perimeter and area of plane figures, volume and …©edulabz interna tional math class viii 1...

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Math Class VIII 1 Question Bank

1. Find the area of the triangle whose sides are 13 cm, 20 cm and 21

cm. Also, find the altitude of the triangle corresponding to the largest

side.

Ans. Let a = 13, b = 20 cm and c = 21 cm.

Now, 13 20 21 54

cm cm 27 cm2 2 2

a b cs

+ + + += = = =

Thus, area of the triangle = ( ) ( ) ( )s s a s b s c− − −

227(27 13) (27 20) (27 21) cm= − − −

227 14 7 6 cm= × × ×

23 3 3 2 7 7 2 3 cm= × × × × × × ×

= 2 × 3 × 3 × 7 cm2 = 126 cm2

Here, largest side = 21 cm

Area of the triangle = 126 cm2

⇒ 1

1262

b h× × = ⇒ 126 2

cm21

h×

=

⇒ h = 12 cmHence, altitude of the triangle to the largest side = 12 cm.

2. Find the area of the triangle whose sides are 50 cm, 48 cm, and 14

cm. Find the height of the triangle corresponding to the side

measuring 48 cm.

Ans.Let a = 50 cm, b = 48 cm and c = 14 cm.

Now,50 48 14 112

cm cm 56 cm2 2 2

a b cs

+ + + += = = =

2

PERIMETER AREA OF PLANE

FIGURES, VOLUME AND

SURFACE AREA OF SOLIDS

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Thus, area of the triangle

( ) ( ) ( )s s a s b s c= − − −

256(56 50) (56 48) (56 14) cm= − − −

256 6 8 42 cm= × × ×

22 2 2 7 2 3 2 2 2 2 3 7 cm= × × × × × × × × × × ×

= 2 × 2 × 2 × 2 × 3 × 7 = cm2 = 336 cm2

Area of the triangle = 336 cm2

⇒ 1

3362

b h× × = ⇒ 1

48 3362

h× × =

⇒ 336 2

cm48

h×

= ⇒ h = 14 cm

Hence, the required height is 14 cm.

3. Find the area of a triangular field whose sides are 17 m, 19 m and

32 m. Find the altitude of the triangle corresponding to the small-

est side. (Given 15 3.873= )

Ans.Let a = 17 m, b = 19 m and c = 32 m

then17 19 32 68

m m 34 m2 2

+ += = =s

Thus, area of the triangle

( ) ( ) ( )s s a s b s c= − − −

234(34 17) (34 19) (34 34) m= − − − 234 17 15 2 m= × × ×

22 17 17 3 5 2 m= × × × × × 22 17 3 5 m= × ×

234 15 m= = 34 × 3.873 m2 = 131.68 m2

Here smallest side (b) = 17 m

Area of the triangle 131.68

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⇒ 1

131.682

b h× × = ⇒1

17 131.682

h× × =

⇒131.68 2

17

×=h ⇒ h = 15.49 m ⇒ h = 15.49 m

Hence, the required height is 15.49 m

4. Find the area of an isosceles triangle in which each of the equal

sides measures 30 cm and the third side is 48 cm long.

Ans.In ABC,∆

AB = AC = 30 cm, BC = 48 cm

AD BC⊥ is drawn which bisects BC at D

Then, BD = DC = 48

2 = 24 cm

In right ABC,∆ we have

AB2 = AD2 + BD2 ⇒ (30)2 = AD2 + (24)2

⇒ AD2 = (30)2 – (24)2 = 900 – 576 = 324 = (18)2 ∴ AD = 18 cm

Hence, area of triangle 1

ABC2

= × base × altitude

2148 18 432 cm

2= × × =

5. The base and the height of a triangle are in the ratio 5 : 3 and its

area is 43.2 m2. Find the base and the height of the triangle.

Ans.Let base of the triangle be 5x and height of the triangle be 3x

Area of the triangle = 43.2 m2 (given)

⇒1

43.22

b h× × = ⇒ 1

5 3 43.22

x x× × =

⇒ 21543.2

2x = ⇒ 2 43.2 2

15x

×=

⇒ x2 = 5.76 ⇒ 5.76x = ⇒ x = 2.4 m

Hence, base of the triangle = (5 × 2.4) m = 12 m

Height of the triangle = (3 × 2.4) m = 7.2 m.

B CD

48 cm

30 c

m 30 cm

A

24 cm 24 cm

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6. Two sides of a triangle are 6 cm and 8 cm. If height of the triangle

corresponding to 6 cm side is 4 cm ; find :

(a) area of the triangle.

(b) height of the triangle corresponding to 8 cm side.

Ans. BC = 6 cm

height AD = 4 cmarea of triangle ABC

= 1

2 base × height =

1

2 × BC × AD

= 1

2 × 6 cm × 4 cm = 12 cm2

Again, area of triangle

ABC = 1

2 × AC × BE

12 = 1

2 × 8 × BE

Thus 12 2

BE8

×=

BE = 3 cm

∴ (a) 12 cm2 (b) 3 cm.

