Question 1.
Represent √3 on the number line.
Solution:


On the number line, take OA = 1 unit. Draw AB = 1 unit perpendicular to OA. Join OB.
Again, on OB, draw BC = 1 unit perpendicular to OB. Join OC.
By Pythagoras Theorem, we obtain OC = √3. Using
compasses, with centre O and radius OC, draw an arc, which intersects the number line at point
D. Thus, OD = √3 and D corresponds to √3.

Question 2.
Represent √3.2 on the number line.
Solution:
First of all draw a line of length 3.2 units such that AB = 3.2 units. Now, from point B, mark a distance of 1 unit. Let this point be ‘C’. Let ‘O’ be the mid-point of the distance AC. Now, draw a semicircle with centre ‘O’ and radius OC. Let us draw a line perpendicular to AC passing through the point ‘B’ and intersecting the semicircle at point ‘D’.
∴ The distance BD = √3.2

Now, to represent √3.2 on the number line. Let us take the line BC as number line and point ‘B’ as zero, point ‘C’ as ‘1’ and so on. Draw an arc with centre B and radius BD, which intersects the number line at point ‘E’.
Then, the point ‘E’ represents √3.2.

Question 3.
Express 1.32 + 0.35 as a fraction in the simplest form.
Solution:
Let . x = 1.32 = 1.3222…..(i)

Multiplying eq. (i) by 10, we have
10x = 13.222…
Again, multiplying eq. (i) by 100, we have
100x = 132.222… …(iii)
Subtracting eq. (ii) from (iii), we have
100x – 10x = (132.222…) – (13.222…)
90x = 119
⇒ x = 11990
Again, y = 0.35 = 0.353535……
Multiply (iv) by 100, we have …(iv)
100y = 35.353535… (v)
Subtracting (iv) from (u), we have
100y – y = (35.353535…) – (0.353535…)
99y = 35
y = 3599

Question 4.
Find the square root of 10 + √24 + √60 + √40.
Solution:

Question 5.
If x = 9 + 4√5, find the value of √x – 1x√.
Solution:
Here,
x = 9 + 4√5
x = 5 + 4 + 2 x 2√5
x = (√5² + (2² + 2 x 2x √5).
x = (√5 + 2)²
√x = √5 + 2
Now, 1x√ = 15√+2

Question 6.
If x = 15√−2 , find the value of x³ – 3² – 5x + 3
Solution:

∴ x – 2 = √5
Squaring both sides, we have
x² – 4x + 4 = 5
x² – 4x – 1 = 0 …(i)
Now, x³ – 3² – 5x + 3 = (x² – 4x – 1) (x + 1) + 4
= 0 (x + 1) + 4 = 4 [using (i)]

Question 7.
Find ‘x’, if 2^x-7 × 5^x-4 = 1250.
Solution:
We have 2^x-7 × 5^x-4 = 1250
⇒ 2^x-7 × 5^x-4 = 2 5 × 5 × 5 × 5
⇒ 2^x-7 × 5^x-4 = 21 × 54
Equating the powers of 2 and 5 from both sides, we have
⇒ x – 7 = 1 and x – 4 = 4
⇒ x = 8 and x = 8
Hence, x = 8 is the required value.

Question 8.
Evaluate:

Solution:

`Question 9.
If x = p+q√+p−q√p+q√−p−q√, then prove that q² – 2px + 9 = 0.
Solution:


Squaring both sides, we have
⇒ q²x² + p² – 2pqx = p² – q²
⇒ q²x² – 2pqx + q² = 0
⇒ q(q² – 2px + q) = 0
⇒ qx² – 2px + q = 0 (∵ q ≠ 0)

Question 10.
If a = 13−11√ and b = 1a, then find a² – b²
Solution:

Important Link

Quick Revision Notes : Number Systems

NCERT Solution : Number Systems

MCQs: Number Systems

Click here for Free Video Lectures

Question 1.
In the given figure, AB || CD, ∠FAE = 90°, ∠AFE = 40°, find ∠ECD.

Solution:
In AFAE,
ext. ∠FEB = ∠A + F
= 90° + 40° = 130°
Since AB || CD
∴ ∠ECD = FEB = 130°
Hence, ∠ECD = 130°.

Question 2.
In the fig., AD and CE are the angle bisectors of ∠A and ∠C respectively. If ∠ABC = 90°, then find ∠AOC.

Solution:
∵ AD and CE are the bisector of ∠A and ∠C

In ∆AOC,
∠AOC + ∠OAC + ∠OCA = 180°
⇒ ∠AOC + 45o = 180°
⇒ ∠AOC = 180° – 45° = 135°.

Question 3.
In the given figure, prove that m || n.

Solution:
In ∆BCD,
ext. ∠BDM = ∠C + ∠B
= 38° + 25° = 63°
Now, ∠LAD = ∠MDB = 63°
But, these are corresponding angles. Hence,
m || n

Question 4.
In the given figure, two straight lines PQ and RS intersect each other at O. If ∠POT = 75°, find the values of a, b, c.

Solution:
Here, 4b + 75° + b = 180° [a straight angle]
5b = 180° – 75° = 105°
b – 105∘5 = 21°
∴ a = 4b = 4 × 21° = 84° (vertically opp. ∠s]
Again, 2c + a = 180° [a linear pair]
⇒ 2c + 84° = 180°
⇒ 2c = 96°
⇒ c = 96∘2 = 48°
Hence, the values of a, b and c are a = 84°, b = 21° and c = 48°.

Question 5.
In figure, if AB || CD. If ∠ABR = 45° and ∠ROD = 105°, then find ∠ODC.

