

C(8, 4) is

(a) 14 sq units

(b) 28 sq units

(c) 8 sq units

(d) 6 sq unitsSolution 32

Question 33

AOBC is a rectangle whose three vertices are A(0, 3), O(0, 0) and B(5, 0). The length of each of its diagonals is

Solution 33

Question 34

If the distance between the points A(4, p) and B(1, 0) is 5 then

(a) p = 4 only

(b) p = -4 only

(c) p = ± 4

(d) p = 0Solution 34

Exercise Ex. 6B

Question 1

Find the coordinates of the point which divides the join of A(-1, 7) and B(4, -3) in the ratio 2 : 3.Solution 1

The end points of AB are A(-1,7) and B(4, -3)

Let the required point be P(x, y)

By section formula, we have

Hence the required point is P(1, 3)Question 2

Find the coordinates of the points which divides the join of A(-5, 11) and B(4, -7) in the ratio 7 : 2.Solution 2

The end points of PQ are P(-5, 11) and Q(4, -7_

By section formula, we have

Hence the required point is (2, -3)Question 3

Solution 3

Question 4

Solution 4

Question 5

Points P, Q, R and S divide the line segment joining the points A(1, 2) and R(6, 7) in five equal parts. Find the coordinates of P, Q and R.Solution 5

Question 6

Points P, Q and R in that order are dividing a line segment joining A(1, 6) and B(5, -2) in four equal parts, find the coordinates of P, Q and R.Solution 6

Points P, Q, R divide the line segment joining the points A(1,6) and B(5, -2) into four equal parts

Point P divide AB in the ratio 1 : 3 where A(1, 6), B(5, -2)

Therefore,  the point P is

Also, R is the midpoint of the line segment joining Q(3, 2) and B(5, -2)

Question 7

The line segment joining the points A(3, -4) and B(1, 2) is trisected at the points P(p, -2) and . Find the values of p and q.Solution 7

Point P divides the join of A(3, -4) and B(1,2) in the ratio 1 : 2.

Coordinates of P are:

Question 8

Find the coordinates of the midpoint of the line segment joining:

(i)A(3, 0) and B(-5, 4)

(ii)P(-11, -8) and Q(8, -2)Solution 8

(i)The coordinates of mid – points of the line segment joining A(3, 0) and B(-5, 4) are 

(ii)Let M(x, y) be the mid – point of AB, where A is (-11, -8) and B is (8, -2). Then,

Question 9

If (2, p) is the midpoint of the line segment joining the points A(6, -5) and B(-2, 11), find the value of p.Solution 9

The midpoint of line segment joining the points A(6, -5) and B(-2, 11) is

Also, given the midpoint of AB is (2, p)

p = 3Question 10

The midpoint of the line segment A(2a, 4) and B(-2, 3b) is C(1, 2a + 1). Find the value of a and b.Solution 10

C(1, 2a + 1) is the midpoint of A(2a, 4) and B(-2, 3b)

Question 11

The line segment joining A(-2, 9) and B(6, 3) is a diameter of a circle with centre C. Find the coordinates of C.Solution 11

Let A(-2, 9) and B(6, 3) be the two points of the given diameter AB and let C(a, b) be the center of the circle

Then, clearly C is the midpoint of AB

By the midpoint formula of the co-ordinates,

Hence, the required point C(2, 6)Question 12

Find the coordinates of a point A, where AB is a diameter of a circle with centre C(2, -3) and the other end of the diameter is B(1, 4).Solution 12

A, B are the end points of a diameter. Let the coordinates of A be (x, y)

The point B is (1, 4)

The center C(2, -3) is the midpoint of AB

The point A is (3, -10)Question 13

In what ratio does the point P(2, 5) divide the join of A(8, 2) and B(-6, 9)?Solution 13

Let P divided the join of A(8, 2), B(-6, 9) in the ratio k : 1

By section formula, the coordinates of p are

Hence, the required ratio of which is (3 : 4)Question 14

Solution 14

Question 15

Find the ratio in which the point P(m, 6) divides the join of A(-4, 3) and B(2, 8). Also, find the value of m.Solution 15

Let P divided the join of line segment A(-4, 3) and B(2, 8) in the ratio k : 1

the point P is

Question 16

Find the ratio in which the point (-3, k) divides the join of A(-5, -4) and B(-2, 3). Also, find the value of k.Solution 16

Let P is dividing the given segment joining A(-5, -4) and B(-2, 3) in the ratio r : 1

Coordinates of point P

Question 17

In what ratio is the line segment joining A(2, -3) and B(5, 6) divided by the x-axis? Also, find the coordinates of the point of division.Solution 17

Let the x- axis cut the join of A(2, -3) and B(5, 6) in the ratio k : 1 at the point P

Then, by the section formula, the coordinates of P are 

But P lies on the x axis so, its ordinate must be 0

So the required ratio is 1 : 2

Thus the x – axis divides AB in the ratio 1 : 2

Putting we get the point P as

Thus, P is (3, 0) and k = 1 : 2Question 18

In what ratio is the line segment joining the points A(-2, -3) and B(3, 7) divided by the y-axis? Also, find the coordinates of the point of division.Solution 18

Let the y – axis cut the join A(-2, -3) and B(3, 7) at the point P in the ratio k : 1

Then, by section formula, the co-ordinates of P are

But P lies on the y-axis so, its abscissa is 0

So the required ratio is which is 2 : 3

Putting we get the point P as

i.e., P(0, 1)

Hence the point of intersection of AB and the y – axis is P(0, 1) and P divides AB in the ratio 2 : 3Question 19

In what ratio does the line x – y – 2 = 0 divide the line segment joining the points A(3, -1) and B(8, 9)?Solution 19

Let the line segment joining A(3, -1) and B(8, 9) is divided byx – y – 2 = 0 in ratio k : 1 at p

Coordinates of P are

Thus the line x – y – 2 = 0 dividesAB in the ratio 2 : 3Question 20

Find the lengths of the medians of a ABC whose vertices are A(0, -1), B(2, 1) and C(0, 3).Solution 20

Let D, E, F be the midpoint of the side BC, CA and AB respectively in ABC

Then, by the midpoint formula, we have

Hence the lengths of medians AD, BE and CF are given by

Question 21

Find the centroid of ABC whose vertices are A(-1, 0), B(5, -2) and C(8, 2).Solution 21

Here 

Let G(x, y) be the centroid of ABC, then

Hence the centroid of ABC is G(4, 0)Question 22

If G(2, -1) is the centroid of a ABC and two of its vertices are A(1, -6) and B(-5,2), find the third vertex of the triangle.Solution 22

Two vertices of ABC are A(1, -6) and B(-5, 2) let the third vertex be C(a, b)

Then, the co-ordinates of its centroid are

But given that the centroid is G(-2, 1)

Hence, the third vertex C of ABC is (-2, 7)Question 23

Find the third vertex of a ABC if two of its vertices are B(-3, 1) and C(0, -2), and its centroid is at the origin.Solution 23

Two vertices of ABC are B(-3, 1) and C(0, -2) and third vertex be A(a, b)

Then the coordinates of its centroid are

Hence the third vertices A of ABC is A(3, 1)Question 24

Show that the points A(3, 1), B(0, -2), C(1, 1) and D(4, 4) are the vertices of a parallelogram ABCD.Solution 24

Let A(3,1), B(0, -2), C(1, 1) and D(4, 4) be the vertices of quadrilateral

Join AC, BD. AC and BD, intersect other at the point O.

We know that the diagonals of a parallelogram bisect each other

Therefore, O is midpoint of AC as well as that of BD

Now midpoint of AC is 

And midpoint of BD is 

Mid point of AC is the same as midpoint of BD

Hence, A, B, C, D are the vertices of a parallelogram ABCDQuestion 25

If the points P(a, -11), Q(5, b), R(2, 15) and S(1, 1) are the vertices of a parallelogram PQRS, find the values of a and b.Solution 25

Let P(a, -11), Q(5, b), R(2, 15) and S(1, 1) are the vertices of a parallelogram PQRS.

Join the diagonals PR and SQ.

They intersect each other at the point O. We know that the diagonals of a parallelogram bisect each other.

Therefore, O is the midpoint of PR as well as that of SQ

Now, midpoint of PR is 

And midpoint of SQ is 

Hence the required values are a = 4 and b = 3Question 26

If three consecutive vertices of a parallelogram ABCD are A(1, -2), B(3, 6) and C(5, 10), find the fourth vertex D.Solution 26

Let A(1, -2), B(3, 6) and C(5, 10) are the given vertices of the parallelogram ABCD

Let D(a, b) be its fourth vertex. Join AC and BD.

Let AC and BD intersect at the point O.

We know that the diagonals of a parallelogram bisect each other.

So, O is the midpoint AC as well as that of BD

Midpoint of AC is 

Midpoint of BD is 

Hence the fourth vertices is D(3, 2)Question 27

In what ratio does y-axis divide the line segment joining the points (-4, 7) and (3, -7)?Solution 27

Question 28

If the point Plies on the line segment joining the points A(3, -5) 2 and B(-7, 9) then find the ratio in which P divides AB. Also, find the value of y.Solution 28

Question 29

Find the ratio in which the line segment joining the points A(3, -3) and B(-2, 7) is divided by x-axis. Also, find the point of division.Solution 29

Question 30

The base QR of an equilateral triangle PQR lies on x-axis. The coordinates of the point Q are (-4, 0) and origin is the midpoint of the base. Find the coordinates of the points P and R.Solution 30

Question 31

The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of point C are (0, -3). The origin is the midpoint of the base. Find the coordinates of the points A and B. Also, find the coordinates of anther point D such that ABCD is a rhombus.Solution 31

Question 32

Find the ratio in which the points p(-1, y) lying on the line segment joining points A(-3, 10) and B(6, -8) divides it. Also, find the value of y.Solution 32

Question 33

ABCD is a rectangle formed by the points A(-1, -1), B(-1, 4), C(5, 4) and D(5, -1). If P, Q, R and S be the midpoints of AB, BC, CD and DA respectively, show that PQRS is a rhombus.Solution 33

Question 34

The midpoint P of the line segment joining the points A(-10, 4) and B(-2, 0) lies on the line segment joining the points C(-9, -4) and D(-4, y). Find the ratio in which P divides CD. Also find the value of y.Solution 34

Exercise Ex. 1A

Question 1

What do you mean by Euclid’s division lemma?Solution 1

For any two given positive integers a and b there exist unique whole numbers q and r such that  

Here, we call ‘a’ as dividend, b as divisor, q as quotient and r as remainder.