7. Two sides of a triangle are 6.4 m and 4.8 m. If height of the tri-

angle corresponding to 4.8 m side is 6 m ; find :

(i) area of the triangle.

(ii) height of the triangle corresponding to 6.4 m side.

Ans. ABC is the triangle in which BC = 4.8 m

AC = 6.4 m and AD = 6 m

Area of triangle

ABC = 1

2 × BC × AD

= 1

2 × 4.8 × 6 = 14.4 m2

A

B DC

E

4 cm

8 cm

6 cm

A

DCB

E

6.4 m

4.8 m

6 m

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BE is height of triangle corresponding to 6.4 m

Thus1

2 AC × BE = 14.4,

1

2 × 6.4 × BE = 14.4

14.4 2BE

6.4

×=

14.4

3.2= = 4.5 m

∴ (i) 14.4 m2 (ii) 4.5 m.

8. The area of an equilateral triangle is 2144 3 cm ; find its perim-

eter.

Ans. Let each side of an equilateral triangle be x cm.

Area of equilateral triangle = 3

4 (side)2

23144 3

4x= = (given)

⇒ 2 4144 3

3x = × ⇒ x2 = 144 × 4

⇒ x2 = 576 ⇒ 576x = = 24 cm.

⇒ Each side = 24 cm

Hence, perimeter of equilateral triangle = 3 × side = 3(24) = 72 cm.

9. The area of an equilateral triangle is numerically equal to its

perimeter. Find its perimeter correct to 2 decimal places.

Ans. Let each side of the equilateral triangle be x

∴ Area of equilateral triangle = 23

4x

Perimeter of equilateral triangle = 3x

Area of equilateral triangle = Perimeter of equilateral triangle

233

4x x=

⇒ 2 43

3x x= ×

3 4 3 3 4 34 3

33 3

× × × ×= = =

×

x xx

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⇒ 2 3(4 )x x= ⇒ 4 3x =

Hence, perimeter 12 3= units

= 12 × (1.732) = 20.784 = 20.78 units.

10. A field is in the shape of a quadrilateral ABCD in which side

AB = 18 m, side AD = 24 m, side BC = 40 m, DC = 50 m and ∠A

= 90°. Find the area of the field.

Ans. Since A 90∠ = °

∴ By Pythagoras theorem,

In ABD,∆

BD2 = AB2 + AD2

2 2 2 2BD AB AD 18 24= + = +

324 576 900= + = = 30 m.

Area of 1

ABD (18) (24)2

∆ = × = (18) × (12) = 216 m2

In BCD,∆ sides are 30, 40, 50

⇒ By Pythagoras theorem 90CBD∠ = °

[∴ DC2 = BD2 + BC2, ∴ (50)2 = (30)2 + (40)2

Thus, area of 21BCD (40) (30) 600 m

2∆ = × × =

Hence, area of quadrilateral ABCD = Area of ABD∆ + area of

BCD∆ = 216 + 600 = 816 m2.

11. The lengths of the sides of a triangle are in the ratio 4 : 5 : 3 and its

perimeter is 96 cm. Find its area.

Ans.Let the sides of the triangle ABC be 4x, 5x and 3x

Let AB = 4x, AC = 5x and BC = 3x

Perimeter of ABC∆ = 4x + 5x + 3x, 12x = 96 (Given)

D C

BA

50 m

40 m24 m

18 m

90°

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96

12x = x = 8

∴ Sides are BC = 3(8) = 24 cm,

AB = 4 (8) = 32 cm,

AC = 5 (8) = 40 cm

∴ (AC)2 = (AB)2 + (BC)2 [∵ (5x)2 = (3x)2 + (4x)2]

Hence, area of 1

ABC (BC) (AB)2

∆ = 1

24 322

= × ×

= 12 × 32 = 384 cm2

12. Calculate the area of the quadrilateral PQRS shown in the adjoin-

ing figure, it being given that PR = 8 cm, RQ = 17 cm,

RPQ 90 ,∠ = ° RS = 3 cm and PRS 90 .∠ = °

SR6 cm

PQ

17 cm

8 c

m

Ans. In PQR,∆

Using Pythagoras theorem, = PQ2 = QR2 – PR2

2 2PQ (QR) (PR)= − 2 2(17) (8)= −

289 64 225 15 cm= − = =

Thus, area of 1

PQR PQ PR2

∆ = × ×21

15 8 cm2

= × × = 60 cm2

Area of 1

PRS SR PR2

∆ = × × 1

6 82

= × × = 24 cm2

A

B C

4x5x

3x

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Hence, area of the quadrilateral PQRS

= area of PQR∆ + area of PRS∆

= (60 + 24) cm2 = 84 cm2.

13. Find the area of a quadrilateral ABCD whose diagonal AC is 25

cm long and the lengths of perpendiculars from opposite vertices

B and D on AC are BE = 3.6 cm and DF = 2.4 cm.

A

F

D

C

B

E

Ans.Area of the quadrilateral ABCD

= area of ABC∆ + area of ACD∆

1 1AC BE AC DF

2 2= × × + × ×

1AC (EB DF)

2= × × +

2125 (3.6 2.4) cm

2= × × +

2125 6 cm = 75 cm .