Solution:

Through O, draw a line ‘l’ parallel to AB.
⇒ line I will also parallel to CD, then
∠1 = 45°[alternate int. angles]
∠1 + ∠2 + 105° = 180° [straight angle]
∠2 = 180° – 105° – 45°
⇒ ∠2 = 30°
Now, ∠ODC = ∠2 [alternate int. angles]
= ∠ODC = 30°

Question 6.
In the figure, ∠X = 72°, ∠XZY = 46°. If YO and ZO are bisectors of ∠XYZ and ∠XZY respectively of ∆XYZ, find ∠OYZ and ∠YOZ.

Solution:
In ∆XYZ, we have
∠X + XY + ∠Z = 180°
⇒ ∠Y + ∠Z = 180° – ∠X
⇒ ∠Y + ∠Z = 180° – 72°
⇒ Y + ∠Z = 108°
⇒ 12 ∠Y + 12∠Z = 12 × 108°
∠OYZ + ∠OZY = 54°
[∵ YO and ZO are the bisector of ∠XYZ and ∠XZY]
⇒ ∠OYZ + 12 × 46° = 54°
∠OYZ + 23° = 54°
⇒ ∠OYZ = 549 – 23° = 31°
In ∆YOZ, we have
∠YOZ = 180° – (∠OYZ + ∠OZY)
= 180° – (31° + 23°) 180° – 54° = 126°

Question 7.
In the given figure, m and n are two plane mirrors perpendicular to each other. Show that incident rays CA is parallel to reflected ray BD.

Solution:
Let normals at A and B meet at P.
As mirrors are perpendicular to each other, therefore, BP || OA and AP || OB.
So, BP ⊥ PA i.e., ∠BPA = 90°
Therefore, ∠3 + ∠2 = 90° [angle sum property] …(i)
Also, ∠1 = ∠2 and ∠4 = ∠3 [Angle of incidence = Angle of reflection]
Therefore, ∠1 + ∠4 = 90° [from (i)) …(ii]
Adding (i) and (ii), we have
∠1 + ∠2 + ∠3 + ∠4 = 180°
i.e., ∠CAB + ∠DBA = 180°
Hence, CA || BD

Question 8
If in ∆ABC, the bisectors of ∠B and ∠C intersect each other at O. Prove that ∠BOC = 90° + 12∠ A.

Solution:
Let ∠B = 2x and ∠C = 2y
∵OB and OC bisect ∠B and ∠C respectively.
∠OBC = 12∠B = 12 × 2x = x
and ∠OCB = 12∠C = 12 × 2y = y
Now, in ∆BOC, we have
∠BOC + ∠OBC + ∠OCB = 180°
⇒ ∠BOC + x + y = 180°
⇒ ∠BOC = 180° – (x + y)
Now, in ∆ABC, we have
∠A + 2B + C = 180°
⇒ ∠A + 2x + 2y = 180°
⇒ 2(x + y) = 12(180° – ∠A)
⇒ x + y = 90° – 12∠A …..(ii)
From (i) and (ii), we have
∠BOC = 180° – (90° – 12∠A) = 90° + 12 ∠A

Question 9.
In figure, if I || m and ∠1 = (2x + y)°, ∠4 = (x + 2y)° and ∠6 = (3y + 20)°. Find ∠7 and ∠8.

Solution:
Here, ∠1 and ∠4 are forming a linear pair
∠1 + ∠4 = 180°
(2x + y)° + (x + 2y)° = 180°
3(x + y)° = 180°
x + y = 60
Since I || m and n is a transversal
∠4 = ∠6
(x + 2y)° = (3y + 20)°
x – y = 20
Adding (i) and (ii), we have
2x = 80 = x = 40
From (i), we have
40 + y = 60 ⇒ y = 20
Now, ∠1 = (2 x 40 + 20)° = 100°
∠4 = (40 + 2 x 20)° = 80°
∠8 = ∠4 = 80° [corresponding ∠s]
∠1 = ∠3 = 100° [vertically opp. ∠s]
∠7 = ∠3 = 100° [corresponding ∠s]
Hence, ∠7 = 100° and ∠8 = 80°

Question 10.
In the given figure, if PQ ⊥ PS, PQ || SR, ∠SQR = 28o and ∠QRT = 65°. Find the values of x, y and z.

Solution:
Here, PQ || SR .
⇒ ∠PQR = ∠QRT
⇒ x + 28° = 65°
⇒ x = 65° – 28° = 37°
Now, in it. ∆SPQ, ∠P = 90°
∴ ∠P + x + y = 180° [angle sum property]
∴ 90° + 37° + y = 180°
⇒ y = 180° – 90° – 37° = 53°
Now, ∠SRQ + ∠QRT = 180° [linear pair]
z + 65° = 180°
z = 180° – 65° = 115°

Important Link

Quick Revision Notes : Lines and Angles

NCERT Solution : Lines and Angles

MCQs :Lines and Angles

Click here for Free Video Lectures

Q.1  In a rectangle, one diagonal is inclined to one of its sides at 25°. Measure the acute angle between the two diagonals.

Solution:

Let ABCD be a rectangle where AC and BD are the two diagonals which are intersecting at point O.

Now, assume ∠BDC = 25° (given)

Now, ∠BDA = 90° – 25° = 65°

Also, ∠DAC = ∠BDA, (as diagonals of a rectangle divide the rectangle into two congruent right triangles)

So, ∠BOA = the acute angle between the two diagonals = 180° – 65° – 65° = 50°

Q.2. Is it possible to draw a quadrilateral whose all angles are obtuse angles?

Solution:

It is known that the sum of angles of a quadrilateral is always 360°. To have all angles as obtuse, the angles of the quadrilateral will be greater than 360°. So, it is not possible to draw a quadrilateral whose all angles are obtuse angles.

