Draw an ∠BAC of measure 50° such that AB = 5 cm and AC = 7 cm. Through C draw a line parallel to AB and through B draw a line parallel to AC, intersecting each other at D. Measure BD and CD
Solution:
Steps of construction:
Draw angle BAC = 50° such that AB = 5 cm and AC = 7 cm.
Cut an arc through C at an angle of 50°
Draw a straight line passing through C and the arc. This line will be parallel to AB since ∠CAB =∠RCA=50°
Alternate angles are equal; therefore the line is parallel to AB.
Again through B, cut an arc at an angle of 50° and draw a line passing through B and this arc and say this intersects the line drawn parallel to AB at D.
∠SBA =∠BAC = 50°, since they are alternate angles. Therefore BD parallel to AC
Also we can measure BD = 7 cm and CD = 5 cm.
Question: 2
Draw a line PQ. Draw another line parallel to PQ at a distance of 3 cm from it.
Solution:
Steps of construction:
Draw a line PQ.
Take any two points A and B on the line.
Construct ∠PBF = 90° and ∠QAE = 90°
With A as centre and radius 3 cm cut AE at C.
With B as centre and radius 3 cm cut BF at D.
Join CD and produce it on either side to get the required line parallel to AB and at a distance of 5 cm from it.
Question: 3
Take any three non-collinear points A, B, C and draw ∠ABC. Through each vertex of the triangle, draw a line parallel to the opposite side.
Solution:
Steps of construction:
Mark three non collinear points A, B and C such that none of them lie on the same line.
Join AB, BC and CA to form triangle ABC.
Parallel line to AC
With A as centre, draw an arc cutting AC and AB at T and U, respectively.
With centre B and the same radius as in the previous step, draw an arc on the opposite side of AB to cut AB at X.
With centre X and radius equal to TU, draw an arc cutting the arc drawn in the previous step at Y.
Join BY and produce in both directions to obtain the line parallel to AC.
Parallel line to AB
With B as centre, draw an arc cutting BC and BA at W and V, respectively.
With centre C and the same radius as in the previous step, draw an arc on the opposite side of BC to cut BC at P.
With centre P and radius equal to WV, draw an arc cutting the arc drawn in the previous step at Q.
Join CQ and produce in both directions to obtain the line parallel to AB.
Parallel line to BC
With B as centre, draw an arc cutting BC and BA at W and V, respectively (already drawn).
With centre A and the same radius as in the previous step, draw an arc on the opposite side of AB to cut AB at R.
With centre R and radius equal to WV, draw an arc cutting the arc drawn in the previous step at S.
Join AS and produce in both directions to obtain the line parallel to BC.
Question: 4
Draw two parallel lines at a distance of 5kms apart.
Solution:
Steps of construction:
Draw a line PQ.
Take any two points A and B on the line.
Construct ∠PBF = 90° and ∠QAE = 90°
With A as centre and radius 5 cm cut AE at C.
With B as centre and radius 5 cm cut BF at D.
Join CD and produce it on either side to get the required line parallel to AB and at a distance of 5 cm from it.
Exercise 17.2
Question: 1
Draw △ABC in which AB = 5.5 cm. BC = 6 cm and CA = 7 cm. Also, draw perpendicular bisector of side BC.
Solution:
Steps of construction:
Draw a line segment AB of length 5.5 cm.
From B, cut an arc of radius 6 cm.
With centre A, draw an arc of radius 7 cm intersecting the previously drawn arc at C.
Join AC and BC to obtain the desired triangle.
With centre B and radius more than half of BC, draw two arcs on both sides of BC.
With centre C and the same radius as in the previous step, draw two arcs intersecting the arcs drawn in the previous step at X and Y.
Join XY to get the perpendicular bisector of BC.
Question: 2
Draw ∆PQR in which PQ = 3 cm, QR. 4 cm and RP = 5 cm. Also, draw the bisector of ∠Q
Solution:
Steps of construction:
Draw a line segment PQ of length 3 cm.
With Q as centre and radius 4 cm, draw an arc.
With P as centre and radius 5 cm, draw an arc intersecting the previously drawn arc at R.
Join PR and OR to obtain the required triangle.
From Q, cut arcs of equal radius intersecting PQ and QR at M and N, respectively.
From M and N, cut arcs of equal radius intersecting at point S.
Join QS and extend to produce the angle bisector of angle PQR.
Verify that angle PQS and angle SQR are equal to 45° each.
Question: 3
Draw an equilateral triangle one of whose sides is of length 7 cm.
Solution:
Steps of construction:
Draw a line segment AB of length 7 cm.
With centre A, draw an arc of radius 7 cm.
With centre B, draw an arc of radius 7 cm intersecting the previously drawn arc at C.
Join AC and BC to get the required triangle.
Question: 4
Draw a triangle whose sides are of lengths 4 cm, 5 cm and 7 cm. Draw the perpendicular bisector of the largest side.
Solution:
Steps of construction:
Draw a line segment PR of length 7 cm.
With centre P, draw an arc of radius 5 cm.
With centre R, draw an arc of radius 4 cm intersecting the previously drawn arc at Q.
Join PQ and QR to obtain the required triangle.
From P, draw arcs with radius more than half of PR on either sides.
With the same radius as in the previous step, draw arcs from R on either sides of PR intersecting the arcs drawn in the previous step at M and N.
MN is the required perpendicular bisector of the largest side.
Question: 5
Draw a triangle ABC with AB = 6 cm, BC = 7 cm and CA = 8 cm. Using ruler and compass alone, draw (i) the bisector AD of ∠A and (ii) perpendicular AL from A on BC. Measure LAD.
Solution:
Steps of construction:
Draw a line segment BC of length 7 cm.
With centre B, draw an arc of radius 6 cm.
With centre C, draw an arc of radius 8 cm intersecting the previously drawn arc at A.
Join AC and BC to get the required triangle.
Angle bisector steps:
From A, cut arcs of equal radius intersecting AB and AC at E and F, respectively.
From E and F, cut arcs of equal radius intersecting at point H.
Join AH and extend to produce the angle bisector of angle A, meeting line BC at D.
Perpendicular from Point A to line BC steps:
From A, cut arcs of equal radius intersecting BC at P and Q, respectively (Extend BC to draw these arcs).
From P and Q, cut arcs of equal radius intersecting at M.
Join AM cutting BC at L.
AL is the perpendicular to the line BC.
Angle LAD is 15°.
Question: 6
Draw △DEF such that DE= DF= 4 cm and EF = 6 cm. Measure ∠E and ∠F.
Solution:
Steps of construction:
Draw a line segment EF of length 6 cm.
With E as centre, draw an arc of radius 4 cm.
With F as centre, draw an arc of radius 4 cm intersecting the previous arc at D.
Join DE and DF to get the desired triangle DEF.
By measuring we get, ∠E= ∠F= 40°..
Question: 7
Draw any triangle ABC. Bisect side AB at D. Through D, draw a line parallel to BC, meeting AC in E. Measure AE and EC.
Solution:
Steps of construction:
We first draw a triangle ABC with each side = 6 cm.
Steps to bisect line AB:
Draw an arc from A on either side of line AB.
With the same radius as in the previous step, draw an arc from B on either side of AB intersecting the arcs drawn in the previous step at P and Q.
Join PQ cutting AB at D. PQ is the perpendicular bisector of AB.
Parallel line to BC:
With B as centre, draw an arc cutting BC and BA at M and N, respectively.
With centre D and the same radius as in the previous step, draw an arc on the opposite side of AB to cut AB at Y.
With centre Y and radius equal to MN, draw an arc cutting the arc drawn in the previous step at X.
Join XD and extend it to intersect AC at E.
DE is the required parallel line.
Exercise 17.3
Question: 1
Draw △ABC in which AB = 3 cm, BC = 5 cm and ∠Q = 70°.
Solution:
Steps of construction:
Draw a line segment AB of length 3 cm.
Draw ∠XBA=70°.
Cut an arc on BX at a distance of 5 cm at C.
Join AC to get the required triangle.
Question: 2
Draw △ABC in which ∠A=70°., AB = 4 cm and AC= 6 cm. Measure BC.
Solution:
Steps of construction:
Draw a line segment AC of length 6 cm.
Draw ∠XAC=70°.
Cut an arc on AX at a distance of 4 cm at B.
Join BC to get the desired triangle.
We see that BC = 6 cm.
Question: 3
Draw an isosceles triangle in which each of the equal sides is of length 3 cm and the angle between them is 45°.
Solution:
Steps of construction:
Draw a line segment PQ of length 3 cm.
Draw ∠QPX=45°.
Cut an arc on PX at a distance of 3 cm at R.
Join QR to get the required triangle.
Question: 4
Draw △ABC in which ∠A = 120°, AB = AC = 3 cm. Measure ∠B and ∠C.
Solution:
Steps of construction:
Draw a line segment AC of length 3 cm.
Draw ∠XAC = 120°.
Cut an arc on AX at a distance of 3 cm at B.
Join BC to get the required triangle.
By measuring, we get ∠B = ∠C = 30°.
Question: 5
Draw △ABC in which ∠C = 90° and AC = BC = 4 cm.
Solution:
Steps of construction:
Draw a line segment BC of length 4 cm.
At C, draw ∠BCY=90°.
Cut an arc on CY at a distance of 4 cm at A.
Join AB. ABC is the required triangle.
Question: 6
Draw a triangle ABC in which BC = 4 cm, AB = 3 cm and ∠B = 45°. Also, draw a perpendicular from A on BC.
Solution:
Steps of construction:
Draw a line segment AB of length 3 cm.
Draw an angle of 45° and cut an arc at this angle at a radius of 4 cm at C.
Join AC to get the required triangle.
With A as centre, draw intersecting arcs at M and N.
With centre M and radius more than half of MN, cut an arc on the opposite side of A.
With N as centre and radius the same as in the previous step, cut an arc intersecting the previous arc at E.
Join AE, it meets BC at D, then AE is the required perpendicular.
Question: 7
Draw a triangle ABC with AB = 3 cm, BC = 4 cm and ∠B = 60°. Also, draw the bisector of angles C and A of the triangle, meeting in a point O. Measure ∠COA.
Solution:
Steps of construction:
Draw a line segment BC = 4 cm.
Draw ∠CBX = 60°.
Draw an arc on BX at a radius of 3 cm cutting BX at A.
Join AC to get the required triangle.
Angle bisector for angle A:
With A as centre, cut arcs of the same radius cutting AB and AC at P and Q, respectively.
From P and Q cut arcs of same radius intersecting at R.
Join AR to get the angle bisector of angle A.
Angle bisector for angle C:
With A as centre, cut arcs of the same radius cutting CB and CA at M and N, respectively.
From M and N, cut arcs of the same radius intersecting at T
Join CT to get the angle bisector of angle C.
Mark the point of intersection of CT and AR as 0.
Angle ∠COA = 120°.
Exercise 17.4
Question: 1
Construct ∆ABC in which BC = 4 cm, ∠B = 50° and ∠C = 70°.
Solution:
Steps of construction:
Draw a line segment BC of length 4 cm.
Draw ∠CBX such that ∠CBX=50°.
Draw ∠BCY with Y on the same side of BC as X such that ∠BCY=70°.
Let CY and BX intersect at A.
ABC is the required triangle.
Question: 2
Draw ∆ABC in which BC = 8 cm, ∠B = 50° and ∠A = 50°.
∠ABC + ∠BCA + ∠CAB = 180°
∠BCA = 180° − ∠CAB − ∠ABC
∠BCA = 180°− 100° = 80°
Solution:
Steps of construction:
Draw a line segment BC of length 8 cm.
Draw ∠CBX such that ∠CBX = 50°.
Draw ∠BCY with Y on the same side of BC as X such that ∠BCY = 80°.
Let CY and BX intersect at A.
Question: 3
Draw ∆ABC in which ∠Q = 80°, ∠R = 55° and QR = 4.5 cm. Draw the perpendicular bisector of side QR.