2= × ×

14. Find the area of the quadrilateral ABCD, given in the adjoining

figure in which AB = 28 cm. BC = 78 cm, CD = 112 cm, BD = 50

cm and DA = 30 cm. (Given 26 5.099)=

D

C

BA

112 cm

50 cm

30 cm

28 cm

78 c

m

Ans. In ABD∆

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50 30 28 10854 cm

2 2s

+ += = =

Thus, area of ABD∆

254(54 50) (54 30) (54 28) cm= − − −

254 4 24 26 cm= × × ×

22 3 3 3 2 2 2 2 2 3 2 13 cm= × × × × × × × × × × ×

2 2 2 2 2 22 2 2 2 3 3 13 cm= × × × × × ×

22 2 2 3 3 13 2 cm= × × × × ×

272 26 cm= = 72 × 5.0990 = 367.13 cm2

In DBC∆

112 78 50 240120

2 2s

+ += = = cm

Thus area of DBC∆

2120(120 112) (120 78) (120 50) cm= − − −

2120 8 42 70 cm= × × ×

22 2 3 10 2 2 2 2 3 7 7 10 cm= × × × × × × × × × × ×

2 2 2 2 2 22 2 2 3 7 10= × × × × ×

= 2 × 2 × 2 × 3 × 7 × 10

= 1680 cm2

Area of the quadrilateral ABCD

= area of ABD∆ + area of DBC∆

= (367.13 + 1680) cm2

= 2047.13 cm2.

15. In the adjoining figure, ABCD is a square and

E, F are mid-points of the sides BC, CD respec-

tively. If CE = 7 cm., find the area of the un-

shaded region.

D CF

A B

E

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Ans. ABCD is a square

∴ AB = BC = CD = DA, CD = BC

⇒1 1

CD BC2 2

= ⇒ CF = CE.

∴ CF = CE = 7 cm.

Thus area of shaded region 21 497 7 cm

2 2= × × =

Area of square = (Side)2 = 14 × 14 = 196 cm2

Area of unshaded portion 49

1962

= − 2343

171.5 cm2

= =

16. In the adjoining figure, find the area of the path which is 1.9 m

wide all inside the rectangle.

22 m

14 m

Ans.Length of rectangle ABCD = 22 m

Breadth of rectangle ABCD = 14 m

D C

B

F

GH

E

A

1.9 m

(22 – 3.8) (14 –

3.8

)

14

m

1.9 m

22 m

Area of rectangle ABCD = 22 × 14 = 308 m2

Length of rectangle EFGH = 22 – 2 × 1.9 = 18.2 m

Breadth of rectangle EFGH = 14 – 2 × 1.9 = 10.2 m

Area of rectangle EFGH = 18.2 × 10.2 = 185.64 m2

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Hence, area of path = area of rectangle ABCD – area of rectangle

EFGH (308 – 185.64) m2 = 122.36 m2

17. Find the area and the perimeter of the following figures. All angles

are right angles and all measurements are in centimetres.

9

3 32

2

8

8

( )i ( )ii

12

7

5

2

3

2

5

7

2

Ans. (i) Area of rectangle ABCD

= 8 × 2 = 16 cm2

Area of rectangle IJKL = 5 × 2 = 10 cm2

[∵ L = 9 – 2 – 2 = 5 cm. B = 8 – 3 – 3 = 2 cm.]

9

8

8

B

G

3 3 C2

2

A

H

D

5

E 3 3 F

5

K

Area of the figure = 16 + 16 + 10 = 42 cm2.

Perimeter of figure = sum of all the sides

= 8 + 2 + 3 + 5 + 3 + 2 + 8 + 2 + 3 + 5 + 3 + 2 = 46 cm.

(ii) Area of rectangle AEHG = 12 × 2 = 24 cm2

Area of rectangle EFCB = 5 × 2 = 10 cm2

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Area of rectangle OLMN = 3 × 2 = 6 cm2

Area of rectangle HIJK = 5 × 2 = 10 cm2

Total area of figure = 24 + 10 + 6 + 10 = 50 cm2

12

E 7 B

J

C

I

5F

2

33

3

N

ML

K

3

O2

5

7HG

2

A

D

Perimeter of figure = Sum of all the sides

= 7 + 2 + 5 + 3 + 3 + 2 + 3 + 3 + 5 + 2 + 7 + 12 = 54 cm.

18. The inner dimensions of a closed wooden box are 2 m by 1.2 m by

0.75 m. The thickness of the wood is 2.5 cm. Find the cost of

wood required to make the box if 1 m3 of wood costs Rs. 5400.

Ans.Thickness of wood = 2.5 cm = 0.025 m

∴ Inner volume = 2 × 1.2 × 0.75 = 1.8 m3

External length = 2 + 0.05 = 2.05 m

External breadth = 1.2 + 0.05 = 1.25 m

External weight = 0.75 + 0.05 = 0.8 m

External volume = l × b × h

= 2.05 × 1.25 × 0.8 = 2.05 m3

Hence, volume of wood required to make box

= (2.05 – 1.8) m3 = 0.25 m3

Cost of wood = 0.25 × Rs 5400 = Rs 1350.