Q.2 Prove that the angle bisectors of a parallelogram form a rectangle. 

Solution:

LMNO is a parallelogram in which bisectors of the angles L, M, N, and O intersect at P, Q, R and S to form the quadrilateral PQRS.
LM || NO (opposite sides of parallelogram LMNO)
L + M = 180 (sum of consecutive interior angles is 180o)
MLS + LMS = 90
In LMS, MLS + LMS + LSM = 180
90 + LSM = 180
LSM = 90
RSP = 90 (vertically opposite angles)
SRQ = 90, RQP = 90 and SPQ = 90
Therefore, PQRS is a rectangle.

Q3. In a trapezium ABCD, AB∥CD. Calculate ∠C and ∠D if ∠A = 55° and ∠B = 70°

Solution:

In a trapezium ABCD, ∠A + ∠D = 180° and ∠B + ∠C = 180°

So, 55° + ∠D = 180°

Or, ∠D = 125°

Similarly,

70° + ∠C = 180°

Or, ∠C = 110°

Q4. Calculate all the angles of a parallelogram if one of its angles is twice its adjacent angle.

Solution:

Let the angle of the parallelogram given in the question statement be “x”.

Now, its adjacent angle will be 2x.

It is known that the opposite angles of a parallelogram are equal.

So, all the angles of a parallelogram will be x, 2x, x, and 2x

As the sum of interior angles of a parallelogram = 360°,

x + 2x + x + 2x = 360°

Or, x = 60°

Thus, all the angles will be 60°, 120°, 60°, and 120°.

Q5. Calculate all the angles of a quadrilateral if they are in the ratio 2:5:4:1.

Solution:

As the angles are in the ratio 2:5:4:1, they can be written as-

2x, 5x, 4x, and x

Now, as the sum of the angles of a quadrilateral is 360°,

2x + 5x + 4x + x = 360°

Or, x = 30°

Now, all the angles will be,

2x =2 × 30° = 60°

5x = 5 × 30° = 150°

4x = 4 × 30° = 120°, and

x = 30°

Q 6 Prove that a diagonal of a parallelogram divide it into two congruent triangles. [CBSE March 2012]
Solution. Given : A parallelogram ABCD and AC is its diagonal.

Question. 7 ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see fig.). Show that :
(i) AAPB ≅ ACQD (ii) AP = CQ            [CBSE March 2012]
Solution.

Q.8

Solution.

Q.9

Solution.

Q.10

Solution.

Important Link

Quick Revision Notes :Quadrilaterals

NCERT Solution : Quadrilaterals

MCQs: Quadrilaterals

Click here for Free Video Lectures

Question.1 In figure, TR ⊥ PS, PQ // TR and PS // QR. If QR = 8 cm, PQ = 3 cm and SP = 12 cm, find arquad. PQRS).[CBSE-14-17DIG1U]
Solution.

Question.2
]
Solution.

Question.3

Solution.

Question.4

Solution.

Question.5

Solution.

Question.6

Solution.

Question. 7 Prove that parallelogram on equal bases and between the same parallels are equal in area. [CBSE March 2012]
Solution.

Question.8

Solution.

Question.9

Solution.

Question.10

Solution.

Question. 21

Important Link

Quick Revision Notes : Areas of Parallelograms and Triangles

NCERT Solution : Areas of Parallelograms and Triangles

MCQs: Areas of Parallelograms and Triangles

Click here for Free Video Lectures

Question 1.
In the figure, O is the centre of a circle passing through points A, B, C and D and ∠ADC = 120°. Find the value of x.

Solution:
Since ABCD is a cyclic quadrilateral
∠ADC + ∠ABC = 180°
[∴ opp. ∠s of a cyclic quad. are supplementary]
120° + ∠ABC = 180°
∠ABC = 180° – 120° = 60°
Now, ∠ACB = 90° [angle in a semicircle]
In rt. ∠ed ∆CB, ∠ACB = 90°
∠CAB + ∠ABC = 90°
x + 60° = 90°
x = 90° -60°
x = 30°

Question 2.
In the given figure, O is the centre of the circle, ∠AOB = 60° and CDB = 90°. Find ∠OBC.

Solution:
Since angle subtended at the centre by an arc is double the angle
subtended at the remaining part of the circle.
∴ ∠ACB = 13 ∠AOB = 13 x 60° = 30°
Now, in ACBD, by using angle sum property, we have
∠CBD + ∠BDC + ∠DCB = 180°
∠CBO + 90° + ∠ACB = 180°
[∵ ∠CBO = ∠CBD and ∠ACB = ∠DCB are the same ∠s]
∠CBO + 90° + 30° = 180°
∠CBO = 180o – 90° – 30° = 60°
or ∠OBC = 60°

Question 3.
In the given figure, O is the centre of the circle with chords AP and BP being produced to R and Q respectively. If ∠QPR = 35°, find the measure of ∠AOB.

Solution:
∠APB = ∠RPQ = 35° [vert. opp. ∠s]
Now, ∠AOB and ∠APB are angles subtended by an arc AB at centre and at the remaining part of the circle.
∴ ∠AOB = 2∠APB = 2 × 35° = 70°

Question 4.
In the figure, PQRS is a cyclic quadrilateral. Find the value of x.

Solution:
In ∆PRS, by using angle sum property, we have
∠PSR + ∠SRP + ∠RPS = 180°
∠PSR + 50° + 35o = 180°
∠PSR = 180° – 85o = 95°
Since PQRS is a cyclic quadrilateral
∴ ∠PSR + ∠PQR = 180°
[∵ opp. ∠s of a cyclic quad. are supplementary]
95° + x = 180°
x = 180° – 95°
x = 85°

Question 5.
In the given figure, ∠ACP = 40° and BPD = 120°, then find ∠CBD.