Solution:
Steps of construction:
Draw a line segment QR = 4.5 cm.
Draw ∠RQX = 80° and ∠QRY = 55°.
Let QX and RY intersect at P so that PQR is the required triangle.
With Q as centre and radius more that 2.25 cm, draw arcs on either sides of QR.
With R as centre and radius more than 2.25 cm, draw arcs intersecting the previous arcs at M and N.
Join MN
MN is the required perpendicular bisector of QR.
Question: 4
Construct ∆ABC in which AB = 6.4 cm, ∠A = 45° and ∠B = 60°
Solution:
Steps of construction:
Draw a line segment AB = 6.4 cm.
Draw ∠BAX = 45°.
Draw ∠ABY with Y on the same side of AB as X such that ∠ABY = 60°.
Let AX and BY intersect at C.
ABC is the required triangle.
Question: 5
Draw ∆ABC in which AC = 6 cm, ∠A = 90° and ∠B = 60°.
∠A + ∠B + ∠C = 180°
Therefore ∠C = 180°− 60°− 90°= 30°
Solution:
Steps of construction:
Draw a line segment AC = 6 cm.
Draw ∠ACX = 30°.
Draw ∠CAY with Y on the same side of AC as X such that ∠CAY = 90°.
Join CX and AY. Let these intersect at B.
ABC is the required triangle where angle ∠ABC = 60°.
Exercise 17.5
Question: 1
Draw a right triangle with hypotenuse of length 5 cm and one side of length 4 cm.
Solution:
Steps of construction:
Draw a line segment QR = 4 cm.
Draw ∠QRX of measure 90°.
With centre Q and radius PQ = 5 cm, draw an arc of the circle to intersect ray RX at P.
Join PQ to obtain the desired triangle PQR.
PQR is the required triangle.
Question: 2
Draw a right triangle whose hypotenuse is of length 4 cm and one side is of length 2.5 cm.
Solution:
Steps of construction:
Draw a line segment QR = 2.5 cm.
Draw ∠QRX of measure 90°.
With centre Q and radius PQ = 4 cm, draw an arc of the circle to intersect ray RX at P.
Join PQ to obtain the desired triangle PQR.
PQR is the required triangle.
Question: 3
Draw a right triangle having hypotenuse of length 5.4 cm, and one of the acute angles of measure 30°
Solution:
Let ABC be the right triangle at A such that hypotenuse BC = 5.4 cm. Let cC = 30°.
Explain the concept of congruence of figures with the help of certain examples.
Solution:
Congruent objects or figures are exact copies of each other or we can say mirror images of each other. The relation of two objects being congruent is called congruence.
Consider Ball A and Ball B. These two balls are congruent.
Now consider the two stars below. Star A and Star B are exactly the same in size, colour and shape. These are congruent stars.
Question: 2
Fill in the blanks:
(i) Two line segments are congruent if ___
(ii) Two angles are congruent if ___
(iii) Two square are congruent if ___
(iv) Two rectangles are congruent if ___
(v) Two circles are congruent if ___
Solution:
(i) They have the same length, since they can superpose on each other.
(ii) Their measures are the same. On superposition, we can see that the angles are equal.
(iii) Their sides are equal. All the sides of a square are equal and if two squares have equal sides, then all their sides are of the same length. Also angles of a square are 90° which is also the same for both the squares.
(iv) Their lengths are equal and their breadths are also equal. The opposite sides of a rectangle are equal. So if two rectangles have lengths of the same size and breadths of the same size, then they are congruent to each other.
(v) Their radii are of the same length. Then the circles will have the same diameter and thus will be congruent to each other.
Question: 3
In Figure, ∠POQ ≅ ∠ROS, can we say that ∠POR ≅ ∠QOS
Solution:
We have,
∠POQ ≅ ∠ROS (1) Also, ∠ROQ ≅ ∠ROQ Therefore adding ∠ROQ to both sides of (1), Weget, ∠POQ + ∠ROQ ≅ ∠ROQ + ∠ROS Therefore, ∠PQR = ∠QOS
Question: 4
In figure, a = b = c, name the angle which is congruent to ∠AOC
Solution:
We have,
∠ AOB = ∠ BOC = ∠ COD
Therefore, ∠ AOB = ∠ COD
Also, ∠ AOB + ∠ BOC = ∠ BOC + ∠ COD
∠ AOC = ∠ BOD
Hence, ∠ BOD is congruent to ∠ AOC
Question: 5
Is it correct to say that any two right angles are congruent? Give reasons to justify your answer.
Solution:
Two right angles are congruent to each other because they both measure 90 degrees.
We know that two angles are congruent if they have the same measure.
Question: 6
In figure, ∠AOC ≅ ∠PYR and ∠BOC ≅ ∠QYR. Name the angle which is congruent to ∠AOB.
Solution:
∠AOC ≅ ∠PYR…. (i) Also, ∠BOC ≅ ∠QYR Now, ∠AOC = ∠AOB + ∠BOC ∠PYR = ∠PYQ +∠QYR By putting the value of ∠AOC and ∠PYR in equation (i), we get, ∠AOB + ∠BOC ≅ ∠PYQ + ∠QYR ∠AOB ≅ ∠PYQ (∠BOC ≅ ∠QYR) Hence, ∠AOB ≅ ∠PYQ
Question: 7
Which of the following statements are true and which are false;
(i) All squares are congruent.
(ii) If two squares have equal areas, they are congruent.
(iii) If two rectangles have equal areas, they are congruent.
(iv) If two triangles have equal areas, they are congruent.
Solution:
(i) False.
All the sides of a square are of equal length.
However, different squares can have sides of different lengths. Hence all squares are not congruent.
(ii) True.
Area of a square = side x side
Therefore, two squares that have the same area will have sides of the same lengths. Hence they will be congruent.
(iii) False Area of a rectangle = length x breadth
Two rectangles can have the same area. However, the lengths of their sides can vary and hence they are not congruent.
Example: Suppose rectangle 1 has sides 8 m and 8 m and area 64 meter square. Rectangle 2 has sides 16 m and 4 m and area 64 meter square. Then rectangle 1 and 2 are not congruent.
(iv) False
Area of a triangle = 12 x base x height
Two triangles can have the same area but the lengths of their sides can vary and hence they cannot be congruent.
Exercise 16.2
Question: 1
In the following pairs of triangle (Figures), the lengths of the sides are indicated along sides. By applying SSS condition, determine which are congruent. State the result in symbolic.
Solution:
1) In Δ ABC and Δ DEF
AB = DE = 4.5 cm (Side)
BC = EF = 6 cm (Side) and
AC = DF = 4 cm (Side)
Therefore, by SSS criterion of congruence, ΔABC ≅ ΔDEF
2)
In Δ ACB and Δ ADB
AC = AD (Side)
BC = BD (Side) and
AB = AB (Side)
Therefore, by SSS criterion of congruence, ΔACB ≅ ΔADB
3) In Δ ABD and Δ FEC,
AB = FE (Side)
AD = FC (Side)
BD = CE (Side)
Therefore, by SSS criterion of congruence, ΔABD ≅ ΔFEC
4) In Δ ABO and Δ DOC,
AB = DC (Side)
AO = OC (Side)
BO = OD (Side)
Therefore, by SSS criterion of congruence, ΔABO ≅ ΔODC
Question: 2
In figure, AD = DC and AB = BC
(i) Is ΔABD ≅ ΔCBD?
(ii) State the three parts of matching pairs you have used to answer (i).
Solution:
Yes ΔABD = ΔCBD by the SSS criterion. We have used the three conditions in the SSS criterion as follows:
AD = DC
AB = BC and
DB = BD
Question: 3
In Figure, AB = DC and BC = AD.
(i) Is ΔABC ≅ ΔCDA?
(ii) What congruence condition have you used?
(iii) You have used some fact, not given in the question, what is that?
Solution:
We have AB = DC
BC = AD
and AC = AC
Therefore by SSS ΔABC ≅ ΔCDA
We have used Side congruence condition with one side common in both the triangles.
Yes, have used the fact that AC = CA.
Question: 4
In ΔPQR ≅ ΔEFD,
(i) Which side of ΔPQR equals ED?
(ii) Which angle of ΔPQR equals angle E?
Solution:
ΔPQR ≅ ΔEFD
(i) Therefore PR = ED since the corresponding sides of congruent triangles are equal.
(ii) ∠QPR = ∠FED since the corresponding angles of congruent triangles are equal.
Question: 5
Triangles ABC and PQR are both isosceles with AB = AC and PO = PR respectively. If also, AB = PQ and BC = QR, are the two triangles congruent? Which condition do you use?
It ∠B = 50°, what is the measure of ∠R?
Solution:
We have AB = AC in isosceles ΔABC
And PQ = PR in isosceles ΔPQR.
Also, we are given that AB = PQ and QR = BC.
Therefore, AC = PR (AB = AC, PQ = PR and AB = PQ)
Hence, ΔABC ≅ ΔPQR
Now
∠ABC = ∠PQR (Since triangles are congruent)However, ΔPQR is isosceles.
Therefore, ∠PRQ = ∠PQR = ∠ABC = 50°
Question: 6
ABC and DBC are both isosceles triangles on a common base BC such that A and D lie on the same side of BC. Are triangles ADB and ADC congruent? Which condition do you use? If ∠BAC = 40° and ∠BDC = 100°, then find ∠ADB.
Solution:
∠BAD = ∠CAD (c.p.c.t)
∠BAD + ∠CAD = 40°/ 2 ∠BAD = 40°
∠BAD = 40°/2 =20°
∠ABC + ∠BCA + ∠BAC = 180° (Angle sum property)
Since ΔABC is an isosceles triangle,
∠ABC = ∠BCA ∠ABC +∠ABC + 40°= 180°
2 ∠ABC = 180° – 40° = 140° ∠ABC = 140°/2 = 70°
∠DBC + ∠ BCD + ∠ BDC = 180° (Angle sum property)
Since ΔABC is an isosceles triangle, ∠ DBC = ∠BCD ∠DBC + ∠DBC + 100o = 180°
Δ ABC and ΔABD are on a common base AB, and AC = BD and BC = AD as shown in Figure. Which of the following statements is true?
(i) ΔABC ≅ ΔABD
(ii) ΔABC ≅ ΔADB
(iii) ΔABC ≅ ΔBAD
Solution:
In ΔABC and ΔBAD we have,
AC = BD (given)
BC = AD (given)
and AB = BA (common)
Therefore by SSS criterion of congruency, ΔABC ≅ ΔBAD
There option (iii) is true.
Question: 8
In Figure, ΔABC is isosceles with AB = AC, D is the mid-point of base BC.
(i) Is ΔADB ≅ ΔADC?
(ii) State the three pairs of matching parts you use to arrive at your answer.
Solution:
We have AB = AC.
Also since D is the midpoint of BC, BD = DC
Also, AD = DA
Therefore by SSS condition,
ΔADB ≅ ΔADC
We have used AB, AC : BD, DC AND AD, DA
Question: 9
In figure, ΔABC is isosceles with AB = AC. State if ΔABC ≅ ΔACB. If yes, state three relations that you use to arrive at your answer.
Solution:
Yes, ΔABC ≅ ΔACB by SSS condition.
Since, ABC is an isosceles triangle, AB = BC, BC = CB and AC = AB
Question: 10
Triangles ABC and DBC have side BC common, AB = BD and AC = CD. Are the two triangles congruent? State in symbolic form, which congruence do you use? Does ∠ABD equal ∠ACD? Why or why not?
Solution:
Yes,
Given,
Δ ABC and Δ DBC have side BC common, AB = BD and AC = CD
By SSS criterion of congruency, ΔABC ≅ ΔDBC
No, ∠ABD and ∠ACD are not equal because AB ≠ AC
Exercise 16.3
Question: 1
By applying SAS congruence condition, state which of the following pairs of triangle are congruent. State the result in symbolic form
Solution:
(i)
We have OA = OC and OB = OD and
∠AOB = ∠COD which are vertically opposite angles. Therefore by SAS condition, ΔAOC ≅ ΔBOD
(ii)
We have BD = DC
∠ADB = ∠ADC = 90° and
Therefore, by SAS condition, ΔADB ≅ ΔADC.