19. In the diagram, ABCD is a rectangle of size 18 cm by 10 cm. In

BEC,∆ E 90∠ = ° and EC = 8 cm. Find the area enclosed by the

pentagon by the pentagon ABECD.

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D

A

C

B18 cm

E 10 cm

8 cm

Ans.Area of rectangle ABCD

= 18 × 10 = 180 cm2

In BEC,∆ BC2 = CE2 + BE2 (By Pythagoras theorem)

(10)2 = 82 + BE2

∴ BE2 = 100 – 64 = 36 ⇒ BE 36= ⇒ BE = 6 cm.

Hence, area of right BEC∆ 21

6 8 24 cm2

= × × =

Area of pentagon ABECD = Area of rectangle ABCD – area of

BEC∆ = (180 – 24) cm2 = 156 cm2.

20. In the adjoining figure, ABCDE is a pentagon which is symmetri-

cal about the line passing through A and the mid-point M of the

side CD. All measurements are in centimetres. Find the perimeter

and the area enclosed by the pentagon correct to two decimal places.

(Given 10 3.16)=

C M D

EB

A

5.5

4.5

4

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Ans. Join BE

C M D

EB

A

5.55.5

4.5

4.5

4.5

4.5

4

In ,ABE∆ AB = 5.5 cm, AE = 5.5 cm

BE = 4.5 + 4.5 = 9 cm.

5.5 5.5 9

102

s+ +

= =

Area of ABE∆ = 10(10 5.5) (10 5.5) (10 9)− − −

10 4.5 4.5 1= × × ×

4.5 10= × = 4.5 × 3.16 = 14.23 cm2

Area of rectangle BCDE = l × b = 9 cm × 4 cm = 36 cm2

Total area of pentagon = (14.23 + 36) cm2 = 50.23 cm2

Perimeter of pentagon = sum of all the sides

= 5.5 + 4 + 4.5 + 4.5 + 4 + 5.5 = 28 cm.

21. Find the area of the shaded region of each of the following figures,

given that all measurements area in centimetres :

16

20

4 4

8

3

4

4

4

3

16

24

5 10

6

5

5

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Ans.(i) Area of rectangle ABCD = 16 × 4 = 64 cm2

16

20

4 48 C

B

8128

N3O

G P 44

4

4

M 3 H

3

A

D

4 E F

J

KL

I

Area of rectangle EFGH = 8 × 12 = 96 cm2

Area of rectangle IJKL = 20 × 4 = 80 cm2

Area of two triangles = 21

2 3 4 12 cm .2

× × =

Total area of the shaded region = Area of rectangle ABCD +

area of rectangle EFGH + area of rectangle IJKL + area of

∆ MNH + area of ∆ PGO

= 64 + 96 + 80 + 6 + 6 = 252 cm2

(ii) Area of 1

ABC2

b h∆ = × × = 21

10 5 25 cm2

× × =

Area of rectangle ACDI = L × B = 16 × 10 = 60 cm2

16

16 8

24

5 10

6

5

5

H G

I

B C

C D E

A

F

Area of rectangle HDEG = 8 × 6 = 48 cm2

Area of ∆ EFG = s(s a) (s b) (s c)− − −

5 + 5 + 6where s = 8, a 6, b 5, c 5

2

= = = =

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= 8(8 6) (8 5)(8 5)− − − = 28 2 3 3 12 cm× × × =

Area of the shaded region = (25 + 160 + 48 + 12) cm2 = 245cm2.

22. A rectangular lawn 75 m by 60 m has two roads each 4 m wide

running in the middle of it, one parallel to length and the other

parallel to breadth. Find the cost of gravelling the roads at Rs. 4.50

per sq. metre.

75 m

60 m 4 m

4m

Ans.Area of the road length wise = (75 × 4) m2 = 300 m2

Area of the road breadth wise = (60 × 4) m2 = 240 m2

Area of the common road = (4 × 4) m2 = 16 m2

∴ Area of the road which to be gravelled

= (300 + 240 – 16) m2 = (540 – 16) m2 = 524 m2

Cost of gravelling l m2 = Rs 4.50

Cost of gravelling 524 m2 = Rs 4.50 × 524 = Rs 2358.

23. The length and breadth of a rectangular park are in the ratio 5 : 2.

A 2.5 m wide path running all around the outside of the park has

an area of 305 m2. Find the dimensions of the park.