Solution:
∠BDP = ∠ACP = 40° [angle in same segment]
Now, in ∆BPD, we have
∠PBD + ∠BPD + ∠BDP = 180°
⇒ ∠PBD + 120° + 40° = 180°
⇒ ∠PBD = 180° – 160o = 20°
or ∠CBD = 20°

Question 6.
In the given figure, if ∠BEC = 120°, ∠DCE = 25°, then find ∠BAC.

Solution:
∠BEC is exterior angle of ∆CDE.
∴ ∠CDE + ∠DCE = ∠BEC
⇒ ∠CDE + 25° = 120°
⇒ ∠CDE = 95°
Now, ∠BAC = ∠CDE [∵ angle in same segment are equal]
⇒ ∠BAC = 95°

Question 7.
In the given figure, PQR = 100°, where P, Q and R are points on a circle with centre O. Find LOPR.

Solution:
Take any point A on the circumcircle of the circle.
Join AP and AR.
∵ APQR is a cyclic quadrilateral.
∴ ∠PAR + ∠PQR = 180° [sum of opposite angles of a cyclic quad. is 180°]
∠PAR + 100° = 180°
⇒ Since ∠POR and ∠PAR are the angles subtended by an arc PR at the centre of the circle and circumcircle of the circle.
∠POR = 2∠PAR = 2 x 80° = 160°
∴ In APOR, we have OP = OR [radii of same circle]
∠OPR = ∠ORP [angles opposite to equal sides]
Now, ∠POR + ∠OPR + ∠ORP = 180°
⇒ 160° + ∠OPR + ∠OPR = 180°
⇒ 2∠OPR = 20°
⇒ ∠OPR = 10°

Question 8.
In figure, ABCD is a cyclic quadrilateral in which AB is extended to F and BE || DC. If ∠FBE = 20° and DAB = 95°, then find ∠ADC.

Solution:
Sum of opposite angles of a cyclic quadrilateral is 180°
∴ ∠DAB + ∠BCD = 180°
⇒ 95° + ∠BCD = 180°
⇒ ∠BCD = 180° – 95° = 85°
∵ BE || DC
∴ ∠CBE = ∠BCD = 85°[alternate interior angles]
∴ ∠CBF = CBE + ∠FBE = 85° + 20° = 105°
Now, ∠ABC + 2CBF = 180° [linear pair]
and ∠ABC + ∠ADC = 180° [opposite angles of cyclic quad.]
Thus, ∠ABC + ∠ADC = ∠ABC + 2CBF
⇒ ∠ADC = CBF
⇒ ∠ADC = 105° [∵ CBF = 105°]

Question 9
Equal chords of a circle subtends equal angles at the centre.

Solution:
Given : In a circle C(O, r), chord AB = chord CD
To Prove : ∠AOB = ∠COD.
Proof : In ∆AOB and ∆COD
AO = CO (radii of same circle]
BO = DO [radii of same circle]
Chord AB = Chord CD (given]
⇒ ∆AOB = ACOD [by SSS congruence axiom]
⇒ ∠AOB = COD (c.p.c.t.]

Question 10.
In the given figure, P is the centre of the circle. Prove that : ∠XPZ = 2(∠X∠Y + ∠YXZ).

Solution:
Arc XY subtends ∠XPY at the centre P and ∠XZY in the remaining part of the circle.
∴ ∠XPY = 2 (∠X∠Y)
Similarly, arc YZ subtends ∠YPZ at the centre P and ∠YXZ in the remaining part of the circle.
∴ ∠YPZ = 2(∠YXZ) ….(ii)
Adding (i) and (ii), we have
∠XPY + ∠YPZ = 2 (∠XZY + ∠YXZ)
∠XP2 = 2 (∠XZY + ∠YXZ)

Important Link

Quick Revision Notes : Circles

NCERT Solution : Circles

MCQs: Circles

Click here for Free Video Lectures

1. Construct an angle of 90° at the initial point of the given ray. [CBSE-15-6DWMW5A]
Answer.

2. Draw a line segment PQ = 8.4 cm. Divide PQ into four equal parts using ruler and compass. [CBSE-14-ERFKZ8H], [CBSE – 14-GDQNI3W], [CBSE-14-17DIG1U]
Answer. Steps of construction :

1. Draw a line segment PQ = 8.4 cm.
2. With P and Q as centres, draw arcs of radius little more than half of PQ. Let his line intersects PQ in M.
3. With M and Q as centres, draw arcs of radius little more than half of MQ. Let this line intersects PQ in N.
4. With P and M as centres, draw arcs of radius little more than half of PM. Let this line intersects PQ in L. Thus, L, M and N divide the line segment PQ in four equal parts.
   

3. Draw any reflex angle. Bisect it using compass. Name the angles so obtained. [CBSE-15-NS72LP7]
Answer.

4. Why we cannot construct a ΔABC, if ∠A=60°, AB — 6 cm, AC + BC = 5 cm but construction of A ABC is possible if ∠A=60°, AB = 6 cm and AC – BC = 5 cm. [CBSE-14-GDQNI3W]
Answer. We know that, by triangle inequality property, construction of triangle is possible if sum of two sides of a triangle is greater than the third side.
Here, AC + BC = 5 cm which is less than AB ( 6 cm)
Thus, ΔABC is not possible.
Also, by triangle inequality property, construction of triangle is possible, if difference of two
sides of a triangle is less than the third side
Here, AC – BC = 5 cm, which is less than AB (6 cm)
Thus, ΔABC is possible.

5. Construct angle of [5212]0 using compass only. [CBSE-14-17DIG1U]
Answer.

SHORT ANSWER QUESTIONS TYPE-I
6. Using ruler and compass, construct 4∠XYZ, if ∠XYZ= 20° [CBSE-14-ERFKZ8H]
Answer.