Dividend = (divisor quotient) + remainderQuestion 2

A number when divided by 61 gives 27 as quotient and 32 as remainder. Find the number.Solution 2

By Euclid’s Division algorithm we have:

Dividend = (divisor × quotient) + remainder

= (61 27) + 32 = 1647 + 32 = 1679Question 3

By what number should 1365 be divided to get 31 as quotient and 32 as remainder?Solution 3

By Euclid’s Division Algorithm, we have:

Dividend = (divisor  quotient) + remainder

Question 4

Using Euclid’s algorithm, find the HCF of:

(i) 405 and 2520

(ii) 504 and 1188

(iii) 960 and 1575Solution 4

(i)

On dividing 2520 by 405, we get

Quotient = 6, remainder = 90

2520 = (405 6) + 90 

Dividing 405 by 90, we get

Quotient = 4,

Remainder = 45

405 = 90 4 + 45

Dividing 90 by 45

Quotient = 2, remainder = 0

90 = 45 2

H.C.F. of 405 and 2520 is 45

(ii) Dividing 1188 by 504, we get

Quotient = 2, remainder = 180

1188 = 504 2+ 180

Dividing 504 by 180

Quotient = 2, remainder = 144

504 = 180 × 2 + 144

Dividing 180 by 144, we get

Quotient = 1, remainder = 36

Dividing 144 by 36

Quotient = 4, remainder = 0

H.C.F. of 1188 and 504 is 36

(iii) Dividing 1575 by 960, we get

Quotient = 1, remainder = 615

1575 = 960 × 1 + 615

Dividing 960 by 615, we get

Quotient = 1, remainder = 345

960 = 615 × 1 + 345

Dividing 615 by 345

Quotient = 1, remainder = 270

615 = 345 × 1 + 270

Dividing 345 by 270, we get

Quotient = 1, remainder = 75

345 = 270 × 1 + 75

Dividing 270 by 75, we get

Quotient = 3, remainder =45

270 = 75 × 3 + 45

Dividing 75 by 45, we get

Quotient = 1, remainder = 30

75 = 45 × 1 + 30

Dividing 45 by 30, we get

Remainder = 15, quotient = 1

45 = 30 × 1 + 15

Dividing 30 by 15, we get

Quotient = 2, remainder = 0

H.C.F. of 1575 and 960 is 15

Question 5

Show that every positive integer is either even or odd.Solution 5

Question 6

Show that any positive odd integer is of the form (6m + 1) or (6m + 3) or (6m + 5), where m is some integer.Solution 6

Question 7

Show that any positive odd integer is of the form (4m + 1) or (4m + 3), where in is some integer.Solution 7

Exercise Ex. 1B

Question 1(i)

Using prime factorization, find the HCF and LCM of:

36, 84

In each case, verify that:

HCF x LCM = product of given numbers.Solution 1(i)

Question 1(ii)

Using prime factorization, find the HCF and LCM of:

23, 31

In each case, verify that:

HCF x LCM = product of given numbers.Solution 1(ii)

Question 1(iii)

Using prime factorization, find the HCF and LCM of:

96, 404

In each case, verify that:

HCF x LCM = product of given numbers.Solution 1(iii)

Question 1(iv)

Using prime factorization, find the HCF and LCM of:

144,198

In each case, verify that:

HCF x LCM = product of given numbers.Solution 1(iv)

Question 1(v)

Using prime factorization, find the HCF and LCM of:

396, 1080

In each case, verify that:

HCF x LCM = product of given numbers.Solution 1(v)

Question 1(vi)

Using prime factorization, find the HCF and LCM of:

1152, 1664

In each case, verify that:

HCF x LCM = product of given numbers.Solution 1(vi)

Question 5(i)

Using prime factorization, find the HCF and LCM of:

8, 9, 25Solution 5(i)

Question 5(ii)

Using prime factorization, find the HCF and LCM of:

12, 15, 21Solution 5(ii)

Question 5(iii)

Using prime factorization, find the HCF and LCM of:

17, 23, 29Solution 5(iii)

Question 5(v)

Using prime factorization, find the HCF and LCM of:

30, 72, 432Solution 5(v)

Question 5(vi)

Using prime factorization, find the HCF and LCM of:

21, 28, 36, 45Solution 5(vi)

Question 6

Is it possible to have two numbers whose HCF is 18 and LCM is 760? Give reason.Solution 6

Question 7

Find the simplest form of:

(iv)
Solution 7

(iv)

Question 8

The HCF of two numbers is 23 and their LCM is 1449. If one of the numbers is 161, find the other.Solution 8

Question 9

The HCF of two numbers is 145 and their LCM is 2175. If one of the numbers is 725, find the other.Solution 9

Question 12

The HCF of two numbers is 18 and their product is 12960. Find their LCM.Solution 12

Question 15

Find the largest number which divides 438 and 606, leaving remainder 6 in each case.Solution 15

Question 16

Find the largest number which divides 320 and 457 leaving remainders 5 and 7 respectively.Solution 16

Subtracting 5 and 7 from 320 and 457 respectively:

 320 – 5 = 315,

457 – 7 = 450

Let us now find the HCF of 315 and 405 through prime factorization:

The required number is 45.Question 17

Find the least number which when divided by 35, 56 and 91 leaves the same remainder 7 in each case.Solution 17

Question 18

Find the smallest number which when divided by 28 and 32 leaves remainders 8 and 12 respectively.Solution 18

Question 19

Find the smallest number which when increased by 17 is exactly divisible by both 468 and 520.Solution 19

Question 20

Find the greatest number of four digits which is exactly divisible by 15, 24 and 36.Solution 20

Question 24

Find the missing numbers in the following factorisation:

Solution 24

By going upward

5 11= 55

55 3= 165

1652 = 330

330 2 = 660Question 25

In a seminar, the number of participants in Hindi, English and mathematics are 60, 84 and 108 respectively. Find the minimum number of rooms required, if in each room, the same number of participants are to be seated and all of them being in the same subject.Solution 25

Question 26

Three sets of English, Mathematics and Science books containing 336, 240 and 96 books respectively have to be stacked in such a way that all the books are stored subjectwise and the height of each stack is the same. How many stacks will be there?Solution 26

Let us find the HCF of 336, 240 and 96 through prime factorization:

Each stack of book will contain 48 books

Number of stacks of the same height

Question 27

Three pieces of timber 42 m, 49 m and 63 m long have to be divided into planks of the same length. What is the greatest possible length of each plank?Solution 27

The prime factorization of 42, 49 and 63 are:

42 = 2 3 7, 49 = 7 7, 63 = 3 3 7

H.C.F. of 42, 49, 63 is 7

Hence, greatest possible length of each plank = 7 mQuestion 28

Find the greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm and 12 m 95 cm.Solution 28

7 m = 700cm, 3m 85cm = 385 cm

12 m 95 cm = 1295 cm

Let us find the prime factorization of 700, 385 and 1295:

Greatest possible length = 35cmQuestion 29

Find the maximum number of students among whom 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and the same number of pencils.Solution 29

Let us find the prime factorization of 1001 and 910:

1001 = 11 7 13

910 = 2 5 7 13

H.C.F. of 1001 and 910 is 7 13 = 91

Maximum number of students = 91Question 30

Find the least number of square tiles required to pave the ceiling of a room 15 m 17 cm long and 9 m 2 cm broad.Solution 30

Question 31

Three measuring rods are 64 cm, 80 cm and 96 cm in length. Find the least length of cloth that can be measured an exact number of times, using any of the rods.Solution 31

Let us find the LCM of 64, 80 and 96 through prime factorization:

L.C.M of 64, 80 and 96

= 

 Therefore, the least length of the cloth that can be measured an exact number of times by the rods of 64cm, 80cm and 96cm = 9.6mQuestion 32

An electronic device makes a beep after every 60 seconds. Another device makes a beep after every 62 seconds. They beeped together at 10 a.m. At what time will they beep together at the earliest?Solution 32

Interval of beeping together = LCM (60 seconds, 62 seconds)

The prime factorization of 60 and 62:

60 = 30 2, 62 = 31 2

L.C.M of 60 and 62 is 30 31 2 = 1860 sec = 31min

electronic device will beep after every 31minutes

After 10 a.m., it will beep at 10 hrs 31 minutesQuestion 33

The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they all change simultaneously at 8 hours, then at what time will they again change simultaneously?Solution 33

Question 34

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10, 12 minutes respectively. In 30 hours, how many times do they toll together?Solution 34

Exercise Ex. 1C

Question 1(i)

Without actual division, show that each of the following rational number is a terminating decimal. Express each in decimal form.

Solution 1(i)

Question 1(ii)

Without actual division, show that each of the following rational number is a terminating decimal. Express each in decimal form.

Solution 1(ii)

Question 1(iii)

Without actual division, show that each of the following rational number is a terminating decimal. Express each in decimal form.

Solution 1(iii)

Question 1(iv)

Without actual division, show that each of the following rational number is a terminating decimal. Express each in decimal form.

Solution 1(iv)

Question 1(v)

Without actual division, show that each of the following rational number is a terminating decimal. Express each in decimal form.

Solution 1(v)

Question 1(vi)

Without actual division, show that each of the following rational number is a terminating decimal. Express each in decimal form.

Solution 1(vi)

Question 2(i)

Without actual division, show that each of the following rational number is a nonterminating repeating decimal.

Solution 2(i)

Question 2(ii)

Without actual division, show that each of the following rational number is a nonterminating repeating decimal.

Solution 2(ii)

Question 2(iii)

Without actual division, show that each of the following rational number is a nonterminating repeating decimal.

Solution 2(iii)

Question 2(iv)

Without actual division, show that each of the following rational number is a nonterminating repeating decimal.

Solution 2(iv)

Question 2(v)

Without actual division, show that each of the following rational number is a nonterminating repeating decimal.

Solution 2(v)

Question 2(vi)

Without actual division, show that each of the following rational number is a nonterminating repeating decimal.

Solution 2(vi)

Question 2(vii)

Without actual division, show that each of the following rational number is a nonterminating repeating decimal.

Solution 2(vii)

Question 2(viii)

Without actual division, show that each of the following rational number is a nonterminating repeating decimal.

Solution 2(viii)

Question 3

Express each of the following as a fraction in simplest form:

Solution 3

Exercise Ex. 1D

Question 1

Define (i) rational numbers, (ii) irrational numbers, (iii) real numbers.Solution 1

Question 2

Classify the following numbers as rational or irrational:

Solution 2

Question 3

Prove that each of the following numbers is irrational:

Solution 3

Question 7

Solution 7

Question 8

Solution 8

Question 10

Solution 10

Question 11

Solution 11

Question 12

Prove that is irrational.Solution 12

Question 13

Solution 13

Question 16

(i) Give an example of two irrationals whose sum is rational.