(iii)
We have AB = DC
∠ABD = ∠CDB and
Therefore, by SAS condition, ΔABD ≅ ΔCBD
(iv)
We have BC = QR
ABC = PQR = 90°
And AB = PQ
Therefore, by SAS condition, ΔABC≅ ΔPQR.
Question: 2
State the condition by which the following pairs of triangles are congruent.
Solution:
(i)
AB = AD
BC = CD and AC = CA
Therefore by SSS condition, ΔABC≅ ΔADC
(ii)
AC = BD
AD = BC and AB = BA
Therefore, by SSS condition, ΔABD ≅ ΔADC
(iii)
AB = AD
∠BAC = ∠DAC and
Therefore by SAS condition, ΔBAC ≅ ΔBAC
(iv)
AD = BC
∠DAC = ∠BCA and
Therefore, by SAS condition, ΔABC ≅ ΔADC
Question: 3
In figure, line segments AB and CD bisect each other at O. Which of the following statements is true?
(i) ΔAOC ≅ ΔDOB
(ii) ΔAOC ≅ ΔBOD
(iii) ΔAOC ≅ ΔODB
State the three pairs of matching parts, you have used to arrive at the answer.
Solution:
We have,
And, CO = OD
Also, AOC = BOD
Therefore, by SAS condition, ΔAOC ≅ ΔBOD
Question: 4
Line-segments AB and CD bisect each other at O. AC and BD are joined forming triangles AOC and BOD. State the three equality relations between the parts of the two triangles that are given or otherwise known. Are the two triangles congruent? State in symbolic form, which congruence condition do you use?
Solution:
We have AO = OB and CO = OD since AB and CD bisect each other at 0.
Also ∠AOC = ∠BOD since they are opposite angles on the same vertex.
Therefore by SAS congruence condition, ΔAOC ≅ ΔBOD
Question: 5
ΔABC is isosceles with AB = AC. Line segment AD bisects ∠A and meets the base BC in D.
(i) Is ΔADB ≅ ΔADC?
(ii) State the three pairs of matching parts used to answer (i).
(iii) Is it true to say that BD = DC?
Solution:
(i) We have AB = AC (Given)
∠BAD = ∠CAD (AD bisects ∠BAC)
Therefore by SAS condition of congruence, ΔABD ≅ ΔACD
(ii) We have used AB, AC; ∠BAD = ∠CAD; AD, DA.
(iii) Now, ΔABD≅ΔACD therefore by c.p.c.t BD = DC.
Question: 6
In Figure, AB = AD and ∠BAC = ∠DAC.
(i) State in symbolic form the congruence of two triangles ABC and ADC that is true.
(ii) Complete each of the following, so as to make it true:
(a) ∠ABC =
(b) ∠ACD =
(c) Line segment AC bisects ___ and ___
Solution:
i) AB = AD (given)
∠BAC = ∠DAC (given)
AC = CA (common)
Therefore by SAS condition of congruency, ΔABC ≅ ΔADC
ii) ∠ABC = ∠ADC (c.p.c.t)
∠ACD = ∠ACB (c.p.c.t)
Question: 7
In figure, AB || DC and AB = DC.
(i) Is ΔACD ≅ ΔCAB?
(ii) State the three pairs of matching parts used to answer (i).
(iii) Which angle is equal to ∠CAD ?
(iv) Does it follow from (iii) that AD || BC?
Solution:
(i) Yes by SAS condition of congruency, ΔDCA ≅ ΔBAC
(ii) We have used AB = DC, AC = CA and ∠DCA = ∠BAC.
(iii) ∠CAD = ∠ACB since the two triangles are congruent.
(iv) Yes this follows from AD // BC as alternate angles are equal. lf alternate angles are equal the lines are parallel
Exercise 16.4
Question: 1
Which of the following pairs of triangle are congruent by ASA condition?
Solution:
i)
We have,
Since ∠ABO = ∠CDO = 45° and both are alternate angles, AB // DC, ∠BAO = ∠DCO (alternate angle, AB // CD and AC is a transversal line)
∠ABO = ∠CDO = 45° (given in the figure) Also, AB = DC (Given in the figure)
Therefore, by ASA ΔAOB ≅ ΔDOC
ii)
In ABC,
Now AB =AC (Given)
∠ABD = ∠ACD = 40° (Angles opposite to equal sides)
BC = QR but none of the angles of ΔABC and ΔPQR are equal.
Therefore, ΔABC and Cong ΔPRQ
Question: 2
In figure, AD bisects A and AD and AD ⊥ BC.
(i) Is ΔADB ≅ ΔADC?
(ii) State the three pairs of matching parts you have used in (i)
(iii) Is it true to say that BD = DC?
Solution:
(i) Yes, ΔADB≅ΔADC, by ASA criterion of congruency.
(ii) We have used ∠BAD = ∠CAD ∠ADB = ∠ADC = 90°
Since, AD ⊥ BC and AD = DA
(iii) Yes, BD = DC since, ΔADB ≅ ΔADC
Question: 3
Draw any triangle ABC. Use ASA condition to construct other triangle congruent to it.
Solution:
We have drawn
Δ ABC with ∠ABC = 65° and ∠ACB = 70°
We now construct ΔPQR ≅ ΔABC has ∠PQR = 65° and ∠PRQ = 70°
Also we construct ΔPQR such that BC = QR
Therefore by ASA the two triangles are congruent
Question: 4
In Δ ABC, it is known that ∠B = C. Imagine you have another copy of Δ ABC
(i) Is ΔABC ≅ ΔACB
(ii) State the three pairs of matching parts you have used to answer (i).
(iii) Is it true to say that AB = AC?
Solution:
(i) Yes ΔABC ≅ ΔACB
(ii) We have used ∠ABC = ∠ACB and ∠ACB = ∠ABC again.
Also BC = CB
(iii) Yes it is true to say that AB = AC since ∠ABC = ∠ACB.
Question: 5
In Figure, AX bisects ∠BAC as well as ∠BDC. State the three facts needed to ensure that ΔACD ≅ ΔABD
Solution:
As per the given conditions, ∠CAD = ∠BAD and ∠CDA = ∠BDA (because AX bisects ∠BAC)
AD = DA (common)
Therefore, by ASA, ΔACD ≅ ΔABD
Question: 6
In Figure, AO = OB and ∠A = ∠B.
(i) Is ΔAOC ≅ ΔBOD
(ii) State the matching pair you have used, which is not given in the question.
(iii) Is it true to say that ∠ACO = ∠BDO?
Solution:
We have
∠OAC = ∠OBD,
AO = OB
Also, ∠AOC = ∠BOD (Opposite angles on same vertex)
Therefore, by ASA ΔAOC ≅ ΔBOD
Exercise 16.5
Question: 1
In each of the following pairs of right triangles, the measures of some part are indicated alongside. State by the application of RHS congruence conditions which are congruent, and also state each result in symbolic form.
Solution:
i)
∠ADC = ∠BCA = 90°
AD = BC and hyp AB = hyp AB
Therefore, by RHS ΔADB ≅ ΔACB
ii)
AD = AD (Common)
hyp AC = hyp AB (Given)
∠ADB + ∠ADC = 180° (Linear pair)
∠ADB + 90° = 180°
∠ADB = 180° – 90° = 90°
∠ADB = ∠ADC = 90°
Therefore, by RHS Δ ADB = Δ ADC
iii)
hyp AO = hyp DOBO = CO ∠B = ∠C = 90°
Therefore, by RHS, ΔAOB≅ΔDOC
iv)
Hyp A = Hyp CABC = DC ∠ABC = ∠ADC = 90°
Therefore, by RHS, ΔABC ≅ ΔADC
v)
BD = DB Hyp AB = Hyp BC, as per the given figure,
∠BDA + ∠BDC = 180°
∠BDA + 90° = 180°
∠BDA= 180°- 90° = 90°
∠BDA = ∠BDC = 90°
Therefore, by RHS, ΔABD ≅ ΔCBD
Question: 2
Δ ABC is isosceles with AB = AC. AD is the altitude from A on BC.
i) Is ΔABD ≅ ΔACD?
(ii) State the pairs of matching parts you have used to answer (i).
(iii) Is it true to say that BD = DC?
Solution:
(i) Yes, ΔABD ≅ ΔACD by RHS congruence condition.
(ii) We have used Hyp AB = Hyp AC
AD = DA
∠ADB = ∠ADC = 90° (AD ⊥ BC at point D)
(iii)Yes, it is true to say that BD = DC (c.p.c.t) since we have already proved that the two triangles are congruent.
Question: 3
ΔABC is isosceles with AB = AC. Also. AD ⊥ BC meeting BC in D. Are the two triangles ABD and ACD congruent? State in symbolic form. Which congruence condition do you use? Which side of ADC equals BD? Which angle of Δ ADC equals ∠B?
Solution:
We have AB = AC …… (i)
AD = DA (common) ……(ii)
And, ∠ADC = ∠ADB (AD ⊥ BC at point D) ……(iii)
Therefore, from (i), (ii) and (iii), by RHS congruence condition, ΔABD ≅ ΔACD, the triangles are congruent.
Therefore, BD = CD.
And ∠ABD = ∠ACD (c.p.c.t)
Question: 4
Draw a right triangle ABC. Use RHS condition to construct another triangle congruent to it.
Solution:
Consider
Δ ABC with ∠B as right angle.
We now construct another triangle on base BC, such that ∠C is a right angle and AB = DC
Also, BC = CB
Therefore, BC = CB
Therefore by RHS, ΔABC ≅ ΔDCB
Question: 5
In figure, BD and CE are altitudes of Δ ABC and BD = CE.
(i) Is ΔBCD ≅ ΔCBE?
(ii) State the three pairs or matching parts you have used to answer (i)
Take three non-collinear points A. B and C on a page of your notebook. Join AB, BC and CA. What figure do you get? Name the triangle. Also, name
(i) The side opposite to ∠B
(ii) The angle opposite to side AB
(iii) The vertex opposite to side BC
(iv) The side opposite to vertex B.
Solution:
(i) AC
(ii) ∠B
(iii) A
(iv) AC
Question: 2
Take three collinear points A, B and C on a page of your note book. Join AB. BC and CA. Is the figure a triangle? If not, why?
Solution:
No, the figure is not a triangle. By definition a triangle is a plane figure formed by three non-parallel line segments
Question: 3
Distinguish between a triangle and its triangular region.
Solution:
A triangle is a plane figure formed by three non-parallel line segments, whereas, its triangular region includes the interior of the triangle along with the triangle itself.
Question: 4
D is a point on side BC of a ∆CAD is joined. Name all the triangles that you can observe in the figure. How many are they?
Solution:
We can observe the following three triangles in the given figure
(i) ∆ABC
(ii) ∆ACD
(iii) ∆ADB
Question: 5
A, B. C and D are four points, and no three points are collinear. AC and BD intersed at O. There are eight triangles that you can observe. Name all the triangles
Solution:
(i) ∆ABC
(ii) ∆ABD
(iii) ∆ABO
(iv) ∆BCD
(v) ∆DCO
(vi) ∆AOD
(vii) ∆ACD
(viii) ∆BCD
Question: 6
What is the difference between a triangle and triangular region?
Solution:
Plane of figure formed by three non-parallel line segments is called a triangle where as triangular region is the interior of triangle ABC together with the triangle ABC itself is called the triangular region ABC
Question: 7
Explain the following terms:
(i) Triangle
(a) Parts or elements of a triangle
(iii) Scalene triangle
(iv) Isosceles triangle
(v) Equilateral triangle
(vi) Acute triangle
(vii) Right triangle
(viii) Obtuse triangle
(ix) Interior of a triangle
(x) Exterior of a triangle
Solution:
(i) A triangle is a plane figure formed by three non-parallel line segments.