7. Construct an equilateral triangle LMN, one of whose side is 5 cm. Bisect ∠ M of the triangle. [CBSE March 2012]
Answer. Steps of construction :

1.  Draw a line segment LM = 5 cm.
2. Taking L as centre and radius 5 cm draw an arc.
3. Taking M as centre and radius draw an other arc intersecting previous arc at N.
4. Join LN and MN. Thus, ΔLMN is the required equilateral triangle.
5. Taking M as centre and any suitable radius, draw an arc intersecting LM at P and MN at Q.
6. Taking P and Q as centres and same radii, draw arcs intersecting at S.
7. Join MS and produce it meet LN at R. Thus, MSR is the required bisector of ∠M.
   

8. Construct a A ABC with BC = 8 cm, ∠B= 45° and AB – AC = 3.1 cm. [CBSE-15-NS72LP7]
Answer.

9. Construct an isosceles triangle whose two equal sides measure 6 cm each and whose base is 5 cm. Draw the perpendicular bisector of its base and show that it passes through the opposite vertex [CBSE-15-6DWMW5A]
Answer. Steps of construction :

1. Draw a line segment AB = 5 cm.
2. With A and B as centres, draw two arcs of radius 6 cm and let they intersect each other in C.
3.  Join AC and BC to get ΔABC.
4. With A and B as centres, draw two arcs of radius little more than half of AB. Let they intersect each other in P and Q. Join PQ and produce, to pass through C.
   

10. Construct a right triangle whose base is 8 cm and sum of the hypotenuse and other side is 16 cm.
Answer. Given : In ΔABC, BC = 8 cm, ∠B= 90° and AB + AC = 16 cm.
Required : To construct ΔABC.
Steps of construction:

1. Draw a line segment BC = 8 cm.
2. At B, Draw ∠CBX = 90°.
3. From ray BX, cut off BE = 16 cm.
4.  Join CE .
5. Draw the perpendicular bisector of EC meeting BE at A.
6. Join AC to obtain the required ΔABC.
   

Important Link

Quick Revision Notes : Constructions

NCERT Solution : Constructions

MCQs: Constructions

Click here for Free Video Lectures

Q.1 Find the Area of a Triangle whose two sides are 18 cm and 10 cm respectively and the perimeter is 42cm.

Solution:

Let us consider the third side of the triangle to be “c”.

Now, the three sides of the triangle are a = 18 cm, b = 10 cm, and “c” cm 

It is given that the perimeter of the triangle = 42cm

So, 

18 + 10 + c = 42

c = 42 – (18 + 10) cm = 14 cm

∴ The semi perimeter of triangle (s) = 42/2 = 21 cm

Using Heron’s formula,

Area of the triangle A = s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−√s(s−a)(s−b)(s−c)

= 21(21–18)(21–10)(21–14)−−−−−−−−−−−−−−−−−−−−−√21(21–18)(21–10)(21–14)

=  21×3×11×7−−−−−−−−−−−−−√21×3×11×7

= 21×21×11−−−−−−−−−−√21×21×11

= 2111−−√2111cm²

Q2: Sides of a Triangle are in the ratio of 14 : 20: 25 and its perimeter is 590cm. Find its area.

Solution:

The ratio of the sides of the triangle is given as 14: 20: 25

Let us consider the common ratio between the sides of the triangle be “a”

∴ The sides are 14a, 20a and 25a

It is also given that the perimeter of the triangle = 590 cm

12a + 17a + 25a = 590 

=> 59a = 590

So, a = 10

Now, the sides of the triangle are 140 cm, 200 cm, 250 cm.

So, the semi perimeter of the triangle (s) = 590/2 = 295 cm

Using Heron’s formula for Area of the triangle

= s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−√s(s−a)(s−b)(s−c)

= 295(295−140)(295−200)(295−250)−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√295(295−140)(295−200)(295−250)

= 295×155×95×45−−−−−−−−−−−−−−−−√295×155×95×45

= 195,474,375−−−−−−−−−−√195,474,375

= 13981.21cm²

Q3: A field is in the Shape of a Trapezium whose parallel sides are 22 m and 10 m. The non-parallel sides are given as 13 m and 14 m. Find the area of the field.

Solution:

Draw a line segment BE line AD. Then, draw a perpendicular on the line segment CD from point B

ImagewillbeuploadedsoonImagewillbeuploadedsoon

Now, it can be seen that the quadrilateral ABED is a parallelogram. So,

AB = DE = 10 m

AD = BE = 13 m

EC = DC– ED 

= 22 – 10 = 12 m

Now, consider the triangle BEC,

Its semi perimeter (s) = (13+ 14 + 12)/2 

= 39/2 m

=19.5m

By using Heron’s formula,

Area of ΔBEC =

= s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−√s(s−a)(s−b)(s−c)

= 19.5(19.5−13)(19.5−14)(19.5−15)−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√19.5(19.5−13)(19.5−14)(19.5−15)

= 19.5×6.5×5.5×4.5−−−−−−−−−−−−−−−−−√19.5×6.5×5.5×4.5

= 3137.06−−−−−−√3137.06

= 56m²

We also know that the area of ΔBEC = (½) × EC × BF

56 cm² = (½) × 12 × BF

BF = 56 x 2 /12cm 

     = 9.3 cm

So, the total area of ABED will be BF × DE i.e. 9.3 × 10 = 93 m²

∴ Area of the field = 93 + 56 = 149 m²

Q4: Find the Area of the Triangular field of sides 55 m, 60 m, and 65 m. Find the cost of laying the grass in the triangular field at the rate of Rs 8 per m².