(ii) Give an example of two irrationals whose product is rational.Solution 16

Question 17

State whether the given statement is true or false:

(i) The sum of two rationals is always rational.

(ii) The product of two rationals is always rational.

(iii) The sum of two irrationals is an irrational.

(iv) The product of two irrationals is an irrational.

(v) The sum of a rational and an irrational is irrational.

(vi) The product of a rational and an irrational is irrational.Solution 17

(i) The sum of two rationals is always rational – True

(ii) The product of two rationals is always rational – True

(iii) The sum of two irrationals is an irrational – False

(iv) The product of two irrationals is an irrational – False

(v) The sum of a rational and an irrational is irrational – True

(vi) The product of a rational and an irrational is irrational – True

Exercise Ex. 1E

Question 1

State Euclid’s division lemma.Solution 1

Question 2

State fundamental theorem of arithmetic.Solution 2

Question 3

Express 360 as product of its prime factors.Solution 3

Question 4

If a and b are two prime numbers then find HCF(a, b).Solution 4

Question 5

If a and b are two prime numbers then find LCM(a, b).Solution 5

Question 6

If the product of two numbers is 1050 and their HCF is 25, find their LCM.Solution 6

Question 7

What is a composite number?Solution 7

A whole number that can be divided evenly by numbers other than 1 or itself.Question 8

If a and b are relatively prime then what is their HCF?Solution 8

Question 9

If the rational number has a terminating decimal expansion, what is the condition to be satisfied by b?Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Show that there is no value of n for which (2ⁿ x 5ⁿ) ends in 5.Solution 12

Question 13

Is it possible to have two numbers whose HCF is 25 and LCM is 520?Solution 13

Question 14

Give an example of two irrationals whose sum is rational.Solution 14

Question 15

Give an example of two irrationals whose product is rational.Solution 15

Question 16

If a and b are relatively prime, what is their LCM?Solution 16

Question 17

The LCM of two numbers is 1200. Show that the HCF of these numbers cannot be 500. Why?Solution 17

Question 18

Express as a rational number in simplest form.Solution 18

Question 19

Express as a rational number in simplest formSolution 19

Question 20

Explain why 0.15015001500015 … is an irrational number.Solution 20

Question 21

Solution 21

Question 22

Write a rational number betweenand 2.Solution 22

Question 23

Explain why is a rational number.Solution 23

Exercise MCQ

Question 1

Which of the following is a pair of co-primes?

(a) (14, 35)

(b) (18, 25)

(c) (31,93)

(d)(32, 62)Solution 1

Question 2

If a = (2²×3³×5⁴) and b = (2³×3²×5) then HCF (a, b) = ?

(a) 90

(b) 180

(c) 360

(d)540Solution 2

Question 3

HCF of (2³×3²×5), (2²×3³×5²) and (2⁴×3×5³×7) is

(a) 30

(b) 48

(c) 60

(d)105Solution 3

Question 4

LCM of (2³×3×5) and (2⁴×5×7) is

(a) 40

(b) 560

(c) 1120

(d)1680Solution 4

Question 5

The HCF of two numbers is 27 and their LCM is 162. If one of the numbers is 54, what is the other number?

(a) 36

(b) 45

(c) 9

(d)81Solution 5

Question 6

The product of two numbers is 1600 and their HCF is 5. The LCM of the numbers is

(a) 8000

(b) 1600

(c) 320

(d)1605Solution 6

Question 7

What is the largest number that divides each one of 1152 and 1664 exactly?

(a) 32

(b) 64

(c) 128

(d)256Solution 7

Question 8

What is the largest number that divides 70 and 125, leaving remainders 5 and 8 respectively?

(a) 13

(b) 9

(c) 3

(d)585Solution 8

Question 9

What is the largest number that divides 245 and 1029, leaving remainder 5 in each case?

(a) 15

(b) 16

(c) 9

(d)5Solution 9

Question 10

Solution 10

Question 11

Euclid’s division lemma states that for any positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy

(a) 1 < r < b

(b) 0 < r ≤  b

(c) 0 ≤ r  < b

(d)0 < r < bSolution 11

Question 12

A number when divided by 143 leaves 31 as remainder. What will be the remainder when the same number is divided by 13?

(a) 0

(b) 1

(c) 3

(d)5Solution 12

Question 13

Which of the following is an irrational number?

(a) 

(b) 3.1416

(c) 

(d) 3.141141114 …Solution 13

Question 14

𝜋 is

(a) an integer

(b) a rational number

(c) an irrational number

(d)none of theseSolution 14

Question 15

(a) an integer

(b) a rational number

(c) an irrational number

(d) none of theseSolution 15

Question 16

2.13113111311113… is

(a) an integer

(b) a rational number

(c) an irrational number

(d)none of theseSolution 16

Question 17

The number 3.24636363 … is

(a) an integer

(b) a rational number

(c) an irrational number

(d)none of theseSolution 17

Question 18

Which of the following rational numbers is expressible as a terminating decimal?

Solution 18

Question 19

(a) one decimal place

(b) two decimal places

(c) three decimal places

(d) four decimal placesSolution 19

Question 20

(a) one decimal place

(b) two decimal places

(c) three decimal places

(d)four decimal placesSolution 20

Question 21

The number 1.732 is

(a) an irrational number

(b) a rational number

(c) an integer

(d)a whole numberSolution 21

Question 22

a and b are two positive integers such that the least prime factor of a is 3 and the least prime factor of b is 5. Then, the least prime factor of (a+b) is

(a) 2

(b) 3

(c) 5

(d)8Solution 22

Question 23

(a) a rational number

(b) an irrational number

(c) a terminating decimal

(d)a nonterminating repeating decimalSolution 23

Question 24

(a) a fraction

(b) a rational number

(c) an irrational number

(d)none of theseSolution 24

Question 25

(a) an integer

(b) a rational number

(c) an irrational number

(a) none of theseSolution 25

Question 26

What is the least number that is divisible by all the natural numbers from 1 to 10 (both inclusive)

(a) 100

(b) 1260

(c) 2520

(d) 5040Solution 26

Exercise FA

Question 1

(a) a terminating decimal

(b) a nonterminating, repeating decimal

(c) a nonterminating and nonrepeating decimal

(d)none of theseSolution 1

Question 2

Which of the following has a terminating decimal expansion?

Solution 2

Question 3

On dividing a positive integer n by 9, we get 7 as remainder. What will be the remainder if (3n – 1) is divided by 9?

(a) 1

(b) 2

(c) 3

(d)4Solution 3

Question 4

Solution 4

Question 5

Show that any number of the form 4ⁿ, n ∊ N can never end with the digit 0.Solution 5

Question 6

The HCF of two numbers is 27 and their LCM is 162. If one of the number is 81, find the other.Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Which of the following numbers are irrational?

Solution 9

Question 10

Solution 10

Question 11

Find the HCF and LCM of 12, 15, 18, 27.Solution 11

Question 12

Give an example of two irrationals whose sum is rational.Solution 12

Question 13

Give prime factorization of 4620.Solution 13

Question 14

Find the HCF of 1008 and 1080 by prime factorization method.Solution 14

Question 15

Solution 15

Question 16

Find the largest number which divides 546 and 764, leaving remainders 6 and 8 respectively.Solution 16

Question 17

Solution 17

Question 18

Show that every positive odd integer is of the form (4q + 1) or (4q + 3) for some integer q.Solution 18

Question 19

Show that one and only one out of n, (n+2) and (n+4) is divisible by 3, where n is any positive integer.Solution 19

Question 20

Solution 20

Exercise Ex. 3A

Question 1

Solve each of the following systems of equations graphically:

2x + 3y = 2

x – 2y = 8Solution 1

Question 3

Solve each of the following systems of equations graphically:

2x + 3y = 8

x – 2y + 3 = 0Solution 3

Question 4

Solve each of the following systems of equations graphically:

2x – 5y + 4 = 0

2x + y – 8 = 0Solution 4

Question 5

Solve each of the following systems of equations graphically:

3x + 2y = 12, 5x – 2y = 4.Solution 5

Since the two graphs intersect at (2, 3),

x = 2 and y = 3.Question 6

Solve each of the following systems of equations graphically:

3x + y + 1 = 0

2x – 3y + 8 = 0Solution 6

Question 7

Solve each of the following systems of equations graphically:

2x + 3y + 5 = 0, 3x – 2y – 12 = 0.Solution 7

Since the two graphs intersect at (2, -3),

x = 2 and y = -3.Question 8

Solve each of the following systems of equations graphically:

2x – 3y + 13 = 0, 3x – 2y + 12 = 0.Solution 8

Since the two graphs intersect at (-2, 3),

x = -2 and y = 3.Question 9

Solve each of the following systems of equations graphically:

2x + 3y – 4 = 0, 3x – y + 5 = 0.Solution 9

Since the two graphs intersect at (-1, 2),

x = -1 and y = 2.Question 10

Solve each of the following systems of equations graphically:

x + 2y + 2 = 0

3x + 2y – 2 = 0Solution 10

Question 11

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the x-axis:

x – y + 3 = 0, 2x + 3y – 4 = 0.Solution 11

Question 12

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the x-axis:

2x – 3y + 4 = 0, x + 2y – 5 = 0.Solution 12

Question 13

Solve the following system of linear equations graphically:

4x – 3y + 4 = 0, 4x + 3y – 20 = 0

Find the area of the region bounded by these lines and the x-axis.Solution 13

Question 14

Solve the following system of linear equation graphically:

x – y + 1 = 0, 3x + 2y – 12 = 0

Calculate the area bounded by these lines and x-axis.Solution 14

Question 15

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the x-axis:

x – 2y + 2 = 0, 2x + y – 6 = 0.Solution 15

Question 16

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:

2x – 3y + 6 = 0, 2x + 3y – 18 = 0.Solution 16

Question 17

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:

4x – y – 4 = 0, 3x + 2y – 14 = 0.Solution 17

Question 18

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:

x – y – 5 = 0, 3x + 5y – 15 = 0.Solution 18

Question 19

Solve the following system of linear equations graphically:

2x – 5y + 4 = 0, 2x + y – 8 = 0

Find the point, where these lines meet the y-axisSolution 19

Question 20

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:

5x – y – 7 = 0, x – y + 1 = 0.Solution 20

Question 21

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:

2x – 3y = 12, x + 3y = 6.Solution 21

Question 22

Show graphically that each of the following given systems of equations has infinitely many solutions:

2x + 3y = 6, 4x + 6y = 12.Solution 22

Since the graph of the system of equations is coincident lines, the system has infinitely many solutions.Question 23

Show graphically that the system of equations 3x – y = 5, 6x – 2y = 10 has infinitely many solutions.Solution 23

Question 24

Show graphically that the system of equations 2x + y = 6, 6x + 3y = 18 has infinitely many solutions.Solution 24

Question 25

Show graphically that each of the following given systems of equations has infinitely many solutions:

x – 2y = 5, 3x – 6y = 15.Solution 25

Since the graph of the system of equations is coincident lines, the system has infinitely many solutions. Question 26

Show graphically that each of the following given systems of equations is inconsistent, i.e., has no solution:

x – 2y = 6, 3x – 6y = 0Solution 26

Since the graph of the system of equations is parallel lines, the system has no solution and hence is inconsistent.Question 27

Show graphically that each of the following given systems of equations is inconsistent, i.e., has no solution:

2x + 3y = 4, 4x + 6y = 12.Solution 27

Since the graph of the system of equations is parallel lines, the system has no solution and hence is inconsistent.Question 28

Show graphically that each of the following given systems of equations is inconsistent, i.e., has no solution:

2x + y = 6, 6x + 3y = 20.Solution 28

Since the graph of the system of equations is parallel lines, the system has no solution and hence is inconsistent.Question 29

Draw the graphs of the following equations on the same graph paper:

2x + y = 2, 2x + y = 6.