(ii) The three sides and the three angles of a triangle are together known as the parts or elements of that triangle.
(iii) A scalene triangle is a triangle in which no two sides are equal.
(iv) An isosceles triangle is a triangle in which two sides are equal. Isosceles triangle
(v) An equilateral triangle is a triangle in which all three sides are equal. Equilateral triangle
(vi) An acute triangle is a triangle in which all the angles are acute (less than 90°).
(vii) A right angled triangle is a triangle in which one angle is right angled, i.e. 90°.
(viii) An obtuse triangle is a triangle in which one angle is obtuse (more than 90°).
(ix) The interior of a triangle is made up of all such points that are enclosed within the triangle.
(x) The exterior of a triangle is made up of all such points that are not enclosed within the triangle.
Question: 8
In Figure, the length (in cm) of each side has been indicated along the side. State for each triangle angle whether it is scalene, isosceles or equilateral:
Solution:
(i) This triangle is a scalene triangle because no two sides are equal.
(ii) This triangle is an isosceles triangle because two of its sides, viz. PQ and PR, are equal.
(iii) This triangle is an equilateral triangle because all its three sides are equal.
(iv) This triangle is a scalene triangle because no two sides are equal.
(v) This triangle is an isosceles triangle because two of its sides are equal.
Question: 9
There are five triangles. The measures of some of their angles have been indicated. State for each triangle whether it is acute, right or obtuse.
Solution:
(i) This is a right triangle because one of its angles is 90°.
(ii) This is an obtuse triangle because one of its angles is 120°, which is greater than 90°.
(iii) This is an acute triangle because all its angles are acute angles (less than 90°).
(iv) This is a right triangle because one of its angles is 90°.
(v) This is an obtuse triangle because one of its angles is 110°, which is greater than 90°.
Question: 10
Fill in the blanks with the correct word/symbol to make it a true statement:
(i) A triangle has _____ sides.
(ii) A triangle has ______ vertices.
(iii) A triangle has _____ angles.
(iv) A triangle has ________ parts.
(v) A triangle whose no two sides are equal is known as ________
(v0 A triangle whose two sides are equal is known as ________
(vii) A triangle whose all the sides are equal is known as ________
(viii) A triangle whose one angle is a right angle is known as _________
(ix) A triangle whose all the angles are of measure less than 90′ is known as _________
(x) A triangle whose one angle is more than 90′ is known as _________
Solution:
(i) three
(ii) three
(iii) three
(iv) six (three sides + three angles)
(v) a scalene triangle
(vi) an isosceles triangle
(vii) an equilateral triangle
(viii) a right triangle
(ix) an acute triangle
(x) an obtuse triangle
Question: 11
In each of the following, state if the statement is true (T) or false (F):
(i) A triangle has three sides.
(ii) A triangle may have four vertices.
(iii) Any three line-segments make up a triangle.
(iv) The interior of a triangle includes its vertices.
(v) The triangular region includes the vertices of the corresponding triangle.
(vi) The vertices of a triangle are three collinear points.
(vii) An equilateral triangle is isosceles also.
(viii) Every right triangle is scalene.
(ix) Each acute triangle is equilateral.
(x) No isosceles triangle is obtuse.
Solution:
(i) True.
(ii) False. A triangle has three vertices.
(iii) False. Any three non-parallel line segments can make up a triangle.
(iv) False. The interior of a triangle is the region enclosed by the triangle and the vertices are not enclosed by the triangle.
(v) True. The triangular region includes the interior region and the triangle itself.
(vi) False. The vertices of a triangle are three non-collinear points.
(vii) True. In an equilateral triangle, any two sides are equal.
(viii) False. A right triangle can also be an isosceles triangle.
(ix) False. Each acute triangle is not an equilateral triangle, but each equilateral triangle is an acute triangle.
(x) False. An isosceles triangle can be an obtuse triangle, a right triangle or an acute triangle
Exercise 15.2
Question: 1
Two angles of a triangle are of measures 150° and 30°. Find the measure of the third angle.
Solution:
Let the third angle be x
Sum of all the angles of a triangle = 180°
105° + 30° + x = 180°
135° + x = 180°
x = 180° – 135°
x = 45°
Therefore the third angle is 45°
Question: 2
One of the angles of a triangle is 130°, and the other two angles are equal. What is the measure of each of these equal angles?
Solution:
Let the second and third angle be x
Sum of all the angles of a triangle = 180°
130° + x + x = 180°
130° + 2x = 180°
2x = 180°– 130°
2x = 50°
x = 50/2
x = 25°
Therefore the two other angles are 25° each
Question: 3
The three angles of a triangle are equal to one another. What is the measure of each of the angles?
Solution:
Let the each angle be x
Sum of all the angles of a triangle =180°
x + x + x = 180°
3x = 180°
x = 180/3
x = 60°
Therefore angle is 60° each
Question: 4
If the angles of a triangle are in the ratio 1: 2: 3, determine three angles.
Solution:
If angles of the triangle are in the ratio 1: 2: 3 then take first angle as ‘x’, second angle as ‘2x’ and third angle as ‘3x’
Sum of all the angles of a triangle=180°
x + 2x + 3x = 180°
6x = 180°
x = 180/6
x = 30°
2x = 30° × 2 = 60°
3x = 30° × 3 = 90°
Therefore the first angle is 30°, second angle is 60° and third angle is 90°
Question: 5
The angles of a triangle are (x − 40) °, (x − 20) ° and (1/2 − 10) °. Find the value of x.
Solution:
Question: 6
The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angles is 10°. Find the three angles.
Solution:
Let the first angle be x
Second angle be x + 10°
Third angle be x + 10° + 10°
Sum of all the angles of a triangle = 180°
x + x + 10° + x + 10° +10° = 180°
3x + 30 = 180
3x = 180 – 30
3x = 150
x = 150/3
x = 50°
First angle is 50°
Second angle x + 10° = 50 + 10 = 60°
Third angle x + 10° +10° = 50 + 10 + 10 = 70°
Question: 7
Two angles of a triangle are equal and the third angle is greater than each of those angles by 30°. Determine all the angles of the triangle
Solution:
Let the first and second angle be x
The third angle is greater than the first and second by 30° = x + 30°
The first and the second angles are equal
Sum of all the angles of a triangle = 180°
x + x + x + 30° = 180°
3x + 30 = 180
3x = 180 – 30
3x = 150
x = 150/3
x = 50°
Third angle = x + 30° = 50° + 30° = 80°
The first and the second angle is 50° and the third angle is 80°
Question: 8
If one angle of a triangle is equal to the sum of the other two, show that the triangle is a right triangle.
Solution:
One angle of a triangle is equal to the sum of the other two
x = y + z
Let the measure of angles be x, y, z
x + y + z = 180°
x + x = 180°
2x = 180°
x = 180/2
x = 90°
If one angle is 90° then the given triangle is a right angled triangle
Question: 9
If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.
Solution:
Each angle of a triangle is less than the sum of the other two
Measure of angles be x, y and z
x > y + z
y < x + z
z < x + y
Therefore triangle is an acute triangle
Question: 10
In each of the following, the measures of three angles are given. State in which cases the angles can possibly be those of a triangle:
(i) 63°, 37°, 80°
(ii) 45°, 61°, 73°
(iii) 59°, 72°, 61°
(iv) 45°, 45°, 90°
(v) 30°, 20°, 125°
Solution:
(i) 63°, 37°, 80° = 180°
Angles form a triangle
(ii) 45°, 61°, 73° is not equal to 180°
Therefore not a triangle
(iii) 59°, 72°, 61° is not equal to 180°
Therefore not a triangle
(iv) 45°, 45°, 90° = 180°
Angles form a triangle
(v) 30°, 20°, 125° is not equal to 180°
Therefore not a triangle
Question: 11
The angles of a triangle are in the ratio 3: 4: 5. Find the smallest angle
Solution:
Given that
Angles of a triangle are in the ratio: 3: 4: 5
Measure of the angles be 3x, 4x, 5x
Sum of the angles of a triangle =180°
3x + 4x + 5x = 180°
12x = 180°
x = 180/12
x = 15°
Smallest angle = 3x
=3 × 15°
= 45°
Question: 12
Two acute angles of a right triangle are equal. Find the two angles.
Solution:
Given acute angles of a right angled triangle are equal
Right triangle: whose one of the angle is a right angle
Measured angle be x, x, 90°
x + x + 180°= 180°
2x = 90°
x = 90/2
x = 45°
The two angles are 45° and 45°
Question: 13
One angle of a triangle is greater than the sum of the other two. What can you say about the measure of this angle? What type of a triangle is this?
Solution:
Angle of a triangle is greater than the sum of the other two
Measure of the angles be x, y, z
x > y + z or
y > x + z or
z > x + y
x or y or z > 90° which is obtuse
Therefore triangle is an obtuse angle
Question: 14
AC, AD and AE are joined. Find
∠FAB + ∠ABC + ∠BCD + ∠CDE + ∠DEF + ∠EFA
∠FAB + ∠ABC + ∠BCD + ∠CDE + ∠DEF + ∠EFA
Solution:
We know that sum of the angles of a triangle is 180°
Find x, y, z (whichever is required) from the figures given below:
Solution:
(i) In ∆ABC and ∆ADE we have:
∠ADE = ∠ABC (corresponding angles)
x = 40°
∠AED = ∠ACB (corresponding angles)
y = 30°
We know that the sum of all the three angles of a triangle is equal to 180°
x + y + z = 180° (Angles of ∆ADE)
Which means: 40° + 30° + z = 180°
z = 180° – 70°
z = 110°
Therefore, we can conclude that the three angles of the given triangle are 40°, 30° and 110°.
(ii) We can see that in ∆ADC, ∠ADC is equal to 90°.
(∆ADC is a right triangle)
We also know that the sum of all the angles of a triangle is equal to 180°.
Which means: 45° + 90° + y = 180° (Sum of the angles of ∆ADC)
135° + y = 180°
y = 180° – 135°.
y = 45°.
We can also say that in ∆ABC, ∠ABC + ∠ACB + ∠BAC is equal to 180°.
(Sum of the angles of ∆ABC)
40° + y + (x + 45°) = 180°
40° + 45° + x + 45° = 180° (y = 45°)
x = 180° –130°
x = 50°
Therefore, we can say that the required angles are 45° and 50°.
(iii) We know that the sum of all the angles of a triangle is equal to 180°.
Therefore, for △ABD:
∠ABD +∠ADB + ∠BAD = 180° (Sum of the angles of ∆ABD)
50° + x + 50° = 180°
100° + x = 180°
x = 180° – 100°
x = 80°
For ∆ABC:
∠ABC + ∠ACB + ∠BAC = 180° (Sum of the angles of ∆ABC)
50° + z + (50° + 30°) = 180°
50° + z + 50° + 30° = 180°
z = 180° – 130°
z = 50°
Using the same argument for ∆ADC:
∠ADC + ∠ACD + ∠DAC = 180° (Sum of the angles of ∆ADC)
y +z + 30° =180°
y + 50° + 30° = 180° (z = 50°)
y = 180° – 80°
y = 100°
Therefore, we can conclude that the required angles are 80°, 50° and 100°.
(iv) In ∆ABC and ∆ADE we have:
∠ADE = ∠ABC (Corresponding angles)
y = 50°
Also, ∠AED = ∠ACB (Corresponding angles)
z = 40°
We know that the sum of all the three angles of a triangle is equal to 180°.
Which means: x + 50° + 40° = 180° (Angles of ∆ADE)
x = 180° – 90°
x = 90°
Therefore, we can conclude that the required angles are 50°, 40° and 90°.
Question: 16
If one angle of a triangle is 60° and the other two angles are in the ratio 1: 2, find the angles
Solution:
We know that one of the angles of the given triangle is 60°. (Given)
We also know that the other two angles of the triangle are in the ratio 1: 2.