Solution:

Given that 

Sides of the triangular field are 50 m, 60 m and 65 m.

Cost of laying grass in a triangular field = Rs 8 per m²

Let a = 55, b = 60, c = 65

Semi- Perimeter s = (a + b + c)/2

⇒ s = (55 + 60 + 65)/2

= 180/2

= 90.

By using Heron’s formula,

Area of ΔBEC =

= s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−√s(s−a)(s−b)(s−c)

= 90(90−55)(90−60)(90−65)−−−−−−−−−−−−−−−−−−−−−−−−√90(90−55)(90−60)(90−65)

= 90×35×30×25−−−−−−−−−−−−−−√90×35×30×25

=2362500−−−−−−−√2362500

= 1537m²

Cost of laying grass = Area of triangle × Cost of laying grass per m²

= 1537×8

= Rs.12296

Q5: The Perimeter of an Isosceles triangle is 42 cm. The ratio of the equal side to its base is 3: 4. Find the area of the triangle.

Solution:

Given that,

The perimeter of the isosceles triangle = 42 cm

It is also given that,

Ratio of equal side to base = 3 : 4

Let the equal side = 3x

So, base = 4x

Perimeter of the triangle = 42

⇒ 3x + 3x + 4x = 42

⇒ 10x = 42

⇒ x = 4.2

Equal side = 3x = 3×4.2 = 12.6

Base = 4x = 4×4.2 = 16.8

The sides of the triangle = 12.6cm, 12.6cm and 16.8cm.

Let a = 12.6, b = 12.6, c = 16.8

s = (a + b + c)/2

⇒ s = (12.6 + 12.6 + 16.8)/2

= 42/2

= 21.

By using Heron’s formula,

Area of ΔBEC =

= s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−√s(s−a)(s−b)(s−c)

= 21(21−12.6)(21−12.6)(21−16.8)−−−−−−−−−−−−−−−−−−−−−−−−−−−−√21(21−12.6)(21−12.6)(21−16.8)

=21×8.4×8.4×4.2−−−−−−−−−−−−−−−−√21×8.4×8.4×4.2

= 6223.39−−−−−−√6223.39

= 78.88cm²

Q.6 Find the area of a triangle whose two sides are 18 cm and 10 cm and the perimeter is 42cm.

Solution:

Assume that the third side of the triangle to be “x”.

Now, the three sides of the triangle are 18 cm, 10 cm, and “x” cm

It is given that the perimeter of the triangle = 42cm

So, x = 42 – (18 + 10) cm = 14 cm

∴ The semi perimeter of triangle = 42/2 = 21 cm

Using Heron’s formula,

Area of the triangle,

= √[s (s-a) (s-b) (s-c)]

= √[21(21 – 18) (21 – 10) (21 – 14)] cm²

= √[21 × 3 × 11 × 7] m²

= 21√11 cm²

Q.7: The sides of a triangle are in the ratio of 12: 17: 25 and its perimeter is 540cm. Find its area.

Solution:

The ratio of the sides of the triangle is given as 12: 17: 25

Now, let the common ratio between the sides of the triangle be “x”

∴ The sides are 12x, 17x and 25x

It is also given that the perimeter of the triangle = 540 cm

12x + 17x + 25x = 540 cm

=> 54x = 540cm

So, x = 10

Now, the sides of the triangle are 120 cm, 170 cm, 250 cm.

So, the semi perimeter of the triangle (s) = 540/2 = 270 cm

Using Heron’s formula,

Area of the triangle

= 9000 cm²

Q.8: A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.

Solution:

First, draw a line segment BE parallel to the line AD. Then, from B, draw a perpendicular on the line segment CD.

Now, it can be seen that the quadrilateral ABED is a parallelogram. So,

AB = ED = 10 m

AD = BE = 13 m

EC = 25 – ED = 25 – 10 = 15 m

Now, consider the triangle BEC,

Its semi perimeter (s) = (13+ 14 + 15)/2 = 21 m

By using Heron’s formula,

Area of ΔBEC =

= 84 m²

We also know that the area of ΔBEC = (½) × CE × BF

84 cm² = (½) × 15 × BF

=> BF = (168/15) cm = 11.2 cm

So, the total area of ABED will be BF × DE, i.e. 11.2 × 10 = 112 m²

∴ Area of the field = 84 + 112 = 196 m²

Q.9: A rhombus-shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?

Solution:

Draw a rhombus-shaped field first with the vertices as ABCD. The diagonal AC divides the rhombus into two congruent triangles which are having equal areas. The diagram is as follows.

Consider the triangle BCD,

Its semi-perimeter = (48 + 30 + 30)/2 m = 54 m

Using Heron’s formula,

Area of the ΔBCD =

= 432 m²

∴ Area of field = 2 × area of the ΔBCD = (2 × 432) m² = 864 m²

Thus, the area of the grass field that each cow will be getting = (864/18) m² = 48 m²

Q.10: Find the cost of laying grass in a triangular field of sides 50 m, 65 m and 65 m at the rate of Rs 7 per m².

Solution:

According to the question,

Sides of the triangular field are 50 m, 65 m and 65 m.

Cost of laying grass in a triangular field = Rs 7 per m2

Let a = 50, b = 65, c = 65

s = (a + b + c)/2

⇒ s = (50 + 65 + 65)/2

= 180/2

= 90.

Area of triangle = √(s(s-a)(s-b)(s-c))

= √(90(90-50)(90-65)(90-65))

= √(90×40×25×25)

= 1500m²

Cost of laying grass = Area of triangle ×Cost per m²

= 1500×7

= Rs.10500

Important Link

Quick Revision Notes : Heron’s Formula

NCERT Solution : Heron’s Formula

MCQs: Heron’s Formula

Click here for Free Video Lectures

Question.1

If a wooden box of dimensions 8 m x 7 m x 6 m is to carry boxes of dimensions 8 cm x 7 cm x 6 cm, then find the maximum number of boxes that can be carried in the wooden box. [CBSE-14-GDQNI3W]
Answer.