Ans.Area of path = 305 m2

Let length of park (l) = 5x, then breadth (b) = 2x

∴ Area = l × b = 5x × 2x = 10 x2

Width of path = 2.5 m

∴ Outer length (L) = 5x + 2 × 2.5 = (5x + 5) m

and breadth (B) = 2x + 2 × 2.5 = 2x + 5 m

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∴ Area of path = L × B – l × b = (5x + 5) (2x + 5) – 5x × 2x

= 10x2 + 25x + 10x + 25 – 10x2

∴ 35x + 25 = 305 ⇒ 35x = 280 ⇒280

835

x = =

∴ Length of the park = 5x = 5 × 8 = 40 m

and breadth of park = 2x = 2 × 8 = 16 m

24. Find the area of the shaded region of each of the given figures :

A

B D

C

E

8 c

m

17 c

m

15 cm

F

A

E

B

6 cm

H

5 cm

G

6 cm

5 cm

CD 3 cm

6 c

m(i) (ii)

(iii) (iv)

D E C

A B

20 cm

25 cm

14

cm

17

cm

G F

H B

A

D

E

C

9 cm

9 c

m

5cm

17 cm

5cm

Ans. (i) In BDC∆

Using Pythagoras Theorem we have

2 2BD (BC) (DC)= − 2 2(15) (9)= −

[ ∵ DC = EC – AB = 17 – 8 = 9 cm]

225 81 144 12 cm= − = =

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∴ Area of 1

BDC BD DC2

∆ = × ×

2 2112 9 cm 54 cm

2= × × =

Area of the rectangle ABDE = AB × BD = 8 × 12 cm2 = 96 cm2

Hence, Area of the shaded region

= area of BDC∆ + area of rectangle ABDE

= (54 + 96) cm2 = 150 cm2

(ii) In BDC∆

Using Pythagoras theorem we have,

2 2BD (BC) (CD)= − 2 2(5) (3)= − 25 9= −

16 4 cm= =

Similarly, AG = 4 cm

Thus, area of BDC∆ 3 21 1BD DC 4 3 cm 6 cm

2 2= × × = × × =

Similarly, area of the 2AGH 6 cm∆ = = 6 cm2

Area of rectangle ABEF = AB × BE = 6 × ( 6 + 4) cm2

= 6 × 10 cm2 = 60 cm2

Hence, area of the shaded region

= ( 6 + 6 + 60) cm2 = 72 cm2

(iii) Area of the rectangle ABCD

= AB × BC = 25 × 14 cm2 = 350 cm2

In AEB∆ , Using Pythagoras theorem, we have

2 2AE (AB) (EB)= − 2 2(25) (20)= −

625 400= − 225= = 15 cm

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Area of the AEB∆ , 21 1

AE EB 15 20 cm2 2

= × × = × × = 150 cm2

Area of the shaded region

= area of rectangle ABCD – area of ABE∆

= (350 – 150) cm2 = 200 cm2

(iv) In AHB∆

AH = AG – GH = 17 – (9 + 5) = 17 – 14 = 3 cm

Area of

1AHB AH HB

2∆ = × ×

2 2 21 153 5 cm cm 7.5 cm

2 2

= × × = =

From C draw perpendicular ar on GF

Then, area of region in two rectangular forms

= [(9 × 5) + (17 – 3) × 5] cm2

= [45 + 14 × 5] cm2

= (45 + 70) cm2 = 115 cm2

Area of DEF∆ 1

DF FE2

= × × 21

5 3 cm2

= × ×

2 215

cm 7.5 cm2

= =

∴ Area of the shaded region

= (7.5 + 115 + 7.5) cm2 = 130 cm2.

25. The parallel sides of a trapezium are 17 cm and 25 cm. If the dis-

tance between them is 13 cm, find the area of the trapezium.

Ans. Area of trapezium

1(AB CD)

2h= + ×

21

42 13 273 sq. cm2

= × × =

D C

A B25 cm

13

cm

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26. The cross-section ABCD of a swimming pool is a trapezium.

A B14 m

1.5 m

D

C

8 m

Its width AB = 14 m, depth at the shallow end is 1.5 m and at the

deep end is 8 m. Find the area of the cross-section.

Ans.Here, two parallel sides of trapezium are AD and BC and distance

between them is 14 m.

Hence, area of trapezium 1

(1.5 8) 142

= + × 21

(9.5 14 ) 66.5 m .2

= × =

27. The cross-section of a canal is shown

in the adjoining diagram. If the canal

is 8 m wide at the top and 6 m wide at

the bottom and the area of the cross-

section is 16.8 m2, calculate its depth.

Ans.Area of trapezium ABCD

1( )

2AB CD h= × + ×

But, area of trapezium = 16.8 m2

∴ 1

16.8 (6 8)2

h= × + ×

⇒ 16.8

7h = = 2.4 h = 2.4 m

28. From the adjoining diagram, calculate

(i) the area of trapezium ACDE

(ii) the area of parallelogram ABDE

(iii) the area of triangle BCD

8 m

6 m

8 m

6 m

D

A B

C

h

E D7 cm

A CB

13 m

6.5 m

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Ans.

(i) Area of trapezium ACDE

1(AC DE)

2h= × + ×

21(13 7) 6.5 65 m

2= × + × =

(ii) Area of parallelogram ABDE = b × h = 7 × 6.5 = 45.5 m2

(iii) Area of 1

BCD BC2

h∆ = × × 21

6 6.5 19.5 m2

= × × =

29. Calculate the area enclosed by the following shapes. All measure-

ments are in cm.