Let one of the other two angles be x.
Therefore, the second one will be 2x.
We know that the sum of all the three angles of a triangle is equal to 180°.
60° + x + 2x = 180°
3x = 180° – 60°
3x = 120°
x = 120/3
x = 40°
2x = 2 × 40
2x = 80°
Hence, we can conclude that the required angles are 40° and 80°.
Question: 17
It one angle of a triangle is 100° and the other two angles are in the ratio 2: 3. find the angles.
Solution:
We know that one of the angles of the given triangle is 100°.
We also know that the other two angles are in the ratio 2: 3.
Let one of the other two angles be 2x.
Therefore, the second angle will be 3x.
We know that the sum of all three angles of a triangle is 180°.
100° + 2x + 3x = 180°
5x = 180° – 100°
5x = 80°
x = 80/5
2x = 2 ×16
2x = 32°
3x = 3×16
3x = 48°
Thus, the required angles are 32° and 48°.
Question: 18
In △ABC, if 3∠A = 4∠B = 6∠C, calculate the angles.
Solution:
We know that for the given triangle, 3∠A = 6∠C
∠A = 2∠C — (i)
We also know that for the same triangle, 4∠B = 6∠C
∠B = (6/4)∠C — (ii)
We know that the sum of all three angles of a triangle is 180°.
Therefore, we can say that:
∠A + ∠B + ∠C = 180° (Angles of ∆ABC) — (iii)
On putting the values of ∠A and ∠B in equation (iii), we get:
2∠C + (6/4)∠C +∠C = 180°
(18/4) ∠C = 180°
∠C = 40°
From equation (i), we have:
∠A = 2∠C = 2 × 40
∠A = 80°
From equation (ii), we have:
∠B = (6/4)∠C = (6/4) × 40°
∠B = 60°
∠A = 80°, ∠B = 60°, ∠C = 40°
Therefore, the three angles of the given triangle are 80°, 60°, and 40°.
Question: 19
Is it possible to have a triangle, in which
(i) Two of the angles are right?
(ii) Two of the angles are obtuse?
(iii) Two of the angles are acute?
(iv) Each angle is less than 60°?
(v) Each angle is greater than 60°?
(vi) Each angle is equal to 60°
Solution:
Give reasons in support of your answer in each case.
(i) No, because if there are two right angles in a triangle, then the third angle of the triangle must be zero, which is not possible.
(ii) No, because as we know that the sum of all three angles of a triangle is always 180°. If there are two obtuse angles, then their sum will be more than 180°, which is not possible in case of a triangle.
(iii) Yes, in right triangles and acute triangles, it is possible to have two acute angles.
(iv) No, because if each angle is less than 60°, then the sum of all three angles will be less than 180°, which is not possible in case of a triangle.
Proof:
Let the three angles of the triangle be ∠A, ∠B and ∠C.
As per the given information,
∠A < 60° … (i)
∠B< 60° … (ii)
∠C < 60° … (iii)
On adding (i), (ii) and (iii), we get:
∠A + ∠B + ∠C < 60°+ 60°+ 60°
∠A + ∠B + ∠C < 180°
We can see that the sum of all three angles is less than 180°, which is not possible for a triangle.
Hence, we can say that it is not possible for each angle of a triangle to be less than 60°.
(v) No, because if each angle is greater than 60°, then the sum of all three angles will be greater than 180°, which is not possible.
Proof:
Let the three angles of the triangle be ∠A, ∠B and ∠C. As per the given information,
∠A > 60° … (i)
∠B > 60° … (ii)
∠C > 60° … (iii)
On adding (i), (ii) and (iii), we get:
∠A + ∠B + ∠C > 60°+ 60°+ 60°
∠A + ∠B + ∠C > 180°
We can see that the sum of all three angles of the given triangle are greater than 180°, which is not possible for a triangle.
Hence, we can say that it is not possible for each angle of a triangle to be greater than 60°.
(vi) Yes, if each angle of the triangle is equal to 60°, then the sum of all three angles will be 180° , which is possible in case of a triangle.
Proof:
Let the three angles of the triangle be ∠A, ∠B and ∠C. As per the given information,
∠A = 60° … (i)
∠B = 60° …(ii)
∠C = 60° … (iii)
On adding (i), (ii) and (iii), we get:
∠A + ∠B + ∠C = 60°+ 60°+ 60°
∠A + ∠B + ∠C =180°
We can see that the sum of all three angles of the given triangle is equal to 180°, which is possible in case of a triangle. Hence, we can say that it is possible for each angle of a triangle to be equal to 60°.
Question: 20
In ∆ABC, ∠A = 100°, AD bisects ∠A and AD perpendicular BC. Find ∠B
Solution:
Consider ∆ABD
∠BAD = 100/2 (AD bisects ∠A)
∠BAD = 50°
∠ADB = 90° (AD perpendicular to BC)
We know that the sum of all three angles of a triangle is 180°.
Thus,
∠ABD + ∠BAD + ∠ADB = 180° (Sum of angles of ∆ABD)
Or,
∠ABD + 50° + 90° = 180°
∠ABD =180° – 140°
∠ABD = 40°
Question: 21
In ∆ABC, ∠A = 50°, ∠B = 100° and bisector of ∠C meets AB in D. Find the angles of the triangles ADC and BDC
Solution:
We know that the sum of all three angles of a triangle is equal to 180°.
Therefore, for the given △ABC, we can say that:
∠A + ∠B + ∠C = 180° (Sum of angles of ∆ABC)
50° + 70° + ∠C = 180°
∠C= 180° –120°
∠C = 60°
∠ACD = ∠BCD =∠C2 (CD bisects ∠C and meets AB in D. )
∠ACD = ∠BCD = 60/2= 30°
Using the same logic for the given ∆ACD, we can say that:
∠DAC + ∠ACD + ∠ADC = 180°
50° + 30° + ∠ADC = 180°
∠ADC = 180°– 80°
∠ADC = 100°
If we use the same logic for the given ∆BCD, we can say that
∠DBC + ∠BCD + ∠BDC = 180°
70° + 30° + ∠BDC = 180°
∠BDC = 180° – 100°
∠BDC = 80°
Thus,
For ∆ADC: ∠A = 50°, ∠D = 100° ∠C = 30°
∆BDC: ∠B = 70°, ∠D = 80° ∠C = 30°
Question: 22
In ∆ABC, ∠A = 60°, ∠B = 80°, and the bisectors of ∠B and ∠C, meet at O. Find
(i) ∠C
(ii) ∠BOC
Solution:
We know that the sum of all three angles of a triangle is 180°.
Hence, for △ABC, we can say that:
∠A + ∠B + ∠C = 180° (Sum of angles of ∆ABC)
60° + 80° + ∠C= 180°.
∠C = 180° – 140°
∠C = 140°.
For △OBC,
∠OBC = ∠B2 = 80/2 (OB bisects ∠B)
∠OBC = 40°
∠OCB =∠C2 = 40/2 (OC bisects ∠C)
∠OCB = 20°
If we apply the above logic to this triangle, we can say that:
∠OCB + ∠OBC + ∠BOC = 180° (Sum of angles of ∆OBC)
20° + 40° + ∠BOC = 180°
∠BOC = 180° – 60°
∠BOC = 120°
Question: 23
The bisectors of the acute angles of a right triangle meet at O. Find the angle at O between the two bisectors.
Solution:
We know that the sum of all three angles of a triangle is 180°.
Hence, for ∆ABC, we can say that:
∠A + ∠B + ∠C = 180°
∠A + 90° + ∠C = 180°
∠A + ∠C = 180° – 90°
∠A + ∠C = 90°
For ∆OAC:
∠OAC = ∠A2 (OA bisects LA)
∠OCA = ∠C2 (OC bisects LC)
On applying the above logic to △OAC, we get:
∠AOC + ∠OAC + ∠OCA = 180° (Sum of angles of ∆AOC)
∠AOC + ∠A2 + ∠C2 = 180°
∠AOC + ∠A + ∠C2 = 180°
∠AOC + 90/2 = 180°
∠AOC = 180° – 45°
∠AOC = 135°
Question: 24
In ∆ABC, ∠A = 50° and BC is produced to a point D. The bisectors of ∠ABC and ∠ACD meet at E. Find ∠E.
Solution:
In the given triangle,
∠ACD = ∠A + ∠B. (Exterior angle is equal to the sum of two opposite interior angles.)
We know that the sum of all three angles of a triangle is 180°.
Therefore, for the given triangle, we can say that:
Question: 25
In ∆ABC, ∠B = 60°, ∠C = 40°, AL perpendicular BC and AD bisects ∠A such that L and D lie on side BC. Find ∠LAD
Solution:
We know that the sum of all angles of a triangle is 180°
Therefore, for ∆ABC, we can say that:
Question: 26
Line segments AB and CD intersect at O such that AC perpendicular DB. It ∠CAB = 35° and ∠CDB = 55°. Find ∠BOD.
Solution:
We know that AC parallel to BD and AB cuts AC and BD at A and B, respectively.
∠CAB = ∠DBA (Alternate interior angles)
∠DBA = 35°
We also know that the sum of all three angles of a triangle is 180°.
In Figure, ∆ABC is right angled at A, Q and R are points on line BC and P is a point such that QP perpendicular to AC and RP perpendicular to AB. Find ∠P
Solution:
In the given triangle, AC parallel to QP and BR cuts AC and QP at C and Q, respectively.
∠QCA = ∠CQP (Alternate interior angles)
Because RP parallel to AB and BR cuts AB and RP at B and R, respectively,
∠ABC = ∠PRQ (alternate interior angles).
We know that the sum of all three angles of a triangle is 180°.
(ii) the interior opposite angles to exterior ∠CBX
Also, name the interior opposite angles to an exterior angle at A.
Solution:
(i) ∠ABC
(ii) ∠BAC and ∠ACB
Also the interior angles opposite to exterior are ∠ABC and ∠ACB
Question: 2
In the fig, two of the angles are indicated. What are the measures of ∠ACX and ∠ACB?
Solution:
In ∆ABC, ∠A = 50° and ∠B = 55°
Because of the angle sum property of the triangle, we can say that
∠A + ∠B + ∠C = 180°
50°+ 55°+ ∠C = 180°
Or
∠C = 75°
∠ACB = 75°
∠ACX = 180°− ∠ACB = 180°−75° = 105°
Question: 3
In a triangle, an exterior angle at a vertex is 95° and its one of the interior opposite angles is 55°. Find all the angles of the triangle.
Solution:
We know that the sum of interior opposite angles is equal to the exterior angle.
Hence, for the given triangle, we can say that:
∠ABC+ ∠BAC = ∠BCO
55° + ∠BAC = 95°
Or,
∠BAC= 95°– 95°
= ∠BAC = 40°
We also know that the sum of all angles of a triangle is 180°.
Hence, for the given △ABC, we can say that:
∠ABC + ∠BAC + ∠BCA = 180°
55° + 40° + ∠BCA = 180°
Or,
∠BCA = 180° –95°
= ∠BCA = 85°
Question: 4
One of the exterior angles of a triangle is 80°, and the interior opposite angles are equal to each other. What is the measure of each of these two angles?
Solution:
Let us assume that A and B are the two interior opposite angles.
We know that ∠A is equal to ∠B.
We also know that the sum of interior opposite angles is equal to the exterior angle.
Hence, we can say that:
∠A + ∠B = 80°
Or,
∠A +∠A = 80° (∠A= ∠B)
2∠A = 80°
∠A = 40/2 =40°
∠A= ∠B = 40°
Thus, each of the required angles is of 40°.
Question: 5
The exterior angles, obtained on producing the base of a triangle both ways are 104° and 136°. Find all the angles of the triangle.
Solution:
In the given figure, ∠ABE and ∠ABC form a linear pair.
∠ABE + ∠ABC =180°
∠ABC = 180°– 136°
∠ABC = 44°
We can also see that ∠ACD and ∠ACB form a linear pair.