Question 2

Question3

. Calculate the edge of the cube if its volume is 1331 cm³. [CBSE-15-6DWMW5A]
Answer.

Question.4

If in a cylinder, radius is doubled and height is halved, then find its curved surface area.
Answer.

Question 5. The radii of two cylinders of the same height are in the ratio 4 :5, then find the ratio of their volumes.
Answer.

Question 6. Find the area of the sheet required to make closed cylindrical vessel of height 1 m and diameter 140 cm.
Answer.

Question 9. Find the volume of cone of radius r/2 and height ‘2h’.
Answer.

10. How many balls, each of radius 2 cm can be made from a solid sphere of lead of radius 8 cm ? [CBSE-14-17DIG1U]
Answer.

Important Link

Quick Revision Notes : Surface Areas and Volumes

NCERT Solution : Surface Areas and Volumes

MCQs: Surface Areas and Volumes

Click here for Free Video Lectures

Question.1

Question.2
(a) Identify the organism shown in the picture and write the common name and scientific name of the organism.
(b) Name its phylum and kingdom.
(c) Which organ of digestive system normally harbours this organism ?

Answer.
(a) It is Ascaris.Common name is Roundworm.Scientific name is Ascaris lumbricoides.
(b) Phylum — Nematoda and Kingdom-Animalia.
(c) Small intestine (an organ of digestive system) normally harbours this organism.

Question.3
(a) Identify three features possessed by all chordates.
(b) Name the classes of vertebrates which have the following characteristics:
(1) Animals that have streamlined body and gills for breathing.
(2) Animals that are found both on land and in water.
(3) Animals that have mammary glands for the production of milk to nourish their young ones.
Answer.
(a) Features of chordates:

• They have notochord.
• They have a dorsal hollow nerve chord.
• They are triploblastic.
• They have paired gill pouches.
• They are coelomate.

(b)

1. Pisces
2. Amphibians
3. Mammals.

Question.4

Question.5

How do poriferan animals differ from coelenterate animals ? [SAII -2011]
Answer.

Question.6

How do annelid animals differ from arthropods ? [SAII -2014]
Answer.

Question.7

What are the differences between amphibians and reptiles ? [SAII – 2011]
Answer.

Question.8 What are the differences between animals belonging to the Aves group and those in the Mammalia
group ? [SAII- 2011]
Answer.

Question.9 What are the advantages of classifying organisms ?
Answer. The importance of classification is :

1.  It provides us an information regarding the diversity of plants and animals.
2.  It provides insight into the origin of organism and interrelationship between them.
3. It makes the study of wide variety of organisms.
4.  It helps in understanding the evolution of organisms. –
5. Various fields of applied biology like agriculture, environmental biology etc., also depends t upon correct identification and classification of pest, disease, vector etc.
6. It serves as a base for the development of other biological sciences like biogeography and
   ecology.

The science of classification thus contributes to a larger extent in advancing knowledge in
most of the other disciplines of biology.

Question.10 What are the major divisions in the plantae ? What is the basis for these divisions ? [SAII – 2014]
Answer.
The major divisions in the kingdom plantae are :
(i) Thallophyta (ii) Bryophyta
(iii) Pteridophyta (iv) Gymnosperms
(v) Angiosperms.
The basis of these divisions are :
(a) Whether the plant body has well differentiated, distinct components.
(b) Whether the differentiated plant body has special tissues for the transport of water and other substances within it.
(c) The ability to bear seeds.
(d) Whether the seeds are enclosed within fruits.

Important Link

Quick Revision Notes : Diversity in Living Organisms

NCERT Solution : Diversity in Living Organisms

MCQs: Diversity in Living Organisms

Click here for Free Video Lectures

Question 1.
Describe the circumstances which were responsible for the Russian Revolution.
Answer:
The circumstances which were responsible for the Russian Revolution as given below :

• The Russian peasantry was in a miserable condition. The farmers could not get even two square meals a day. Their land holdings were very small and they had to pay heavy taxes.
• The Russian as well as the foreign capitalist industrialists exploited the workers by taking 12-14 hours of work and paying very low wages to them. The workers had no right to form trade unions or seek reforms. They led a miserable life.
• The Tsar Nicholas II was a despotic and autocratic ruler. He enjoyed unlimited powers and rights. The people of the higher strata enjoyed great privileges. The bureaucracy was corrupt and inefficient. The common people who suffered most, were fed up with the absolute rule of the Tsar and wanted to get rid of him.
• Karl Marx propagated ‘Scientific Socialism’. He strongly opposed capitalism which meant untold exploitation of the common men.

Question 2.
Explain in brief Lenin’s contribution to the Russian Revolution of 1917.
Answer:

• Lenin had played an important part in the Russian Revolution of 1917. It is true that after the fall of Tsar, Lenin led the revolutionaries. Really, it was the beginning of the revolution.
• The Provisional Government, under the leadership of Kerenskii, could not implement the demand of the people and failed.
• Under Lenin’s leadership, the Bolshevik Party put forward clear policies to end the war, transfer the land to the peasants and advance the slogan ‘All power to the Soviets’.
• He had described the Russian empire as a Prison of Nations and had declared that . no genuine democracy could be established unless all the non-Russian people were given equal rights.