Find the coordinates of the vertices of the trapezium formed by these lines. Also, find the area of the trapezium so formed.Solution 29

Exercise Ex. 3B

Question 1

Solve for x and y

x + y = 3

4x – 3y = 26Solution 1

Question 2

Solve for x and y:

Solution 2

Question 3

Solve for x and y

2x + 3y = 0

3x + 4y = 5Solution 3

Question 4

Solve for x and y

2x – 3y = 13

7x – 2y = 20Solution 4

Question 5

Solve for x and y

3x – 5y – 19 = 0

-7x + 3y + 1 = 0Solution 5

Question 6

Solve for x and y:

2x – y + 3 = 0, 3x – 7y + 10 = 0.Solution 6

Question 7

Solve for x and y:

Solution 7

Question 8

Solve for x and y

Solution 8

Question 9

Solve for x and y

Solution 9

Question 10

Solve for x and y

Solution 10

Question 11

Solve for x and y

Solution 11

Question 12

Solve for x and y

Solution 12

Question 13

Solve for x and y:

0.4x + 0.3y = 1.7, 0.7x – 0.2y = 0.8.Solution 13

Question 14

Solve for x and y:

0.3x + 0.5y = 0.5, 0.5x + 0.7y = 0.74.Solution 14

Question 15

Solve for x and y

7(y + 3) – 2(x + 2) = 14

4(y – 2) + 3(x – 3) = 2Solution 15

Question 16

Solve for x and y

6x + 5y = 7x + 2y + 1 = 2(x + 6y – 1)Solution 16

Question 17

Solve for x and y

Solution 17

Question 18

×Solve for x and y

Solution 18

Putting the given equations become

5u + 6y = 13—(1)

3u + 4y = 7 —-(2)

Multiplying (1) by 4 and (2) by 6, we get

20u + 24y = 52—(3)

18u + 24y = 42—(4)

Subtracting (4) from (3), we get

2u = 10 u = 5

Putting u = 5 in (1), we get

5 × 5 + 6y = 13

6y = 13 – 25

6y = -12

y = -2

Question 19

Solve for x and y

Solution 19

The given equations are and 

Putting 

x + 6v = 6 —-(1)

3x – 8v = 5—(2)

Multiplying (1) by 4 and (2) by 3

4x + 24v = 24—(3)

9x – 24v = 15 —(4)

Adding (3) and (4)

13x = 24 + 15 = 39

Puttingx = 3 in (1)

3 + 6v = 6

6v = 6 – 3 = 3

solution is x = 3, y = 2Question 20

Solve for x and y

Solution 20

Putting in the given equation

2x – 3v = 9 —(1)

3x + 7v = 2 —(2)

Multiplying (1) by7 and (2) by 3

14x – 21v = 63 —(3)

9x + 21v = 6 —(4)

Adding (3) and (4), we get

Putting x= 3 in (1), we get

2 × 3 – 3v = 9

         -3v = 9 – 6

     -3v= 3

       v = -1

the solution is x = 3, y = -1Question 21

Solve for x and y

Solution 21

Question 22

Solve for x and y

Solution 22

Putting in the equation

9u – 4v = 8 —(1)

13u + 7v = 101—(2)

Multiplying (1) by 7 and (2) by 4, we get

63u – 28v = 56—(3)

52u + 28v = 404—(4)

Adding (3) and (4), we get

Putting u = 4 in (1), we get

9 × 4 – 4v = 8

36 – 4v = 8

-4v = 8 – 36

-4v = -28

Question 23

Solve for x and y

Solution 23

Putting in the given equation, we get

5u – 3v = 1 —(1)

Multiplying (1) by 4 and (2) by 3, we get

20u – 12v = 4—-(3)

27u + 12v = 90—(4)

Adding (3) and (4), we get

Putting u = 2 in (1), we get

(5 × 2) – 3v = 1

10 – 3v = 1

-3v = 1 – 10 -3v = -9 

 v = 3

Question 24

Solve for x and y:

Solution 24

Question 25

Solve for x and y

4x + 6y = 3xy

8x + 9y = 5xy;Solution 25

4x + 6y = 3xy

Putting in (1) and (2), we get

4v + 6u = 3—(3)

8v + 9u = 5—(4)

Multiplying (3) by 9 and (4) by 6, we get

36v + 54u = 27 —(5)

48v + 54u = 30 —(6)

Subtracting (3) from (4), we get

12v = 3

Putting in (3), we get

the solution is x = 3, y = 4Question 26

Solve for x and y:

Solution 26

Question 27

Solve for x and y:

Solution 27

Question 28

Solve for x and y

Solution 28

Putting 

3u + 2v = 2—-(1)

9u – 4v = 1—-(2)

Multiplying (1) by 2 and (2) by 1. We get

6u + 4v = 4—-(3)

9u – 4v = 1—-(4)

Adding (3) and (4), we get

Adding (5) and (6), we get

Putting in (5). We get

the solution is Question 29

Solve for x and y

Solution 29

The given equations are

Putting 

Adding (1) and (2)

Putting value of u in (1)

Hence the required solution isx = 4, y = 5Question 30

Solve for x and y

Solution 30

Putting in the equation, we get

44u + 30v = 10—-(1)

55u + 40v = 13—-(2)

Multiplying (1) by 4 and (2) by 3, we get

176u + 120v = 40—(3)

165u + 120v = 39—(4)

Subtracting (4) from (3), we get

Putting in (1) we get

Adding (5) and (6), we get

Putting x = 8 in (5), we get

8 + y = 11 y = 11 – 8 = 3

the solution is x = 8, y = 3Question 31

Solve for x and y:

Solution 31

Question 32

Solve for x and y

71x + 37y = 253

37x + 71y = 287Solution 32

The given equations are

71x + 37y = 253—(1)

37x + 71y = 287—(2)

Adding (1) and (2)

108x + 108y = 540

108(x + y) = 540

—-(3)

Subtracting (2) from (1)

34x – 34y = 253 – 287 = -34

34(x – y) = -34

—(4)

Adding (3) and (4)

2x = 5 – 1= 4

Subtracting (4) from (3)

2y = 5 + 1 = 6

solution is x = 2, y = 3Question 33

Solve for x and y

217x + 131y = 913

131x + 217y = 827Solution 33

217x + 131y = 913—(1)

131x + 217y = 827—(2)

Adding (1) and (2), we get

348x + 348y = 1740

348(x + y) = 1740

x + y = 5—-(3)

Subtracting (2) from (1), we get

86x – 86y = 86

86(x – y) = 86

   x – y = 1—(4)

Adding (3) and (4), we get

2x = 6

x = 3

putting x = 3 in (3), we get

3 + y = 5

y = 5 – 3 = 2

solution is x = 3, y = 2Question 34

Solve for x and y:

23x – 29y = 98, 29x – 23y = 110.Solution 34

Question 35

Solve for x and y:

Solution 35

Question 36

Solve for x and y:

Solution 36

Question 37

Solve for x and y

where Solution 37

The given equations are

Multiplying (1) by 6 and (2) by 20, we get

Multiplying (3) by 6 and (4) by 5, we get

18u + 60v = -54—(5)

125u – 60v = —(6)

Adding (5) and (6), we get

Question 38

Solve for x and y:

Solution 38

Question 39

Solve for x and y:

Solution 39

Question 40

Solve for x and y:

x + y = a + b, ax – by = a² – b².Solution 40

Question 41

Solve for x and y

Solution 41

Question 42

Solve for x and y:

px + py = p – q, qx – py = p + q.Solution 42

Question 43

Solve for x and y:

Solution 43

Question 44

Solve for x and y

6(ax + by) = 3a + 2b

6(bx – ay) = 3b – 2aSolution 44

6(ax + by) = 3a + 2b

6ax + 6by = 3a + 2b —(1)

6(bx – ay) = 3b – 2a

6bx – 6ay = 3b- 2a —(2)

6ax + 6by = 3a + 2b —(1)

6bx – 6ay = 3b – 2a —(2)

Multiplying (1) by a and (2) by b

Adding (3) and (4), we get

Substituting  in(1), we get

Hence, the solution is  Question 45

Solve for x and y:

ax – by = a² + b², x + y = 2a.Solution 45

Question 46

Solve for x and y

bx – ay + 2ab = 0Solution 46

Question 47

Solve for x and y

x + y = 2abSolution 47

Taking L.C.M, we get

Multiplying (1) by 1 and (2) by 

Subtracting (4) from (3), we get

Substituting x = ab in (3), we get

solution is x = ab, y = abQuestion 48

Solve for x and y:

x + y = a + b, ax – by = a² – b².Solution 48

Question 49

Solve for x and y:

a²x + b²y = c², b²x + a²y = d².Solution 49

Question 50

Solve for x and y:

Solution 50

Exercise Ex. 3C

Question 1

Solve for x and y by method of cross multiplication:

x + 2y + 1 = 0

2x – 3y – 12 = 0Solution 1

x + 2y + 1 = 0 —(1)

2x – 3y – 12 = 0 —(2)

By cross multiplication, we have

Hence, x = 3 and y = -2 is the solutionQuestion 2

Solve for x and y by method of cross multiplication:

3x – 2y + 3 = 0

4x + 3y – 47 = 0Solution 2

3x – 2y + 3 = 0

4x + 3y – 47 = 0

By cross multiplication we have

the solution is x = 5, y = 9Question 3

Solve for x and y by method of cross multiplication:

6x – 5y – 16 = 0

7x – 13y + 10 = 0Solution 3

6x – 5y – 16 = 0

7x – 13y + 10 = 0

By cross multiplication we have

the solution is x = 6, y = 4Question 4

Solve for x and y by method of cross multiplication:

3x + 2y + 25 = 0

2x + y + 10 = 0Solution 4

3x + 2y + 25 = 0

2x + y + 10 = 0

By cross multiplication, we have

the solution is x = 5,y = -20Question 5

Solve for x and y by method of cross multiplication:

2x +5y = 1

2x + 3y = 3Solution 5

2x + 5y – 1 = 0 —(1)

2x + 3y – 3 = 0—(2)

By cross multiplication we have

the solution is x = 3, y = -1Question 6

Solve for x and y by method of cross multiplication:

2x + y – 35 = 0

3x + 4y – 65 = 0Solution 6

2x + y – 35 = 0

3x + 4y – 65 = 0

By cross multiplication, we have

Question 7

Solve each of the following systems of equations by using the method of cross multiplication:

7x – 2y = 3, 22x – 3y = 16.Solution 7

Question 8

Solve for x and y by method of cross multiplication:

Solution 8

Question 9

Solve for x and y by method of cross multiplication:

Solution 9

Taking 

u + v – 7 = 0

2u + 3v – 17 = 0

By cross multiplication, we have

the solution is Question 10

Solve for x and y by method of cross multiplication:

Solution 10

Let in the equation

5u – 2v + 1 = 0

15u + 7v – 10 = 0

Question 11

Solve for x and y by method of cross multiplication:

Solution 11

Question 12

Solve for x and y by method of cross multiplication:

2ax + 3by – (a + 2b) = 0

3ax+ 2by – (2a + b) = 0Solution 12

2ax + 3by – (a + 2b) = 0

3ax+ 2by – (2a + b) = 0

By cross multiplication, we have

Question 13

Solve each of the following systems of equations by using the method of cross multiplication:

Solution 13

Exercise Ex. 3D

Question 1

Show that the following system of equations has a unique solution:

3x + 5y = 12, 5x + 3y = 4

Also, find the solution of the given system of equations.Solution 1

Question 2

Show that each of the following systems of equations has a unique solution and solve it:

2x – 3y = 17, 4x + y = 13.Solution 2

Question 3

Show that the following system of equations has a unique solution:

Also, find the solution of the given system of equationsSolution 3

Question 4

Find the value of k for which each of the following systems of equations has a unique solution:

2x + 3y – 5 = 0, kx – 6y – 8 = 0.Solution 4

Question 5

Find the value of k for which each of the following systems of equations has a unique solution:

x – ky = 2, 3x + 2y + 5 = 0.Solution 5

Question 6

Find the value of k for which each of the following systems of equations has a unique solution:

5x – 7y – 5 = 0, 2x + ky – 1 = 0.Solution 6

Question 7

Find the value of k for which each of the following systems of equations has a unique solution:

4x + ky + 8 = 0, x + y + 1 = 0.Solution 7

Question 8

Find the value of k for which each of the following systems of equations has a unique solution:

4x – 5y = k , 2x – 3y = 12Solution 8

4x – 5y – k = 0, 2x – 3y – 12 = 0

These equations are of the form

Thus, for all real value of k the given system of equations will have a unique solutionQuestion 9

Find the value of k for which each of the following systems of equations has a unique solution:

kx + 3y = (k – 3),12x + ky = kSolution 9

kx + 3y – (k – 3) = 0

12x + ky – k = 0

These equations are of the form

Thus, for all real value of k other than , the given system of equations will have a unique solutionQuestion 10

Show that the system of equations

2x – 3y = 5, 6x – 9y = 15

has an infinite number of solutions.Solution 10

2x – 3y – 5 = 0, 6x – 9y – 15 = 0

These equations are of the form

Hence the given system of equations has infinitely many solutionsQuestion 11

Show that the system of equations  Solution 11

Question 12

For what value of k, the system of equations

kx + 2y = 5, 3x – 4y = 10

has (i) a unique solution (ii) no solution?Solution 12

kx + 2y – 5 = 0

3x – 4y – 10 = 0

These equations are of the form

This happens when 

Thus, for all real value of k other that , the given system equations will have a unique solution

(ii) For no solution we must have 

Hence, the given system of equations has no solution if Question 13

For what value of k, the system of equations

x + 2y = 5, 3x + ky + 15 = 0

has (i) a unique solution (ii) no solution?Solution 13

x + 2y – 5= 0

3x + ky + 15 = 0

These equations are of the form of

Thus for all real value of k other than 6, the given system ofequation will have unique solution

(ii) For no solution we must have

k = 6

Hence the given system will have no solution when k = 6.Question 14

For what value of k, the system of equations

x + 2y = 3, 5x + ky + 7 = 0

has (i) a unique solution (ii) no solution?

Is there any value of k for which the given system has an infinite number of solutions?Solution 14

x + 2y – 3 = 0, 5x + ky + 7 = 0

These equations are of the form

(i)For a unique solution we must have

Thus, for all real value of k other than 10

The given system of equation will have a unique solution.

(ii)For no solution we must have

Hence the given system of equations has no solution if 

For infinite number of solutions we must have

This is never possible since 

There is no value of k for which system of equations has infinitely many solutionsQuestion 15

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

2x + 3y = 7

(k – 1)x + (k + 2)y = 3kSolution 15

2x + 3y – 7 = 0

(k – 1)x + (k + 2)y – 3k = 0

These are of the form

This hold only when

Now the following cases arises

Case : I

Case: II

Case III

For k = 7, there are infinitely many solutions of the given system of equationsQuestion 16

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

2x + (k – 2)y =k

6x + (2k – 1)y = (2k + 5)Solution 16

2x + (k – 2)y – k = 0

6x + (2k – 1)y – (2k + 5) = 0

These are of the form

For infinite number of solutions, we have

This hold only when

Case (1)

Case (2)

Case (3)

Thus, for k = 5 there are infinitely many solutionsQuestion 17

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

kx + 3y = (2k +1)

2(k + 1)x + 9y = (7k + 1)Solution 17

kx + 3y – (2k +1) = 0

2(k + 1)x + 9y – (7k + 1) = 0

These are of the form

For infinitely many solutions, we must have

This hold only when

Now, the following cases arise

Case – (1)

Case (2)

Case (3)

Thus, k = 2, is the common value for which there are infinitely many solutionsQuestion 18

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

5x + 2y = 2k

2(k + 1)x + ky = (3k + 4)Solution 18

5x + 2y – 2k = 0

2(k +1)x + ky – (3k + 4) = 0

These are of the form

For infinitely many solutions, we must have

These hold only when

Case I

Thus, k = 4 is a common value for which there are infinitely by many solutions.Question 19

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

(k – 1)x – y = 5

(k + 1)x + (1 – k)y = (3k + 1)Solution 19

(k – 1)x – y – 5 = 0

(k + 1)x + (1 – k)y – (3k + 1) = 0

These are of the form

For infinitely many solution, we must now

k = 3 is common value for which the number of solutions is infinitely manyQuestion 20

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

(k – 3)x + 3y = k, kx + ky = 12.Solution 20

Question 21

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:

(a – 1)x + 3y = 2

6x + (1 – 2b)y = 6Solution 21

(a – 1)x + 3y – 2 = 0

6x + (1 – 2b)y – 6 = 0

These equations are of the form

For infinite many solutions, we must have

Hence a = 3 and b = -4Question 22

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:

(2a – 1)x + 3y = 5

3x + (b – 1)y = 2Solution 22

(2a – 1)x + 3y – 5 = 0

3x + (b – 1)y – 2 = 0

These equations are of the form

These holds only when

Question 23

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:

2x – 3y = 7

(a + b)x + (a + b – 3)y = (4a + b)Solution 23

2x – 3y – 7 = 0

(a + b)x + (a + b – 3)y – (4a + b) = 0

These equation are of the form

For infinite number of solution

Putting a = 5b in (2), we get

Putting b = -1 in (1), we get

Thus, a = -5, b = -1Question 24

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:

2x + 3y=7, (a + b + 1)x +(a + 2b + 2)y = 4(a + b)+ 1.Solution 24

Question 25

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:

2x + 3y = 7, (a + b)x + (2a – b)y = 21.Solution 25

Question 26

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:

2x + 3y = 7, 2ax + (a + b)y = 28.Solution 26

Question 27

Find the value of k for which each of the following systems of equations no solution:

8x + 5y = 9, kx + 10y = 15.Solution 27

Question 28

Find the value of k for which each of the following systems of equations no solution:

kx + 3y = 3, 12x + ky = 6.Solution 28

Question 29

Find the value of k for which each of the following systems of equations no solution:

Solution 29

Question 30

Find the value of k for which each of the following systems of equations no solution:

kx + 3y = k – 3, 12x + ky = k.Solution 30

Question 31

Find the value of k for which the system of equations

5x – 3y = 0;2x + ky = 0

has a nonzero solution.Solution 31

We have 5x – 3y = 0 —(1)

2x + ky = 0—(2)

Comparing the equation with

These equations have a non – zero solution if 

Exercise Ex. 3E

Question 1

5 chairs and 4 tables together cost Rs.5600, while 4 chairs and 3 tables together cost Rs.4340. Find the cost of a chair and that of a table.Solution 1

Question 2

23 spoons and 17 forks together cost Rs.1770, while 17 spoons and 23 forks together cost Rs.1830. Find the cost of a spoon and that of a fork.Solution 2

Question 3

A lady has only 25-paisa and 50-paisa coins in her purse. If she has 50 coins in all Rs.19.50, how many coins of each kind does she have?Solution 3

Question 4

The sum of two numbers is 137 and their difference is 43. Find the numbers.Solution 4

Let the two numbers be x and y respectively.

Given:

x + y = 137 —(1)

x – y = 43 —(2)

Adding (1) and (2), we get

2x = 180

Putting x = 90 in (1), we get

90 + y = 137

y = 137 – 90

  = 47

Hence, the two numbers are 90 and 47.Question 5

Find two numbers such that the sum of twice the first and thrice the second is 92, and four times the first exceeds seven times the second by 2.Solution 5

Let the first and second number be x and y respectively.

According to the question:

2x + 3y = 92 —(1)

4x – 7y = 2 —(2)

Multiplying (1) by 7 and (2) by 3, we get

14 x+ 21y = 644 —(3)

12x – 21y = 6 —(4)

Adding (3) and (4), we get

Putting x = 25 in (1), we get

2 × 25 + 3y = 92

50 + 3y = 92

 3y = 92 – 50

y = 14

Hence, the first number is 25 and second is 14Question 6

Find two numbers such that the sum of thrice the first and the second is 142, and four times the first exceeds the second by 138.Solution 6

Let the first and second numbers be x and y respectively.