∠ACD + ∠ACB = 180°
∠AUB = 180°– 104°
∠ACB = 76°
We know that the sum of interior opposite angles is equal to the exterior angle.
Therefore, we can say that:
∠BAC + ∠ABC = 104°
∠BAC = 104°– 44° = 60°
Thus,
∠ACE = 76° and ∠BAC = 60°
Question: 6
In Fig, the sides BC, CA and BA of a ∆ABC have been produced to D, E and F respectively. If ∠ACD = 105° and ∠EAF = 45°; find all the angles of the ∆ABC
Solution:
In a ∆ABC, ∠BAC and ∠EAF are vertically opposite angles.
Hence, we can say that:
∠BAC = ∠EAF = 45°
Considering the exterior angle property, we can say that:
∠BAC + ∠ABC = ∠ACD = 105°
∠ABC = 105°– 45° = 60°
Because of the angle sum property of the triangle, we can say that:
∠ABC + ∠ACS +∠BAC = 180°
∠ACB = 75°
Therefore, the angles are 45°, 65° and 75°.
Question: 7
In Figure, AC perpendicular to CE and C ∠A: ∠B: ∠C= 3: 2: 1. Find the value of ∠ECD.
Solution:
In the given triangle, the angles are in the ratio 3: 2: 1.
Let the angles of the triangle be 3x, 2x and x.
Because of the angle sum property of the triangle, we can say that:
3x + 2x + x = 180°
6x = 180°
Or,
x = 30° … (i)
Also, ∠ACB + ∠ACE + ∠ECD = 180°
x + 90° + ∠ECD = 180° (∠ACE = 90°)
∠ECD = 60° [From (i)]
Question: 8
A student when asked to measure two exterior angles of ∆ABC observed that the exterior angles at A and B are of 103° and 74° respectively. Is this possible? Why or why not?
Solution:
Here,
Internal angle at A + External angle at A = 180°
Internal angle at A + 103° =180°
Internal angle at A = 77°
Internal angle at B + External angle at B = 180°
Internal angle at B + 74° = 180°
Internal angle at B = 106°
Sum of internal angles at A and B = 77° + 106° =183°
It means that the sum of internal angles at A and B is greater than 180°, which cannot be possible.
Question: 9
In Figure, AD and CF are respectively perpendiculars to sides BC and AB of ∆ABC. If ∠FCD = 50°, find ∠BAD
Solution:
We know that the sum of all angles of a triangle is 180°
Therefore, for the given ∆FCB, we can say that:
∠FCB + ∠CBF + ∠BFC = 180°
50° + ∠CBF + 90°= 180°
Or,
∠CBF = 180° – 50°– 90° = 40° … (i)
Using the above rule for ∆ABD, we can say that:
∠ABD + ∠BDA + ∠BAD = 180°
∠BAD = 180° – 90°– 40° = 50° [from (i)]
Question: 10
In Figure, measures of some angles are indicated. Find the value of x.
Solution:
Here,
∠AED + 120° = 180° (Linear pair)
∠AED = 180°– 120° = 60°
We know that the sum of all angles of a triangle is 180°.
Therefore, for △ADE, we can say that:
∠ADE + ∠AED + ∠DAE = 180°
60°+ ∠ADE + 30° =180°
Or,
∠ADE = 180°– 60°– 30° = 90°
From the given figure, we can also say that:
∠FDC + 90° = 180° (Linear pair)
∠FDC = 180°– 90° = 90°
Using the above rule for △CDF, we can say that:
∠CDF + ∠DCF + ∠DFC = 180°
90° + ∠DCF + 60° =180°
∠DCF = 180°−60°− 90°= 30°
Also,
∠DCF + x = 180° (Linear pair)
30° + x = 180°
Or,
x = 180°– 30° = 150°
Question: 11
In Figure, ABC is a right triangle right angled at A. D lies on BA produced and DE perpendicular to BC intersecting AC at F. If ∠AFE = 130°, find
(i) ∠BDE
(ii) ∠BCA
(iii) ∠ABC
Solution:
(i) Here,
∠BAF + ∠FAD = 180° (Linear pair)
∠FAD = 180°- ∠BAF = 180°– 90° = 90°
Also,
∠AFE = ∠ADF + ∠FAD (Exterior angle property)
∠ADF + 90° = 130°
∠ADF = 130°− 90° = 40°
(ii) We know that the sum of all the angles of a triangle is 180°.
ABC is a triangle in which ∠B = ∠C and ray AX bisects the exterior angle DAC. If ∠DAX = 70°. Find ∠ACB.
Solution:
Here,
∠CAX = ∠DAX (AX bisects ∠CAD)
∠CAX =70°
∠CAX +∠DAX + ∠CAB =180°
70°+ 70° + ∠CAB =180°
∠CAB =180° –140°
∠CAB =40°
∠ACB + ∠CBA + ∠CAB = 180° (Sum of the angles of ∆ABC)
∠ACB + ∠ACB+ 40° = 180° (∠C = ∠B)
2∠ACB = 180°– 40°
∠ACB = 140/2
∠ACB = 70°
Question: 13
The side BC of ∆ABC is produced to a point D. The bisector of ∠A meets side BC in L. If ∠ABC= 30° and ∠ACD = 115°, find ∠ALC
Solution:
∠ACD and ∠ACL make a linear pair.
∠ACD+ ∠ACB = 180°
115° + ∠ACB =180°
∠ACB = 180°– 115°
∠ACB = 65°
We know that the sum of all angles of a triangle is 180°.
Therefore, for ∆ABC, we can say that:
∠ABC + ∠BAC + ∠ACB = 180°
30° + ∠BAC + 65° = 180°
Or,
∠BAC = 85°
∠LAC = ∠BAC/2 = 85/2
Using the above rule for ∆ALC, we can say that:
∠ALC + ∠LAC + ∠ACL = 180°
Question: 14
D is a point on the side BC of ∆ABC. A line PDQ through D, meets side AC in P and AB produced at Q. If ∠A = 80°, ∠ABC = 60° and ∠PDC = 15°, find
(i) ∠AQD
(ii) ∠APD
Solution:
∠ABD and ∠QBD form a linear pair.
∠ABC + ∠QBC =180°
60° + ∠QBC = 180°
∠QBC = 120°
∠PDC = ∠BDQ (Vertically opposite angles)
∠BDQ = 75°
In ∆QBD:
∠QBD + ∠QDB + ∠BDQ = 180° (Sum of angles of ∆QBD)
120°+ 15° + ∠BQD = 180°
∠BQD = 180°– 135°
∠BQD = 45°
∠AQD = ∠BQD = 45°
In ∆AQP:
∠QAP + ∠AQP + ∠APQ = 180° (Sum of angles of ∆AQP)
80° + 45° + ∠APQ = 180°
∠APQ = 55°
∠APD = ∠APQ
Question: 15
Explain the concept of interior and exterior angles and in each of the figures given below. Find x and y
Solution:
The interior angles of a triangle are the three angle elements inside the triangle.
The exterior angles are formed by extending the sides of a triangle, and if the side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles.
Using these definitions, we will obtain the values of x and y.
(I) From the given figure, we can see that:
∠ACB + x = 180° (Linear pair)
75°+ x = 180°
Or,
x = 105°
We know that the sum of all angles of a triangle is 180°.
Therefore, for ∆ABC, we can say that:
∠BAC+ ∠ABC +∠ACB = 180°
40°+ y +75° = 180°
Or,
y = 65°
(ii) x + 80°= 180° (Linear pair)
= x = 100°
In ∆ABC:
x+ y+ 30° = 180° (Angle sum property)
100° + 30° + y = 180°
= y = 50°
(iii) We know that the sum of all angles of a triangle is 180°.
Therefore, for ∆ACD, we can say that:
30° + 100° + y = 180°
Or,
y = 50°
∠ACB + 100° = 180°
∠ACB = 80° … (i)
Using the above rule for ∆ACD, we can say that:
x + 45° + 80° = 180°
= x = 55°
(iv) We know that the sum of all angles of a triangle is 180°.
Therefore, for ∆DBC, we can say that:
30° + 50° + ∠DBC = 180°
∠DBC = 100°
x + ∠DBC = 180° (Linear pair)
x = 80°
And,
y = 30° + 80° = 110° (Exterior angle property)
Question: 16
Compute the value of x in each of the following figures
Solution:
(i) From the given figure, we can say that:
∠ACD + ∠ACB = 180° (Linear pair)
Or,
∠ACB = 180°– 112° = 68°
We can also say that:
∠BAE + ∠BAC = 180° (Linear pair)
Or,
∠BAC = 180°– 120° = 60°
We know that the sum of all angles of a triangle is 180°.
Therefore, for ∆ABC:
x + ∠BAC + ∠ACB = 180°
x = 180°– 60°– 68° = 52°
= x = 52°
(ii) From the given figure, we can say that:
∠ABC + 120° = 180° (Linear pair)
∠ABC = 60°
We can also say that:
∠ACB+ 110° = 180° (Linear pair)
∠ACB = 70°
We know that the sum of all angles of a triangle is 180°.
Therefore, for ∆ABC:
x + ∠ABC + ∠ACB = 180°
x = 50°
(iii) From the given figure, we can see that:
∠BAD = ∠ADC = 52° (Alternate angles)
We know that the sum of all the angles of a triangle is 180°.
Therefore, for ∆DEC:
x + 40°+ 52° = 180°
= x = 88°
(iv) In the given figure, we have a quadrilateral whose sum of all angles is 360°.
Thus,
35° + 45° + 50° + reflex ∠ADC = 360°
Or,
reflex ∠ADC = 230°
230° + x = 360° (A complete angle)
= x = 130°
Exercise 15.4
Question: 1
In each of the following, there are three positive numbers. State if these numbers could possibly be the lengths of the sides of a triangle:
(i) 5, 7, 9
(ii) 2, 10.15
(iii) 3, 4, 5
(iv) 2, 5, 7
(v) 5, 8, 20
Solution:
(i) Yes, these numbers can be the lengths of the sides of a triangle because the sum of any two sides of a triangle is always greater than the third side. Here, 5 + 7 > 9, 5 + 9 > 7, 9 + 7 > 5
(ii) No, these numbers cannot be the lengths of the sides of a triangle because the sum of any two sides of a triangle is always greater than the third side, which is not true in this case.
(iii) Yes, these numbers can be the lengths of the sides of a triangle because the sum of any two sides of triangle is always greater than the third side. Here, 3 + 4 > 5, 3 + 5 > 4, 4 + 5 > 3
(iv) No, these numbers cannot be the lengths of the sides of a triangle because the sum of any two sides of a triangle is always greater than the third side, which is not true in this case. Here, 2 + 5 = 7
(v) No, these numbers cannot be the lengths of the sides of a triangle because the sum of any two sides of a triangle is always greater than the third side, which is not true in this case. Here, 5 + 8 < 20
Question: 2
In Fig, P is the point on the side BC. Complete each of the following statements using symbol ‘=’,’ > ‘or ‘< ‘so as to make it true:
(i) AP… AB+ BP
(ii) AP… AC + PC
(iii) AP…. 1/2(AB + AC + BC)
Solution:
(i) In triangle APB, AP < AB + BP because the sum of any two sides of a triangle is greater than the third side.
(ii) In triangle APC, AP < AC + PC because the sum of any two sides of a triangle is greater than the third side.
(iii) AP < 12(AB + AC + BC) In triangles ABP and ACP, we can see that:
AP < AB + BP…(i) (Because the sum of any two sides of a triangle is greater than the third side)
AP < AC + PC…(ii) (Because the sum of any two sides of a triangle is greater than the third side)
On adding (i) and (ii), we have:
AP + AP < AB + BP + AC + PC
2AP < AB + AC + BC (BC = BP + PC)
AP < (AB – FAC + BC)
Question: 3
P is a point in the interior of △ABC as shown in Fig. State which of the following statements are true (T) or false (F):
(i) AP + PB < AB
(ii) AP + PC > AC
(iii) BP + PC = BC
Solution:
(i) False
We know that the sum of any two sides of a triangle is greater than the third side: it is not true for the given triangle.