14

1016

( )i7

5

2

4

3

( )ii

9

2

2

7

7

9

( )iii

Ans. (i) ABCD is a trapezium

A B14

D

C

1016

Hence, Area of trapezium ABCD 1

(10 16) 142

= × + × = 182 cm2

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(ii) Area of trapezium ABCD

1(AB CD) AD

2= + ×

21(5 3) 9 36 cm

2= + × =

Area of rectangle GAFE = 2 × 5 = 10

cm2

Total area of the figure = (36 + 10) cm2 =

46 cm2

(iii) Area of rectangle ABCD = 9 × 2 = 18 cm2

Area of rectangle EFGH = 9 × 2 = 18 cm2

Area of parallelogram BIHJ = 2 × 5 = 10

cm2

Total area of the figure = (18 + 18 + 10)

cm2 = 46 cm2

30. From the adjoining sketch, calculate

(i) the length AD

(ii) the area of trapezium ABCD

(iii) the area of triangle BCD

Ans. (i) In ABD,∆ we have

BD2 = AD2 + AB2 ⇒ AD2 = BD2 – AB2 = (41)2 – (40)2

= 1681 – 1600 = 81

∴ AD 81 9 cm.= =

(ii) Area of trapezium ABCD 1

(AB CD) AD2

= + ×

21

(40 15) 9 247.5 cm2

= × + × =

D C

E F

2

4

3

G A 7 B

5 5

52

A EB F

D HC G

12

7J2

2

2

7

9

D C

BA

41 cm

15 cm

40 cm

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(iii) Area of triangle BCD = Area of trapezium ABCD – Area

of ABD∆ 21247.5 40 9 cm

2

= − × ×

= (247.5 – 180) cm2 = 67.5 cm2

31. The two parallel sides and the distance between them are in the

ratio 3 : 4 : 2. If the area of the trapezium is 175 cm2 ; find its

height.

Ans.Let the two parallel sides and the distance between them be 3x, 4x

and 2x cm respectively,

Area = 1

2 × (sum of parallel sides) × (distance between parallel sides)

1(3 4 ) 2 175

2x x x= + × = (given

⇒ 7x × x = 175 ⇒ 7x2 = 175 ⇒ x2 = 25 ⇒ x = 5

∴ Height i.e., distance between parallel sides = 2x = 10 cm

32. A parallelogram has sides of 15 cm and 12 cm; if the distance

between the 15 cm sides is 6 cm; find the distance between 12 cm

sides.

Ans. Base AB = 15 cm

Distance between 15 cm sides

i.e., Height DP = 6 cm

∴ Area of parallelogram = base ×

height = AB × DP = 15 × 6 = 90 cm2

Let BQ be distance between 12 cm sides

∴ AD × BQ = area of parallelogram ABCD

∴ 12 × BQ = 90 90

BQ12

= = 7.5 cm

33. A parallelogram has sides of 20 cm and 30 cm. If the distance

between its shorter sides is 15 cm; find the distance between the

longer sides.

D C

A P B

Q

6 cm

15 cm

12 cm

15 cm

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Ans. Let ABCD be the parallelogram in which BC = 30 cm andCD = 20 cm

Distance between shorter sides,

i.e., CQ = 15 cm

∴ Area of parallelo-

gram = AB × CQ = 20 × 15

= 300 cm2

Again BC × AP =

Area of parallelogram

30 × AP = 300

300

AP30

= = 10 cm.

Hence, distance between larger sides is 10 cm.

34. In a parallelogram ABCD; AB = 10 cm, AD = 24 and BD = 26 cm,

find the area of parallelogram.

Ans. Firstly, we find area of ABD∆

Sides are a = 10 cm, b = 26 cm,

and c = 24 cm

2

a b cs

+ +=

10 26 24

2

+ +=

6030

2= =

Thus, area of ABD∆ ( ) ( ) ( )s s a s b s c= − − −

30(30 10) (30 26) (30 24)= − − −

30 20 4 6= × × × 5 6 4 5 4 6= × × × × ×

5 5 6 6 4 4= × × × × × = 120 cm2

∵ Diagonal bisects the parallelogram in two equal parts.

A D

CPB

Q

20 cm

20 cm

15 cm

30 cm

30 cm

B

A

C

D

10 cm 26 cm

24 cm

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∴ Area (parallelogram ABCD) = 2 × area ( )ABD∆

= 2 × 120 = 240 cm2

35. The adjacent sides of a parallelogram are 21 cm and 28 cm. If its one

diagonal is 35 cm; find the area of

the parallelogram.

Ans. Firstly, we find area of ABC.∆

Sides are, a = 28 cm, b = 35 cm

and c = 21 cm

2

a b cs

+ +=

28 35 21

2s

+ +=

8442

2= =

Area of ABD ( ) ( ) ( )s s a s b s c∆ = − − −

42(42 28) (42 35) (42 21)= − − − 42 14 7 21= × × ×

2 21 2 7 7 21= × × × × × 2 2 21 21 7 7= × × × × ×

= 2 × 21 × 7 = 294 cm2

Thus, diagonal of parallelogram divides it into two equal parts.