Question 3.
What were the main objectives of the Russian Revolutionaries ?
Answer:
The main objectives of the Russian Revolutionaries were :

• The Tsar had thrown Russia into the First World War to fulfil his imperialistic desires. It was the demand of the revolutionaries that Russia should withdraw from the war. So, it withdrew from the First World War in 1917 after the Revolution.
• After the Revolution, the land was given to the tillers. The landlords had to give the land to the government. Kolkhoz and Sovkhoj farms were established. In Kolkhoz farms, the peasants worked collectively.
• The revolutionaries had demanded an improvement in the conditions of the industrial workers. They demanded better wages, good working conditions and removal of exploitation. After 1917, the industries were nationalised and the dream of workers was fulfilled.
• The next aim of the revolutionaries was that the non-Russians should be given equal status. Lenin believed that without this status these people could never become real Russians.

Question 4.
How Lenin’s name became inseparable from the Russian Revolution ?
Answer:
Lenin’s name became inseparable from the Russian Revolution :

• After completing his education, he joined the Communist Revolutionary Party and started spreading revolutionary ideas among the workers. He favoured the workers. He also favoured the setting up of the new society based on the principles of socialism of Karl Marx.
• He set up a Communist Government in place of the despotic rule in Russia. Therefore, Lenin’s name became inseparable from the Russian Revolution.
• Lenin united the peasants and workers under the Bolshevik Party and directed the revolution against the Provisional Government.
• Efforts were made to set up a Socialist Government on the basis of principles of Karl Marx. The private property was confiscated. Lenin took the land from the landlords and distributed it among the peasants. The Government nationalised all the factories and handed over their management to the workers. All debts were remitted. The property of the Church was also confiscated.

Question 5.
What was the impact of the Russian Revolution on Russia ?
Answer:
The impact of the Russian Revolution on Russia were :

• The Revolution put an end to autocratic monarchy in Russia. The Tsarist empire was transformed into a new state known as the Union of Soviet Socialist Republics or the Soviet Union.
• The most important result of the Bolshevik Revolution was the establishment of a Socialist Government in Russia. All the means of production were brought under state control. Banks, mines, factories, railways, telephones, etc. all were declared as government property and the property of the Church was nationalised. Work became an essential requirement for every person. The non-working person was not entitled to vote.
• The condition of the Russian mass had become miserable due to the First World War. The prime need of the Russian mass was food, not expansion.
• As a result of the Bolshevik Revolution, the government took all the means of production under its control and nationalised all small and big industries. Hence, within a few years Russia emerged as a powerful industrial state. With the growth of industrial and agricultural production, poverty started disappearing and the country moved on to the path of prosperity.

Question 6.
What was the global impact of the Russian Revolution ?
Answer:
The global impact of the Russian Revolution were :

• The Bolshevik Revolution helped in the spread of Socialist and Communist ideas all over the world. Communist Governments were established in many European countries.
• Most of the Bolshevik leaders believed that a series of revolutions will sweep other countries of the world along with revolution in Russia. Many non-Russians from outside the USSR participated in the conference of the people of the east and the Bolshevik-founded Comintern, an international union of Pro-Bolshevik socialist parties.
• The Bolshevik government ‘granted freedom to all its colonies immediately after coming to power. Thus, the new Soviet State came forward as a friend of the subjugated people and proved to be a source of great inspiration to the freedom movements of various Asian and African countries.
• By the end of the 20th century, the international reputation of the USSR as a socialist country had declined through it was recognised that socialist ideals still enjoyed respect among its people.

Question 7.
How did Russia’s participation in the World War cause the fall of the Tsar ?
Answer:
(a) The war was initially popular, and people rallied around Tsar Nicholas II.
(b) As the war continued, support became thin and Tsar’s popularity declined. Anti-German sentiments became high.
(c) The Tsarina Alexandra’s German origins and poor advisers, especially a monk called Rasputin, made the autocracy unpopular.
(d) Defeats were shocking and demoralising. Russia’s armies lost badly in Germany and Austria between 1914 and 1916. There were over 7 million casualties by 1917.
(e) The destruction of crops and buildings led to over 3 million refugees in Russia. The situation discredited the government and the Tsar. Soldiers did not wish to fight such a war.

Question 8.
Explain the main effects of the First World War on the industries in Russia.
Answer:
Effects of the First World War on the industries in Russia were :

• Russian industries were very few and the country was cut off from other suppliers of industrial goods by German control of the Baltic Sea.
• Industrial equipment disintegrated more rapidly in Russia than elsewhere in Europe.
• By 1916 railway lines began to break down. Able bodied men were called up to the war.
• As a result, there were labour shortages and small workshops producing essential commodities were shut down.
• Large supplies of grain were sent to feed the army. For the people in the cities, bread and flour became scarce. By the winter of 1916, riots at bread shops were common.

Question 9.
Differentiate between the ideas of the liberals and radicals in Europe.
Answer:
(a) The liberals did not believe in universal franchise. In contrast, radicals wanted a nation in which government was based on most of a country’s population.
(b) Liberals felt men of prosperity mainly should have the vote. They did not want the vote for women. On the other hand, the radicals supported women’s suffragette movements and opposed the privileges of great landowners and wealthy factory owners.
(c) Radicals were not against the existence of private property but disliked concentration of property in the hands of a few.

Question 10.
Which basic principles, ideas and values had the Russian Revolution for rest of the world ?
Answer:
The basic principles, ideas and values had the Russian Revolution for rest of the world :
(a) Economic equality
(b) Social Equality
(c) Socialism
(d) Anti-capitalism.
(e) International fraternity of all the peasants, craftsmen and workers.

Important Link

Quick Revision Notes :Socialism in Europe and the Russian Revolution

NCERT Solution : Socialism in Europe and the Russian Revolution

MCQs: Socialism in Europe and the Russian Revolution

Click here for Free Video Lectures