(ii) True
We know that the sum of any two sides of a triangle is greater than the third side: it is true for the given triangle.
(iii) False
We know that the sum of any two sides of a triangle is greater than the third side: it is not true for the given triangle.
Question: 4
O is a point in the exterior of △ABC. What symbol ‘>’,’<’ or ‘=’ will you see to complete the statement OA+OB….AB? Write two other similar statements and show that
OA + OB + OC > 1/2(AB + BC +CA)
Solution:
Because the sum of any two sides of a triangle is always greater than the third side, in triangle OAB, we have:
OA + OB > AB — (i)
OB + OC > BC — (ii)
OA + OC > CA — (iii)
On adding equations (i), (ii) and (iii) we get:
OA + OB + OB + OC + OA + OC > AB + BC + CA
2(OA + OB + OC) > AB + BC + CA
OA + OB + OC > (AB + BC + CA)/2
Question: 5
In ∆ABC, ∠B = 30°, ∠C = 50°. Name the smallest and the largest sides of the triangle.
Solution:
Because the smallest side is always opposite to the smallest angle, which in this case is 30°, it is AC. Also, because the largest side is always opposite to the largest angle, which in this case is 100°, it is BC.
Exercise 15.5
Question: 1
State Pythagoras theorem and its converse.
Solution:
The Pythagoras Theorem: In a right triangle, the square of the hypotenuse is always equal to the sum of the squares of the other two sides.
Converse of the Pythagoras Theorem: If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle, with the angle opposite to the first side as right angle.
Question: 2
In right ∆ABC, the lengths of the legs are given. Find the length of the hypotenuse
(i) a = 6 cm, b = 8 cm
(ii) a = 8 cm, b = 15 cm
(iii) a = 3 cm, b = 4 cm
(iv) a = 2 cm, b =1.5 cm
Solution:
According to the Pythagoras theorem,
(Hypotenuse)2 = (Base)2 + (Height)2
(i) c2 = a2 + b2
c2 = 62 + 82
c2 = 36 + 64 = 100
c = 10 cm
(ii) c2 = a2 + b2
c2 = 82 + 152
c2 = 64 + 225 = 289
c = 17cm
(iii) c2 = a2 + b2
c2 = 32 + 42
c2 = 9 + 16 = 25
c = 5 cm
(iv) c2 = a2 + b2
c2 = 22 + 1.52
c2 = 4 + 2.25 = 6.25
c = 2.5 cm
Question: 3
The hypotenuse of a triangle is 2.5 cm. If one of the sides is 1.5 cm. find the length of the other side.
Solution:
Let the hypotenuse be ” c ” and the other two sides be ” b ” and ” c”.
Using the Pythagoras theorem, we can say that:
c2 = a2 + b2
2.52 = 1.52 + b2
b2 = 6.25 −2.25 = 4
c = 2 cm
Hence, the length of the other side is 2 cm.
Question: 4
A ladder 3.7 m long is placed against a wall in such a way that the foot of the ladder is 1.2 m away from the wall. Find the height of the wall to which the ladder reaches.
Solution:
Let the hypotenuse be h.
Using the Pythagoras theorem, we get:
3.72 = 1.22 + h2
h2 = 13.69 – 1.44 = 12.25
h = 3.5 m
Hence, the height of the wall is 3.5 m.
Question: 5
If the sides of a triangle are 3 cm, 4 cm and 6 cm long, determine whether the triangle is right-angled triangle.
Solution:
In the given triangle, the largest side is 6 cm.
We know that in a right angled triangle, the sum of the squares of the smaller sides should be equal to the square of the largest side.
Therefore,
32 + 42 = 9 + 16 = 25
But,
62 = 36
32 + 42 not equal to 62
Hence, the given triangle is not a right angled triangle.
Question: 6
The sides of certain triangles are given below. Determine which of them are right triangles.
(i) a = 7 cm, b = 24 cm and c= 25 cm
(ii) a = 9 cm, b = 16 cm and c = 18 cm
Solution:
(i) We know that in a right angled triangle, the square of the largest side is equal to the sum of the squares of the smaller sides.
Here, the larger side is c, which is 25 cm.
c2 = 625
We have:
a2+ b2 = 72 + 242 = 49 + 576 = 625 = c2
Thus, the given triangle is a right triangle.
(ii) We know that in a right angled triangle, the square of the largest side is equal to the sum of the squares of the smaller sides.
Here, the larger side is c, which is 18 cm.
c2 = 324
We have:
a2+ b2 = 92+162 = 81 + 256 = 337 not equal to c2
Thus, the given triangle is not a right triangle.
Question: 7
Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m. Find the distance between their tops.
(Hint: Find the hypotenuse of a right triangle having the sides (11 – 6) m = 5 m and 12 m)
Solution:
The distance between the tops of the poles is the distance between points A and B.
We can see from the given figure that points A, B and C form a right triangle, with AB as the hypotenuse.
On using the Pythagoras Theorem in ∆ABC, we get:
(11−6)2 + 122 = AB2
AB2 = 25 + 144
AB2 = 169
AB = 13
Hence, the distance between the tops of the poles is 13 m.
Question: 8
A man goes 15 m due west and then 8 m due north. How far is he from the starting point?
Solution:
Let O be the starting point and P be the final point.
By using the Pythagoras theorem, we can find the distance OP.
OP2 = 152 + 82
OP2 = 225 + 64
OP2 = 289
OP = 17
Hence, the required distance is 17 m.
Question: 9
The foot of a ladder is 6 m away from a wall and its top reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its top reach?
Solution:
Given Let the length of the ladder be L m.
By using the Pythagoras theorem, we can find the length of the ladder.
62 + 82 = L2
L2 = 36 + 64 =100
L = 10
Thus, the length of the ladder is 10 m.
When the ladder is shifted:
Let the height of the ladder after it is shifted be H m.
By using the Pythagoras theorem, we can find the height of the ladder after it is shifted.
82 + H2 = 102
H2 = 100 – 64 = 36
H = 6
Thus, the height of the ladder is 6 m.
Question: 10
A ladder 50 dm long when set against the wall of a house just reaches a window at a height of 48 dm. How far is the lower end of the ladder from the base of the wall?
Solution:
Let the distance of the lower end of the ladder from the wall be x m.
On using the Pythagoras theorem, we get:
x2 + 482 = 502
x2 = 502 − 482 = 2500 – 2304 = 196
H = 14 dm
Hence, the distance of the lower end of the ladder from the wall is 14 dm.
Question: 11
The two legs of a right triangle are equal and the square of the hypotenuse is 50. Find the length of each leg.
Solution:
Let the length of each leg of the given triangle be x units.
Using the Pythagoras theorem, we get:
x2 + x2 = (Hypotenuse)2
x2 + x2 = 50
2x2 = 50
x2 = 25
x = 5
Hence, we can say that the length of each leg is 5 units.
Question: 12
Verity that the following numbers represent Pythagorean triplet:
(i) 12, 35, 37
(ii) 7, 24, 25
(iii) 27, 36, 45
(iv) 15, 36, 39
Solution:
We will check for a Pythagorean triplet by checking if the square of the largest side is equal to the sum of the squares of the other two sides.
(i) 372 =1369
122 + 352 = 144 + 1225 = 1369
122 + 352 = 372
Yes, they represent a Pythagorean triplet.
(ii) 252 = 625
72 + 242 = 49 + 576 = 625
72 + 242 = 252
Yes, they represent a Pythagorean triplet.
(iii) 452 = 2025
272 + 362 = 729 + 1296 = 2025
272 + 362 = 452
Yes, they represent a Pythagorean triplet.
(iv) 392 = 1521
152 + 362 = 225 + 1296 = 1521
152 + 362 = 392
Yes, they represent a Pythagorean triplet.
Question: 13
In ∆ABC, ∠ABC = 100°, ∠BAC = 35° and BD perpendicular to AC meets side AC in D. If BD = 2 cm, find ∠C, and length DC.
Solution:
We know that the sum of all angles of a triangle is 180°
Therefore, for the given ∆ABC, we can say that:
∠ABC + ∠BAC + ∠ACB = 180°
100° + 35° + ∠ACB = 180°
∠ACB = 180° –135°
∠ACB = 45°
∠C = 45°
If we apply the above rule on ∆BCD, we can say that:
∠BCD + ∠BDC + ∠CBD = 180°
45° + 90° + ∠CBD = 180° (∠ACB = ∠BCD and BD parallel to AC)
∠CBD = 180°– 135°
∠CBD = 45°
We know that the sides opposite to equal angles have equal length.
Thus, BD = DC
DC = 2 cm
Question: 14
In a ∆ABC, AD is the altitude from A such that AD = 12 cm. BD = 9 cm and DC = 16 cm. Examine if ∆ABC is right angled at A.
Solution:
In ∆ADC,
∠ADC = 90° (AD is an altitude on BC)
Using the Pythagoras theorem, we get:
122 + 162 = AC2
AC2 = 144 + 256 = 400
AC = 20 cm
In ∆ADB,
∠ADB = 90° (AD is an altitude on BC)
Using the Pythagoras theorem, we get:
122 + 92 = AB2
AB2 = 144 + 81 = 225
AB = 15 cm
In ∆ABC,
BC2 = 252 = 625
AB2 + AC2 = 152 + 202 = 625
AB2 + AC2 = BC2
Because it satisfies the Pythagoras theorem, we can say that ∆ABC is right angled at A.
Question: 15
Draw a triangle ABC, with AC = 4 cm, BC = 3 cm and ∠C = 105°. Measure AB. Is (AC)2 + (BC)2? If not which one of the following is true: (AB)2 > (AC)2 + (BC)2 or (AB)2 < (AC)2 + (BC)2
Solution:
Draw ∆ABC.
Draw a line BC = 3 cm.
At point C, draw a line at 105° angle with BC.
Take an arc of 4 cm from point C, which will cut the line at point A.
Now, join AB, which will be approximately 5.5 cm.
AC2 + BC2 = 42 + 32 = 9 +16 = 25
AB2 = 5.52 = 30.25
AB2 not equal to AC2+ BC2
Here,
AB2 > AC2 + BC2
Question: 16
Draw a triangle ABC, with AC = 4 cm, BC = 3 cm and ∠C = 80°. Measure AB. Is (AC)2 + (BC)2? If not which one of the following is true: (AB)2 > (AC)2 + (BC)2 or (AB)2 < (AC)2 + (BC)2
Solution:
Draw ∆ABC.
Draw a line BC = 3 cm.
At point C, draw a line at 80° angle with BC.
Take an arc of 4 cm from point C, which will cut the line at point A.
(i) . Principal = Rs . 2000 , Rate of interest = 5% per annum , and Time = 5 years
(ii) . Principal = Rs . 500 , Rate of interest = 12.5% per annum , and Time = 4 years
(iii) . Principal = Rs . 4500 , Rate of interest = 4% per annum , and Time = 6 months
(iv) . Principal =Rs . 12000 , Rate of interest = 18% per annum , and Time = 4 months
(v) . Principal = Rs . 1000 , Rate of interest = 10% per annum , and Time = 73 days
Question: 2
Find the interest on Rs . 500 for a period of 4 years at the rate of 8% per annum . Also , find the amount to be paid at the end of the period .
Solution:
Question: 3
A sum of Rs . 400 is lent at the rate of 5% per annum . Find the interest at the end of 2 years .
Solution:
Question: 4
A sum of Rs . 400 is lent for 3 years at the rate of 6% per annum . Find the interest .
Solution:
Question: 5
A person deposits Rs . 25000 in a firm who pays an interest at the rate of 20 % per annum . Calculate the income he gets from it annually .
Solution:
Question: 6
A man borrowed Rs . 8000 from a bank at 8% per annum . Find the amount he has to pay after 412 years .