Hence, area of parallelogram = 2 × area of ABC∆

= 2 × 294 = 588 cm2.

36. PQRS is a parallelogram with PQ = 26 cm and QR = 20 cm. If the

distance between its longer sides is 12.5 cm, find :

(i) the area of the parallelogram ;

(ii) the distance between its shorter sides.

Ans. (i) Case I

Base = (b) = 26 cm

Altitude (h) = 12.5 cm

∴ Area of PQRS = b × h = 26

× 12.5 cm = 325 cm2

A

B

D

C

21 cm 35 cm

28 cm

S

P

R

QL

20 cm

26 cm

20 cmM 12.5 cm

26 cm

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(ii) Case II

Base (shorter side) = 20 cm

Hence, height = Area 325

16.25 cmBase 20

= =

37. ABCD is a parallelogram having adjacent sides AB = 16 cm and BC

= 14 cm. If its area is 168 cm2, find the distance between its longer

sides and that between its shorter sides.

Ans. Area of parallelogram ABCD = 168 cm2

Longer side AB = 16 cm

Shorter side BC = 14 cm

Let DL AB⊥ and BM AD⊥

∴ Length Area 168

DL 10.5 cmBase 16

= = =

and length of Area Area 168

BMAD BC 14

= = = = 12 cm. [∵ AD = BC]

38. In the adjoining figure, ABCD is a parallelogram in which AB = 28

cm, BC = 26 cm and diagonal AC = 30 cm. Find

(i) the area of parallelogram ABCD ;

(ii) the distance between AB and DC ;

(iii) the distance between CB and DA.

D

A

C

B

M

26 cm30 c

m

28 cm L

Ans. In parallelogram ABCD

(i) In ∆ABC, AB = 28 cm, BC = 26 cm and AC = 30 cm

D

A

C

BL

14 cm

16 cm

14 cmM

16 cm

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Area of triangle ABC = 1

area (ABCD)2

(∵ A diagonal bisects the parallelogram into two triangles of

equal area)

Now, sides of triangle ABC are 28 cm, 26 cm, 30 cm

∴28 26 30 84

422 2 2

a b cs

+ + + += = = =

∴ Area of ABC ( ) ( ) ( )s s a s b s c∆ = − − −

42(42 28) (42 26) (42 30)= − − −

42 14 16 12= × × × 42 14 16 12= × × ×

2 3 7 7 2 2 2 2 2 2 2 3= × × × × × × × × × × ×

2 2 2 2 2 22 2 2 2 3 7= × × × × ×

= 2 × 2 × 2 × 2 × 3 × 7 = 336 cm2

Hence, area of parallelogram ABCD = 2 × 336 = 672 cm2

(ii) Distance between AB and DC

Area 672

CL cmAB 28

= = = 24 cm

(iii) Distance between CB and DA

Area 672CM

AD 26= = = 25.84 cm

39. The parallel sides of a trapezium are 20 cm and 10 cm, Its non-

parallel sides are both equal, each being 13 cm. Find the area of

the trapezium.

D

A E B

C

20 cm

10 cm

13

cm

13

cm

13 cm

F

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Ans. Let ABCD is a trapezium

Draw CE || AD and CF AB⊥

∴ CE = AD = 13 cm

and AE = DC = 10 cm

EB = AB – AE = 20 – 10 = 10 cm

13 13 10 36

182 2 2

a b cs

+ + + += = = =

∴ Area of ( ) ( ) ( )ECB s s a s b s c∆ = − − −

18(18 13) (18 13) (18 10)= − − −

18 5 5 8 2 3 3 5 5 2 2 2= × × × = × × × × × × ×

= 2 × 2 × 3 × 5 = 60 cm2

Again, area of 260 cmECB∆ =

⇒1

602

FB CF× × = ⇒1

10 602

CF× × =

⇒ 5 × CF = 60 ⇒60

5CF = ⇒ CF = 12 cm.

Hence, area of trapezium ABCD 1

(AB CD) CF2

= × + ×

21 1(20 10) 12 30 12 180 cm

2 2= + × = × × =

40. Find the area of the figure ABCDEFGH, given below, it being given

that AC = 17 m, BC = 8m, EF = 9m, CD = 6 m, GL EF⊥ and GL =

3.6 m.

H C

A B

17 m

F L E

DG

9 m

3.6 m

6 m

8 m

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Ans. Area of the trapezium EFGD

1(EF GD) GL

2= + ×

21(9 6) 3.6 m

2= × + ×

2115 3.6 m

2= × × = 27 m2

Using Pythagoras theorem, we have

2 2AB (AC) (BC)= − 2 2(17) (8)= −

289 64= − 225= = 15 cm.