According to the question:

3x + y = 142 —(1)

4x – y = 138 —(2)

Adding (1) and (2), we get

Putting x = 40 in (1), we get

3 × 40 + y = 142

y = 142 – 120

y = 22

Hence, the first and second numbers are 40 and 22.Question 7

If 45 is subtracted from twice the greater of two numbers, it results in the other number. If 21 is subtracted from twice the smaller number, it results in the greater number. Find the number.Solution 7

Let the greater number be x and smaller be y respectively.

According to the question:

2x – 45 = y

2x – y = 45—(1)

and

2y – x = 21

 -x + 2y = 21—(2)

Multiplying (1) by 2 and (2) by 1

4x – 2y = 90—(3)

-x + 2y = 21 —(4)

Adding (3) and (4), we get

3x = 111

Putting x = 37 in (1), we get

2 × 37 – y = 45

 74 – y = 45

y = 29

Hence, the greater and the smaller numbers are 37 and 29.Question 8

If three times the larger of two numbers is divided by the smaller, we get 4 as the quotient and 8 as the remainder. If five times the smaller is divided by the larger, we get 3 as the quotient and 5 as the remainder. Find the numbers.Solution 8

Let the larger number be x and smaller be y respectively.

We know,

Dividend = Divisor × Quotient + Remainder

3x = y × 4 + 8

3x – 4y = 8 —(1)

And

5y = x × 3 + 5

-3x + 5y = 5 —(2)

Adding (1) and (2), we get

y = 13

putting y = 13 in (1)

Hence, the larger and smaller numbers are 20 and 13 respectively.Question 9

If 2 is added to each of two given numbers, their ratio becomes 1: 2. However, if 4 is subtracted from each of the given numbers, the ratio becomes 5: 11. Find the numbers.Solution 9

Let the required numbers be x and y respectively.

Then,

Therefore,

2x – y =-2—(1)

11x – 5y = 24 —(2)

Multiplying (1) by 5 and (2) by 1

10x – 5y = -10—(3)

11x – 5y = 24—(4)

Subtracting (3) and (4) we get

x = 34

putting x = 34 in (1), we get

 2 × 34 – y = -2

 68 – y = -2

-y = -2 – 68

y = 70

Hence, the required numbers are 34 and 70.Question 10

The difference between two numbers is 14 and the difference between their squares is 448. Find the numbers.Solution 10

Let the numbers be x and y respectively.

According to the question:

x – y = 14 —(1)

From (1), we get

x = 14 + y —(3)

putting x = 14 + y in (2), we get

Putting y = 9 in (1), we get

x – 9 = 14

  x = 14 + 9 = 23

Hence the required numbers are 23 and 9Question 11

The sum of the digits of a two digit number is 12. The number obtained by interchanging its digits exceeds the given number by 18. Find the number.Solution 11

Let the ten’s digit be x and units digit be y respectively.

Then,

x + y = 12—(1)

Required number = 10x + y

Number obtained on reversing digits = 10y + x

According to the question:

10y + x – (10x + y) = 18

10y + x – 10x – y = 18

9y – 9x = 18

y – x = 2 —-(2)

Adding (1) and (2), we get

Putting y = 7 in (1), we get

x + 7 = 12

x = 5

Number= 10x + y

                  = 10 × 5 + 7

                  = 50 + 7

                  = 57

Hence, the number is 57.
Question 12

A number consisting of two digits is seven times the sum of its digits. When 27 is subtracted from the number, the digits are reversed. Find the number.Solution 12

Let the ten’s digit of required number be x and its unit’s digit be y respectively

Required number = 10x + y

 10x + y = 7(x + y)

     10x + y = 7x + 7y

       3x – 6y = 0—(1)

Number found on reversing the digits = 10y + x

(10x + y) – 27 = 10y + x

      10x – x + y – 10y = 27

        9x – 9y = 27

         (x – y) = 27

           x – y = 3—(2)

Multiplying (1) by 1 and (2) by 6

3x – 6y = 0—(3)

6x – 6y = 18 —(4)

Subtracting (3) from (4), we get

Putting x = 6 in(1), we get

3 × 6 – 6y = 0

18 – 6y = 0

Number = 10x + y

            = 10 × 6 + 3

            = 60 + 3

            = 63

Hence the number is 63.
Question 13

The sum of the digits of a two-digit number is 15. The number of obtained by interchanging the digits exceeds the given number by 9. Find the number.Solution 13

Let the ten’s digit and unit’s digits of required number be x and y respectively.

Then,

x + y = 15—(1)

Required number = 10x + y

Number obtained by interchanging the digits = 10y + x

10y + x – (10x + y) = 9

10y + x – 10x – y = 9

       9y – 9x = 9

Add (1) and (2), we get

Putting y = 8 in (1), we get

x + 8 = 15

x = 15 – 8 = 7

Required number = 10x + y

                         = 10 × 7 + 8

                         = 70 + 8

                         = 78

Hence the required number is 78.
Question 14

A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.Solution 14

Let the ten’s and unit’s of required number be x and y respectively.

Then,required number =10x + y

According to the given question:

 10x + y = 4(x + y) + 3

 10x + y = 4x + 4y + 3

6x – 3y = 3

  2x – y = 1 —(1)

And

10x + y + 18 = 10y + x

9x – 9y = -18

 x – y = -2—(2)

Subtracting (2) from (1), we get

x = 3

Putting x = 3 in (1), we get

2 × 3 – y = 1

y = 6 – 1 = 5

x = 3, y = 5

Required number = 10x + y

                         = 10 × 3 + 5

                         = 30 + 5

                         = 35

Hence, required number is 35.
Question 15

A number consists of two digits. When it is divided by the sum of its digits, the quotient is 6 with no remainder. When the number is diminished by 9, the digits are reversed. Find the number.Solution 15

Let the ten’s digit and unit’s digit of required number be x and y respectively.

We know,

Dividend = (divisor × quotient) + remainder

According to the given questiion:

10x + y = 6 × (x + y) + 0

10x – 6x + y – 6y = 0

4x – 5y = 0 —(1)

Number obtained by reversing the digits is 10y + x

 10x + y – 9 = 10y + x

9x – 9y = 9

9(x – y) =9

(x – y) = 1—(2)

Multiplying (1) by 1 and (2) by 5, we get

4x – 5y = 0 —(3)

5x – 5y = 5 —(4)

Subtracting (3) from (4), we get

x = 5

Putting x = 5 in (1), we get

x =5 and y= 4

Hence, required number is 54.Question 16

A two – digit number is such that the product of its digits is 35. If 18 is added to the number, the digits interchanged their places. Find the number.Solution 16

Let the ten’s and unit’s digits of the required number be x and y respectively.

Then, xy = 35

Required number = 10x + y

Also,

(10x + y) + 18 = 10y + x

9x – 9y = -18

9(y – x) = 18—(1)

y – x = 2

Now, 

Adding (1) and (2),

2y = 12 + 2 = 14

y = 7

Putting y = 7 in (1),

7 – x = 2

x = 5

Hence, the required number = 5 × 10 + 7

                                        = 57Question 17

A two-digit number is such that the product of its digit is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.Solution 17

Let the ten’s and unit’s digits of the required number be x and y respectively.

Then, xy = 18

Required number = 10x + y

Number obtained on reversing its digits = 10y + x

(10x + y) – 63 = (10y + x)

9x – 9y = 63

x – y = 7—(1)

Now,

Adding (1) and (2), we get

Putting x = 9 in (1), we get

9 – y = 7

y = 9 – 7

y ₌2

x = 9, y = 2

Hence, the required number = 9 × 10 + 2

                                        = 92.Question 18

The sum of a two-digit number and the number obtained by reversing the order of its digits is 121, and the two digits differ by 3. Find the number.Solution 18

Question 19

The sum of the numerator and denominator of a fraction is 8. If 3 is added to both of the numerator and the denominator, the fraction becomes Find the fraction.Solution 19

Let the numerator and denominator of fraction be x and y respectively.

According to the question:

x + y = 8—(1)

And

Multiplying (1) be 3 and (2) by 1

3x + 3y = 24—(3)

4x – 3y = -3 —(4)

Add (3) and (4), we get

Putting x = 3 in (1), we get

3 + y= 8

y = 8 – 3

y = 5

x = 3, y = 5

Hence, the fraction is Question 20

If 2 is added to the numerator of a fraction, it reduces to and if 1 is subtracted from the denominator, it reduces to . Find the fractionSolution 20

Let the numerator and denominator be x and y respectively.

Then the fraction is .

Subtracting (1) from (2), we get

x = 3

Putting x = 3 in (1), we get

2 × 3 – 4

-y = -4 -6

y = 10

x = 3 and y = 10

Hence the fraction is Question 21

The denominator of a fraction is greater than its numerator by 11. If 8 is added to both its numerator and denominator, it becomes Find the fraction.Solution 21

Let the numerator and denominator be x and y respectively.

Then the fraction is _.

According to the given question:

y = x + 11

y- x = 11—(1)

and

-3y + 4x = -8 —(2)

Multiplying (1) by 4 and (2) by 1

4y – 4x = 44—(3)

-3y + 4x = -8—(4)

Adding (3) and (4), we get

y = 36

Putting y = 36 in (1), we get

y – x = 11

36 – x = 11 

  x = 25

x = 25, y = 36

Hence the fraction is Question 22

Find a fraction which becomes when 1 is subtracted from the numerator and 2 is added to the denominator, and the fraction becomes when 7 is subtracted from the numerator and 2 is subtracted from the denominator.Solution 22

Let the numerator and denominator be x and y respectively.

Then the fraction is 

Subtracting (1) from (2), we get

x = 15

Putting x = 15 in (1), we get

2 × 15 – y = 4

 30 – y = 4

 y = 26

x = 15 and y = 26

Hence the given fraction is Question 23

The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator are increased by 3, they are in the ratio 2 : 3. Determine the fraction.Solution 23

Question 24

The sum of two numbers is 16 and the sum of their reciprocals is Find the numbers.Solution 24

Let the two numbers be x and y respectively.

According to the given question:

x + y = 16—(1)

And

—(2)

From (2),

xy = 48

We know,

Adding (1) and (3), we get

2x = 24

x = 12

Putting x = 12 in (1),

y = 16 – x

   = 16 – 12

   = 4

The required numbers are 12 and 4Question 25

There are two classrooms A and B. If 10 students are sent from A to B, the number of students in each room becomes the same. If 20 students are sent from B to A, the number of students in A becomes double the number of students in B. Find the number of students in each room.Solution 25

Let the number of student in class room A and B be x and y respectively.