Solution:
Question: 7
Rakesh lent out Rs . 8000 for 5 years at 15 % per annum and borrowed Rs . 6000 for 3 years at 12% per annum . How much did he gain or lose ?
Solution:
Question: 8
Anita deposits Rs . 1000 in a savings bank account . The bank pays interest at the rate of 5 % per annum . What amount can Anita get after 1 year ?
Solution:
Question: 9
Nalini borrowed Rs . 550 from her friend at 8% per annum . She returned the amount after six months . How much did she pay ?
Solution:
Question: 10
Rohit borrowed Rs . 60000 from a bank at 9% per annum for 2 years . He lent this sum of money to Rohan at 10% per annum for 2 years . How much did Rohit earn from this transaction ?
Solution:
Question: 11
Romesh borrowed Rs . 2000 at 2% per annum and Rs . 1000 at 5% per annum . He cleared his debt after 2 years by giving Rs . 2800 and a watch . What is the cost of watch ?
Solution:
Question: 12
Mr.Garg lent Rs . 15000 to his friend . He charged 15% per annum on Rs. 12500 and 18% on the rest . How much interest does he earn in 3 years ?
Solution:
Question: 13
Shikha deposited Rs . 2000 in a bank which pays 6% simple interest . She withdrew Rs . 700 at the end of first year .What will be her balance after 3 years ?
Solution:
Question: 14
Reema took a loan of Rs . 8000 from a money lender , who charged interest at the rate of 18% per annum . After 2 years , Reema paid him Rs . 10400 and wrist watch to clear the debt . What is the price of the watch ?
Solution:
Question: 15
Mr. Sharma deposited Rs. 20000 as a fixed deposit in a bank at 10% per annum . If 30% is deducted as income tax on the interest earned , find his annual income .
Gain % = {(Gain/(CP + overhead expenses)} x 100 = {260/(2300 + 300} x 100
= {260/2600} x 100 = 10%
(ii) CP = Rs. 3500, overhead expenses = Rs. 150, loss = Rs. 146
Loss % = {(Loss/(CP + overhead expenses)} x 100 = {146/(3500+ 150)} x 100
= {146/3650} x 100 = 14600/3650 = 4%
Question: 6
A grain merchant sold 600 quintals of rice at a profit of 7%. If a quintal of rice cost him Rs 250 and his total overhead charges for transportation, etc. were Rs 1000 find his total profit and the selling price of 600 quintals of rice.
Solution:
Cost of 1 quintal of rice = Rs. 250
Cost of 600 quintals of rice = 600 x 250 = Rs. 150000
Shikha purchased a wrist watch for Rs 840 and sold it to her friend Vidhi for Rs 910. Find her gain percent.
Solution:
The cost price of the wristwatch that Shikha purchased, CP = Rs. 840
The price at which she sold it, SP = Rs. 910
Gain = SP – CP = (910 – 840) = Rs. 70
Gain % = (Gain/CP) x 100 = (70/840) x 100 = 7000/840 = 8.3%
Question: 11
A business man makes a 10% profit by selling a toy costing him Rs 120. What is the selling price?
Solution:
CP = Rs. 12
Profit % = 10
We now that
SP = {(100 + profit %)/100} x CP = {(100+ 10)/100} x 120
= {(110/100)} x 120 = 1.1 x 120
= Rs. 132
Question: 12
Harish purchased 50 dozen bananas for Rs 135. Five dozen bananas could not be sold because they were rotten. At what price per dozen should Harish sell the remaining bananas so that he makes a profit of 20%?
Solution:
Cost price of 50 dozens bananas that Harish purchased, CP = Rs. 135
Bananas left after removing 5 dozen rotten bananas = 45 dozens
Effective CP of one dozen bananas = Rs. 135/45 = Rs. 3
Calculating the price at which Harish should sell each dozen bananas to make a profit of 20% (or 1/5), we get
Profit % = (Gain/CP) x100
To get a gain of 20% we give profit % = 20 and substitute
20 = (gain/135) x100
Gain = 270/10 = 27
We know ; SP = CP + Gain
SP = 27 + 135
SP = 162
Now that SP is for 45 Dozens of bananas
Calculating for one dozen
→ 162/45
=Rs. 3.6
Harish should sell the bananas at Rs. 3.60 a dozen in order to make a profit of 20%.
Question: 13
A woman bought 50 dozen eggs at Rs 6.40 a dozen. Out of these 20 eggs were found to be broken. She sold the remaining eggs at 55 paise per egg. Find her gain or loss percent.
Solution:
Cost of one dozen eggs = Rs. 6.40
Cost of 50 dozen eggs = 50 x 6.40 = Rs. 320
Total number of eggs = 50 x 12 = 600
Number of eggs left after removing the broken ones = 600 – 20 = 580
SP of 1 egg = 55 paise
So, SP of 580 eggs = 580 x 55 = 31900 paise = Rs. 31900/100 = Rs. 319
Loss = CP – SP = Rs. (320-319) = Re. 1
Loss % = (Loss/CP) x 100 = (1/320) x 100 = 0.31%
Question: 14
Jyotsana bought 400 eggs at Rs 8.40 a dozen. At what price per hundred must she sell them so as to earn a profit of 15%?
Solution:
Cost of eggs per dozen = Rs. 8.40
Cost of 1 egg = 8.40/12 = Rs. 0.7
Cost of 400 eggs = 400 x 0.7 = Rs. 280
Calculating the price at which Jyotsana should sell the eggs to earn a profit of 15%,
So, Jyotsana must sell the 400 eggs for Rs. 322 in order to earn a profit of 15%. Therefore, the SP per one hundred eggs = Rs. 322/4 = Rs. 80.50.
Question: 15
A shopkeeper makes a profit of 15% by selling a book for Rs 230. What is the C.P. and the actual profit ?
Solution:
Given that the SP of a book = Rs. 230
Profit % = 15
Since
CP = (SP x 100) + (100 + profit %)
CP = (230x 100) + (100 + 15)
CP = 23000 + 115 = Rs. 200
Also, Profit = SP – CP = Rs. (230 – 200) = Rs. 30
Actual profit = Rs. 30
Question: 16
A bookseller sells all his books at a profit of 10%. If he buys a book from the distributor at Rs 200, how much does he sell it for ?
Solution:
Given
Profit % = 10% CP = Rs. 200
Since
SP = {(100 + profit %)/100} x CP = {(100 + 10)/100} x 200
= {110/100} x 200 = Rs. 220
The bookseller sells the book for Rs. 220.
Question: 17
A floweriest buys 100 dozen roses at Rs 2 a dozen. By the time the flowers are delivered, 20 dozen roses are mutilated and are thrown away. At what price should he sell the rest if he needs to make a 20% profit on his purchase ?
Solution:
Cost of 1 dozen roses = Rs. 2
Number of roses bought by the florist = 100 dozens
Thus, cost price of 100 dozen roses = 2 x 100 = Rs. 200
Roses left after discarding the mutilated ones = 80 dozens
Calculating the price at which the florist should sell the 80 dozen roses in order to make a profit of 20%, we have
Profit % = ((SP-CP)/CP) x100 = ((SP-200)/200)x100
40 = SP – 200
SP = Rs. 240
Therefore, the SP of the roses should be Rs. 240/80 = Rs. 3 per dozen.
Question: 18
By selling an article for Rs 240, a man makes a profit of 20%.What is his C.P. ? What would his profit percent be if he sold the article for Rs 275 ?
Express each of the following per cents as fractions in the simplest forms:
(i) 45%
(ii) 0.25%
(iii) 150%
(iv) 6(1/4) %
Solution:
Question: 2
Express each of the following fractions as a per cent:
(i) (3/4) %
(ii) (53/100) %
(iii) 1(3/5) %
(iv) (7/20) %
Solution:
Exercise 11.2
Question: 1
Express each of the following ratios as per cents
(i) 4:5
(ii) 1:5
(iii) 11: 125
Solution:
Question: 2
Express each of the following per cents as ratios in simplest forms
(i) 2.5%
(ii) 0.4%
(iii) 13(3/4)%
Solution:
Exercise 11.3
Question: 1
Express each of the following per cents as decimals:
(i) 12.5%
(ii) 75%
(iii) 128.8%
Solution:
Question: 2
Express each of the following decimals as per cent:
(i) 0.004
(ii) 0.24
(iii) 0.02
(iv) 0.275
Solution:
Question: 3
Write each of the following as whole numbers or mixed numbers:
(i) 136%
(ii) 250%
(iii) 300%
Solution:
Exercise 11.4
Question: 1
Find each of the following:
(i) 7% of Rs.7150
(ii) 40% of 400Kg
(iii) 20% of 15.125litres
(iv) 3(1/3)% of 90 km
(v) 2.5% of 600metres
Solution:
Question: 2
Find the number whose 12(1/2) % is 64.
Solution:
Question: 3
What is the number, 6(1/4)% of which is 2?
Solution:
Question: 4
If 6 is 50% of a number, what is the number?
Solution:
Exercise 11.5
Question: 1
What per cent of
(i) 24 is 6?
(ii) Rs.125 is Rs.10?
(iii) 4km is 160 metres?
(iv) Rs. 8 is 25 paise?
(v) 2 days is 8 hours?
(vi) 1 lire is 175 ml
Solution:
Question: 2
What per cent is equivalent to 3/8?
Solution:
Question: 3
Find the following:
(i) 8 is 4% of which number
(ii) 6 is 60% of which number
(iii) 6 is 30% of which number
(iv) 12 is 25% of which number
Solution:
Question: 4
Convert each of the following pairs into percentages and find out which is more?
(i) 25 marks out of 30, 35 marks out of 40
(ii) 100 runs scored off 110 balls, 50 runs scored off 55 balls
Solution:
Question: 5
Find 20% more than Rs.200.
Solution:
Question: 6
Find 10% less than Rs.150
Solution:
Exercise 11.6
Question: 1
Ashu had 24 pages to write. By the evening, he had completed 25% of his work. How many pages were left?
Solution:
Question: 2
A box contains 60 eggs. Out of which 16(2/3)% are rotten ones. How many eggs are rotten?
Solution:
Question: 3
Rohit obtained 45 marks out of 80. What per cent marks did he get?
Solution:
Question: 4
Mr Virmani saves 12% of his salary. If he receives Rs 15900 per month as salary, find his monthly expenditure.
Solution:
Question: 5
A lawyer willed his 3 sons Rs 250000 to be divided into portions 30%, 45% and 25%. How much did each of them inherit?
Solution:
Question: 6
Rajdhani College has 2400 students, 40% of whom are girls. How many boys are there in the college?
Solution:
Question: 7
Aman obtained 410 marks out of 500 in CBSE XII examination while his brother Anish gets 536 marks out of 600 in IX class examination. Find whose performance is better?
Solution:
Question: 8
Rahim obtained 60 marks out of 75 in Mathematics. Find the percentage of marks obtained by Rahim in Mathematics.
Solution:
Question: 9
In an orchard, 16(2/3) % of the trees are apple trees. If the number of trees in the orchard is 240, 3 find the number of other type of trees in the orchard.
Solution:
Question: 10
Ram scored 553 marks out of 700 and Gita scored 486 marks out of 600 in science. Whose performance is better?
Solution:
Question: 11
Out of an income of Rs 15000, Nazima spends Rs 10200. What per cent of her income does she save?
Solution:
Question: 12
45% of the students in a school are boys. If the total number of students in the school is 880, find the number of girls in the school.
Solution:
Question: 13
Mr. Sidhana saves 28% of his income. If he saves As 840 per month, find his monthly income.
Solution:
Question: 14
In an examination, 8% of the students fail. What percentage of the students pass? If 1650 students appeared in the examination, how many passed?
Solution:
Question: 15
In an examination, 92% of the candidates passed and 46 failed. How many candidates appeared?
Take three non-collinear points A. B and C on a page of your notebook. Join AB, BC and CA. What figure do you get? Name the triangle. Also, name
Therefore in ∆ABC, we have