When 10 students are transferred from A to B:

x – 10 = y + 10

 x – y = 20—(1)

When 20 students are transferred from B to A:

 2(y – 20) = x + 20

 2y – 40 = x + 20

-x + 2y = 60—(2)

Adding (1) and (2), we get

y = 80

Putting y = 80 in (1), we get

x – 80 = 20

 x = 100

Hence, number of students of A and B are 100 and 80 respectively.Question 26

Taxi charges in a city consist of fixed charges and the remaining depending upon the distance travelled in kilometres. If a man travels 80 km, he pays Rs.1330, and travelling 90 km, he pays Rs.1490. Find the fixed charges and rate per km.Solution 26

Question 27

A part of monthly hostel charges in a college hostel are fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 25 days, he has to pay Rs.4500, whereas a student B who takes food for 30 days, has to pay Rs.5200. Find the fixed charges per month and the cost of the food per day.Solution 27

Question 28

A man invested an amount at 10% per annum and another amount at 8% per annum simple interest. Thus, he received Rs. 1350 as an annual interest. Had he interchanged the amounts invested, he would have received Rs.45 less as interest. What amounts did he invest at different rates?Solution 28

Question 29

The monthly incomes of A and B are in the ratio 5 : 4 and their monthly expenditures are in the ratio 7 : 5. If each saves Rs.9000 per month, find the monthly income of eachSolution 29

Question 30

A man sold a chair and a table together for Rs.1520, thereby making a profit of 25% on chair and 10% on table. By selling them together for Rs.1535, he would have made a profit of 10% on the chair and 25% on the table. Find the cost price of each.Solution 30

Question 31

Points A and B are 70km apart on a highway. A car starts from A and another starts from B simultaneously. If they travel in the same direction, they meet in 7 hours. But, if they travel towards each other, they meet in 1 hour. Find the speed of each car.Solution 31

Let P and Q be the cars starting from A and B respectively and let their speeds be x km/hr and y km/hr respectively.

Case- I

When the cars P and Q move in the same direction.

Distance covered by the car P in 7 hours = 7x km

Distance covered by the car Q in 7 hours = 7y km

Let the cars meet at point M.

AM = 7x km and BM = 7y km

AM – BM = AB

 7x – 7y = 70

7(x – y) = 70

x – y = 10 —-(1)

Case II

When the cars P and Q move in opposite directions.

Distance covered by P in 1 hour = x km

Distance covered by Q in 1 hour = y km

In this case let the cars meet at a point N.

AN = x km and BN = y km

AN + BN = AB

 x + y = 70—(2)

Adding (1) and (2), we get

2x = 80

 x = 40

Putting x = 40 in (1), we get

40 – y = 10

 y = (40 – 10) = 30

x = 40, y = 30

Hence, the speeds of these cars are 40 km/ hr and 30 km/ hr respectively.Question 32

A train covered a certain distance at a uniform speed. If the train had been 5kmph faster, it would have taken 3 hours less than the scheduled time. And, if the train were slower by 4 kmph, it would have taken 3hours less than the scheduled time. Find the length of the journey.Solution 32

Let the original speed be x km/h and time taken be y hours

Then, length of journey = xy km

Case I:

Speed = (x + 5)km/h and time taken = (y – 3)hour

Distance covered = (x + 5)(y – 3)km

(x + 5) (y – 3) = xy

xy + 5y -3x -15 = xy

5y – 3x = 15 —(1)

Case II:

Speed (x – 4)km/hr and time taken = (y + 3)hours

Distance covered = (x – 4)(y + 3) km

(x – 4)(y + 3) = xy

xy -4y + 3x -12 = xy

 3x – 4y = 12 —(2)

Multiplying (1) by 4 and (2) by 5, we get

20y – 12x = 60 —(3)

-20y + 15x = 60 —(4)

Adding (3) and (4), we get

3x = 120

or x = 40

Putting x = 40 in (1), we get

5y – 3 × 40 = 15 

  5y = 135  

  y = 27

Hence, length of the journey is (40 × 27) km = 1080 kmQuestion 33

Abdul travelled 300 km by train and 200 km by taxi taking 5 hours 30 minutes. But, if he travels 260 km by train and 240 km by taxi, he takes 6 minutes longer. Find the speed of the train and that of the taxi.Solution 33

Question 34

Places A and B are 160 km apart on a highway. One car starts from A and another from B at the same time. If they travel in the same direction, they meet in 8 hours. But, if they travel towards each other, they meet in 2 hours. Find the speed of each car.Solution 34

Question 35

A sailor goes 8 km downstream in 40 minutes and returns in 1 hour. Find the speed of the sailor in still water and the speed of the current.Solution 35

Question 36

A boat goes 12km upstream and 40km downstream in 8hours. It can go 16km upstream and 32 downstream in the same times. Find the speed of the boat in still water and the speed of the stream.Solution 36

Let the speed of the boat in still water be x km/hr and speed of the stream be y km/hr.

Then,

Speed upstream = (x – y)km/hr

Speed downstream = (x + y) km/hr

Time taken to cover 12 km upstream = 

Time taken to cover 40 km downstream = 

Total time taken = 8hrs

Again, time taken to cover 16 km upstream = 

Time taken to taken to cover 32 km downstream = 

Total time taken = 8hrs

Putting 

12u + 40v = 8 

   3u + 10v = 2 —(1)

and

16u + 32v = 8 

   2u + 4v = 1—(2)

Multiplying (1) by 4 and (2) by 10, we get

12u + 40v = 8—(3)

20u + 40v = 10 —(4)

Subtracting (3) from (4), we get

Putting in (3), we get

On adding (5) and (6), we get

2x = 12 

  x = 6

Putting x = 6 in (6) we get

6 + y = 8

y = 8 – 6 = 2

x = 6, y = 2

Hence, the speed of the boat in still water = 6 km/hr and speedof the stream = 2km/hrQuestion 37

2 men and 5 boys can finish a piece of work in 4 days, while 3 men and 6 boys can finish it in 3days. Find the time taken by one man alone to finish the work and that taken by one boy alone to finish the work.Solution 37

Let man’s 1 day’s work be and 1 boy’s day’s work be 

Also let  and 

Multiplying (1) by 6 and (2) by 5 we get

Subtracting (3) from (4), we get

Putting in (1), we get

x = 18, y = 36

The man will finish the work in 18 days and the boy will finish the work in 36 days when they work alone.Question 38

The length of a room exceeds its breadth by 3 meters. If the length is increased by 3 meters and the breadth is decreased by 2 meters, the area remains the same. Find the length and breadth of the room.Solution 38

Let the length = x meters and breadth = y meters

Then,

x = y + 3

 x – y = 3 —-(1)

Also,

(x + 3)(y – 2) = xy

 3y – 2x = 6 —-(2)

Multiplying (1) by 2 and (2) by 1

-2y + 2x = 6 —(3)

3y – 2x = 6 —(4)

Adding (3) and (4), we get

y = 12

Putting y = 12 in (1), we get

x – 12 = 3

 x= 15

x = 15, y = 12

Hence length = 15 metres and breadth = 12 metresQuestion 39

The area of a rectangle gets reduced by 8 , when its length is reduced by 5m and its breadth is increased by 3m. If we increase the length by 3m and breadth by 2m, the area is increased by 74 . Find the length and breadth of the rectangle.Solution 39

Let the length of a rectangle be x meters and breadth be y meters.

Then, area = xy sq.m

Now,

xy – (x – 5)(y + 3) = 8

xy – [xy – 5y + 3x -15] = 8

xy – xy + 5y – 3x + 15 = 8 

 3x – 5y = 7 —(1)

And

(x + 3)(y + 2) – xy = 74

 xy + 3y +2x + 6 – xy = 74

2x + 3y = 68—(2)

Multiplying (1) by 3 and (2) by 5, we get

9x – 15y = 21—(3)

10x + 15y = 340—(4)

Adding (3) and (4), we get

Putting x = 19 in (3) we get

x = 19 meters, y = 10 meters

Hence, length = 19m and breadth = 10mQuestion 40

The area of a rectangle gets reduced by 67 square metres, when its length is increased by 3 m and breadth is decreased by 4 m. If the length. is reduced by 1 m and breadth is increased by 4 m, the area is increased by 89 square metres. Find the dimensions of the rectangle.Solution 40

Question 41

A railway half ticket costs half the full fare and the reservation charge is the same on half ticket as on full ticket. One reserved first class ticket from Mumbai to Delhi costs Rs.4150 while one full and one half reserved first class tickets cost Rs.6255. What is the basic first class full fare and what is the reservation charge?Solution 41

Question 42

Five years hence, a man’s age will be three times the age of his son. Five years ago, the man was seven times as old as his son. Find their present ages.Solution 42

Question 43

Two years ago, a man was five times as old as his son. Two years later, his age will be 8 more than three times the age of the son. Find the present ages of man and his son.Solution 43

Let the present ages of the man and his son be x years and y years respectively.

Then,

Two years ago:

(x – 2) = 5(y – 2)

x – 2 = 5y – 10

  x – 5y = -8 —(1)

Two years later:

(x + 2) = 3(y + 2) + 8

x + 2 = 3y + 6 + 8

 x – 3y = 12 —(2)

Subtracting (2) from (1), we get

-2y = -20

  y = 10

Putting y = 10 in (1), we get

x – 5 10 = -8

x – 50 = -8

 x = 42

Hence the present ages of the man and the son are 42 years and 10 respectively.Question 44

If twice the son’s age in years is added to the mother’s age, the sum is 70 years. But, if twice the mother’s age is added to the son’s age, the sum is 95years. Find the age of the mother and that of the son.Solution 44

Let the present ages of  the mother and her son be x and y respectively.

According to the given question:

x + 2y = 70—(1)

and

2x + y = 95—(2)

Multiplying (1) by 1 and (2) by 2, we get

x + 2y = 70 —(3)

4x + 2y = 190—(4)

Subtracting (3) from (4), we get

Putting x = 40 in (1), we get

40 + 2y = 70

2y = 30

 y = 15

x = 40, y = 15

Hence, the ages of the mother and the son are 40 years and 15 years respectively.Question 45

The present age of a woman is 3 years more than three times the age of her daughter; three years hence, the woman’s age will be 10 years more than twice the age of her daughter. Find their present ages.Solution 45

Let the present ages of woman and daughter be x and y respectively.

Then,

Their present ages:

x = 3y + 3

x – 3y = 3—(1)

Three years later:

 (x + 3) = 2(y + 3) + 10

x + 3 = 2y + 6 + 10

 x – 2y = 13—(2)

Subtracting (2) from (1), we get

 y = 10

Putting y = 10 in (1), we get

x – 3 × 10 = 3

 x = 33

x = 33, y = 10

Hence, present ages of woman and daughter are 33 and 10 years.Question 46

On selling a tea set at 5% loss and a lemon set at 15% gain, a crockery seller gains Rs.7. If he sells the tea set at 5% gain and the lemon set at 10% gain, he gains Rs.13. Find the actual price of each of the tea set and the lemon set.Solution 46

Question 47

A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Mona paid Rs.27 for a book kept for 7 days, while Tanvy paid Rs. 21 for the book she kept for 5 days. Find the fixed charge and the charge for each extra day.Solution 47

Question 48

A chemist has one solution containing 50% acid and a second one containing 25% arid. How much of each should be used to make 10 litres of a 40% arid solution?Solution 48

Question 49

A jeweller has bars of 18-carat gold and 12-carat gold. How much of each must be melted together to obtain a bar of 16-carat gold, weighing 120 g? (Given: Pure gold is 24-carat)Solution 49

Question 50

90% and 97% pure acid solutions are mixed to obtain 21 litres of 95% pure acid solution. Find the quantity of each type of adds to be mixed to form the mixture.Solution 50

Question 51

The larger of the two supplementary angles exceeds the smaller by 18^o.  Find them.Solution 51

Question 52

Solution 52

Question 58

Solution 58

Exercise Ex. 3F

Question 1

Write the number of solutions of the following pair of linear equations:

x + 2y – 8 = 0, 2x + 4y = 16.Solution 1

Question 2

Find the value of k for which the following pair of linear equations have infinitely many solutions:

2x + 3y = 7, (k – 1) x + (k + 2) y = 3k.Solution 2

Question 3

For what value of k does the following pair of linear equations have infinitely many solutions?

10x + 5y – (k – 5) = 0 and 20x + 10y – k = 0.Solution 3

Question 4

For what value of k will the following pair of linear equations have no ‘solution?

2x + 3y = 9, 6x + (k – 2)y = (3k – 2).Solution 4

Question 5

Write the number of solutions of the following pair of linear equations:

x + 3y – 4 = 0 and 2x + 6y – 7 = 0.Solution 5

Question 6

Write the value of k for which the system of equations 3x + ky = 0, 2x – y = 0 has a unique solution.Solution 6

Question 7

The difference between two numbers is 5 and the difference between their squares is 65. Find the numbers.Solution 7

Question 8

The cost of 5 pens and 8 pencils is Rs.120, while the cost of 8 pens and 5 pencils is Rs.153. Find the cost of 1 pen and that of I pencil.Solution 8

Question 9

The sum of two numbers is 80. The larger number exceeds four times the smaller one by 5. Find the numbers.Solution 9

Question 10

A number consists of two digits whose sum is 10. If 18 is subtracted from the number, its digits are reversed. Find the number.Solution 10

Question 11

A man purchased 47 stamps of 20 p and 25 p for Rs.10. Find the number of each type of stamps.Solution 11

Question 12

A man has some hens and cows. If the number of heads be 48 and number of feet be 140, how many cows are there?Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

If 12x + 17y = 53 and 17x + 12y = 63 then find the value of (x +y).Solution 15

Question 16

Find the value of k for which the system 3x + 5y = 0, kx + 10y = 0 has a nonzero solution.Solution 16

Question 17

Find k for which the system kx – y = 2 and 6x – 2y = 3 has a unique solution.Solution 17

Question 18

Find k for which the system 2x + 3y – 5 = 0, 4x + ky – 10 = 0 has an infinite number of solutions.Solution 18

Question 19

Show that the system 2x + 3y – 1 = 0, 4x + 6y – 4 = 0 has no solution.Solution 19

Question 20

Find k for which the system x + 2y = 3 and 5x + ky + 7 = 0 is inconsistent.Solution 20

Question 21

Solution 21

Exercise MCQ

Question 1

If 2x + 3y = 12 and 3x – 2y = 5 then

(a) x = 2, y = 3

(b) x = 2, y = -3

(c) x = 3, y = 2

(d) x = 3, y = -2Solution 1

Question 2

(a) x = 4, y = 2

(b) x = 5, y = 3

(c) x = 6, y = 4

(d) x = 7, y = 5Solution 2

Question 3

(a) x = 2, y = 3

(b) x = -2, y = 3

(c) x = 2, y = -3

(d) x = -2, y = -3Solution 3

Question 4

Solution 4

Question 5

(a) x = 1, y = 1

(b) x = -1, y = -1

(c) x = 1, y = 2

(d) x = 2, y = 1Solution 5

Question 6

Solution 6

Question 7

If 4x+6y=3xy and 8x+9y=5xy then

(a) x=2, y=3

(b) x=1, y=2

(c) x=3, y=4

(d) x=1, y=-1Solution 7

Question 8

If 29x+37y=103 and 37x+29y=95 then

(a) x=1, y=2

(b) x=2, y=1

(c) x=3, y=2

(d) x=2, y=3Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

The system kx – y = 2 and 6x – 2y = 3  has a unique solution only when

(a) k = 0

(b) k ≠ 0

(c) k = 3

(d) k ≠ 3Solution 11

Question 12

The system x – 2y = 3 and 3x + ky = 1 has a unique solution only when

(a) k = -6

(b) k ≠ -6

(c) k = 0

(d) k ≠ 0Solution 12

Question 13

The system x+2y=3 and 5x+ky+7=0 has no solution, when

(a) k=10

(b) k≠10

(c) 

(d) K=-21Solution 13

Question 14

If the lines given by 3x + 2ky = 2 and 2x + 5y + 1 = 0 are parallel then the value of k is

Solution 14

Question 15

For what value of k do the equations kx – 2y = 3 and

3x + y = 5 represent two lines intersecting at a unique point?

(a) k=3

(b) k=-3

(c) k=6

(d) all real values except -6Solution 15

Question 16

The pair of equations x + 2y + 5 = 0 and -3x – 6y + 1 = 0 has

(a) a unique solution

(b) exactly two solutions

(c) infinitely many solutions

(d) no solutionSolution 16

Question 17

The pair of equations 2x + 3y = 5 and 4x + 6y = 15 has

(a) a unique solution

(b) exactly two solutions

(c) infinitely many solutions

(d) no solutionSolution 17

Question 18

If a pair of linear equations is consistent then their graph lines will be

(a) parallel

(b) always coincident

(c) always intersecting

(d) intersecting or coincidentSolution 18

Question 19

If a pair of linear equations is inconsistent then their graph lines will be

(a) parallel

(b) always coincident

(c) always intersecting

(d) intersecting or coincidentSolution 19

Question 20

In a ΔABC, ∠C = 3 ∠B = 2 (∠A + ∠B), then ∠B = ?

(a) 20° 

(b) 40° 

(c) 60° 

(d) 80° Solution 20

Question 21

In a cyclic quadrilateral ABCD, it is being given that

∠A = (x + y + 10) °, ∠B = (y + 20) °,

∠C = (x + y – 30)° and ∠D = (x + y)°. Then, ∠B = ?

(a) 70° 

(b) 80° 

(c) 100° 

(d) 110° Solution 21

Question 22

The sum of the digits of a two-digit number is 15. The number obtained by interchanging the digits exceeds the given number by 9. The number is

(a) 96

(b) 69

(c) 87

(d) 78Solution 22

Question 23

Solution 23

Question 24

5 years hence, the age of a man shall be 3 times the age of his son while 5 years earlier the age of the man was 7 times the age of his son. The present age of the man is

(a) 45 years

(b) 50 years

(c) 47 years

(d) 40 yearsSolution 24

Question 25

The graphs of the equations 6x – 2y + 9 = 0 and

3x – y + 12 = 0 are two lines which are

(a) coincident

(b) parallel

(c) intersecting exactly at one point

(d) perpendicular to each otherSolution 25

Question 26

The graphs of the equations 2x+3y-2=0 and x-2y-8=0 are two lines which are

(a) coincident

(b) parallel

(c) intersecting exactly at one point

(d) perpendicular to each otherSolution 26

Question 27

(a) coincident

(b) parallel

(c) intersecting exactly at one point

(d) perpendicular to each otherSolution 27

Exercise FA

Question 1

The graphic representation of the equations x+2y=3 and 2x+4y+7=0 gives a pair of

(a) parallel lines

(b) intersecting lines

(c) coincident lines

(d) none of theseSolution 1

Question 2

If 2x – 3y = 7 and (a + b) x – (a + b – 3) y = 4a+b have an infinite number of solutions then

(a) a= 5, b = 1

(b) a = -5, b = 1

(c) a = 5, b = -1

(d) a = -5, b = -1Solution 2

Question 3

The pair of equations 2x+y=5, 3x+2y=8 has

(a) a unique solution

(b) two solutions

(c) no solution

(d) infinitely many solutionsSolution 3

Question 4

If x = -y and y > 0, which of the following is wrong?

(a) x²y > 0

(b) x + y = 0

(c) xy < 0

(d) Solution 4

Question 5

Solution 5

Question 6

For what values of k is the system of equations kx + 3y = k – 2, 12x + ky = k inconsistent?Solution 6

Question 7

Solution 7

Question 8

Solve the system of equations x – 2y = 0, 3x + 4y = 20.Solution 8

Question 9

Show that the paths represented by the equations x – 3y = 2 and -2x + 6y = 5 are parallel.Solution 9

Question 10

The difference between two numbers is 26 and one number is three times the other. Find the numbers.Solution 10

Question 11

Solve : 23x+29y=98, 29x+23y=110.Solution 11

Question 12

Solve : 6x+3y=7xy and 3x+9y = 11xy.Solution 12

Question 13

Find the value of k for which the system of equations 3x+y=1 and kx+2y=5 has (i) a unique solution, (ii) no solution.Solution 13

Question 14

In a ΔABC, ∠C =3∠B =2(∠A+∠B). Find the measure of each one of the ∠A, ∠B and ∠C. Solution 14

Question 15

5 pencils and 7 pens together cost Rs. 195 while 7 pencils and 5 pens together cost Rs. 153. Find the cost of each one of the pencil and the pen.Solution 15

Question 16

Solve the following system of equations graphically :

2x-3y=1, 4x-3y+1=0.Solution 16

Since the intersection of the lines is the point with coordinates (-1, -1), x = -1 and y = -1.Question 17

Find the angles of a cyclic quadrilateral ABCD in which ∠A =(4x+20)°, ∠B=(3x-5)°, ∠C=(4y)° and ∠D=(7y+5)° Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20