Area of rectangle ABCH = AB × BC

= 15 × 8 cm2 = 120 cm2

Hence, area of the figure ABCDEFGH = area of trapezium EFGD +

area of rectangle ABCH

= (27 + 120) cm2 = 147 cm2

41. Find the area of the shaded region in the figure given below :

3.5 m

3 m

2.5

m

6 m

4.6 m

Ans.From figure,

Area of the upper rectangular part = 3.5 × 3 = 10.5 m2

In trapezium part, distance between parallel sides

= (4.6 – 3) m = 1.6 m

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Area of trapezium 21 1

(6 3.5) 1.6 9.5 1.6 7.6 cm2 2

= × + × = × × =

Area of lower rectangular part = (6 × 2.5) m2 = 15 m2

Hence, Area of the shaded region

= (10.5 + 7.6 + 15) m2 = 33.1 m2

42. Find the area of figure ABCDE, it being given that :

AE || BD, AF BD,⊥ CG BD⊥

AE = 12 cm, BD = 16 cm, AF = 6.5 cm and CG = 8.5 cm

F

A E

B

C

DG

Ans. This figure consists of two figures one is a triangle and other is atrapezium

∴ Area of 1

BCD BD CG2

∆ = × 21

16 8.5 cm2

= × × = 68 cm2

Area of trapezium ABDE

1(BD AE) AF

2= + ×

21(16 12) 6.5 cm

2= × + ×

2128 6.5 91 cm

2= × × =

Hence, area of the figure ABCDE = Area of ∆BCD + Area of the

pezium ABDE = 68 + 91 = 159 cm2.

43. Find the area of the field ABCDEFA in which BP AD,⊥

CR AD,⊥ FQ AD,⊥ ES AD,⊥ and AP = 20 m, AQ = 35 m,

AR = 58 m, AS = 65 m, AD = 75 m, BP = 15 m, CR = 20 m, ES = 15

m and RQ = 10m

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F

E

AP

B

Q

R

SD

C

Ans. This figure of a field contains triangles and trapeziums

Area of 1

APB AP BP2

∆ = × 21

20 15 150 m2

= × × =

Area of 1

RCD RD RC2

∆ = × ×

1 1(AD AR) RC (75 58) 20

2 2= − × = × − ×

2117 20 170 m

2= × × =

Area of 1

ESD SD ES2

∆ = ×

1= (AD- AS) × ES

2

1(75 65) 15

2= − ×

2110 15 75 m

2= × × =

Area of 1

AQF AQ FQ2

∆ = × 21

35 10 175 m2

= × × =

Area of trapezium BPRC

1(BP CR) PR

2= + ×

1(15 20) (AR AP)

2= + × −

1(15 20) (58 20)

2= × + × −

2135 38 665 m

2= × × =

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and area of trapezium QFES

1( )

2FQ ES QS= + ×

1( ) ( )

2FQ ES AS AQ= + × −

1(10 15) (65 35)

2= + × −

2125 30 375 m

2= × × =

Hence, area of the total field ABCDEFA

= (150 + 170 + 75 + 175 + 665 + 375) m2

= 1610 m2.

44. Cotton thread is wound on a reel which has diameter of 2.8 cm.

There are 1200 turns of the thread on the reel. What is the total

length, in metre of the thread on the reel ?

Ans.Circumference of reel = 2 rπ 22

2 1.4 8.8 cm.7

× × =

Length taken by thread in one turn = 8.8 cm.

⇒ Total Length taken by thread in 1200 turns

= 8.8 × 1200 = 10560 cm 10560

m 105.6 m100

=

45. How many times will the wheel of a car rotate in a journey of 88

km, given that the diameter of the wheel is 56 cm.

Ans.Distance covered by wheel of car = 88 km = 88 × 1000 × 100 cm.

Diameter of wheel is 56 cm

∴ radius = 56 cm

2 = 28 cm.

⇒ Distance covered by wheel of car in one turn = 2 rπ

222 28 176 cm.

7× × =

Number of times wheel rotates

88 1000 1001000 50

176

× ×= = × = 50,000

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46. A Tonga is being driven at 11 km/hr. If each wheel of the Tonga is 1.4

m in diameter, find the number of revolutions made by each wheel

per minute.

Ans. Distance covered by wheel of Tonga in one revolution = 2 rπ

22 1.42 4.4 m

7 2= × × =

Distance covered by tonga in 60 minutes = 11 km. = 11,000 m.

Distance covered by tonga in 1 minute 11,000

60=

Number of revolutions in one minute

11000 11000 10 2504.4

60 60 44 6= ÷ = × =

125 241

3 3= =

47. From a square cardboard of side 21cm, a circle of maximum area is

cut out. Find the area of the cardboard left.

Ans. Circle of maximum area can be cut out from a

square if diameter of circle is equal to side of the

square.

∴ Area of square = (21 × 21) cm2 = 441 cm2

Area of circle 22 1 21 693

7 2 2 2

2= × × =

21

2r

=

∵

Area of remaining cardboard

2 2693 882 693441 cm cm

2 2

− = − =

2189cm

2= = 94.5 cm2

48. A wire is in the form of a square of side 27.5 cm. It is straightened

and bent into the shape of a circle. Find the area of the circle.

Ans.Perimeter of square = Circumference of circle

4 side 2 r× = π ⇒ 22

4 27.5 27

r× = × ×

21 cm

d

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⇒ 4 27.5 7 2 2.5 7

2 22 2r

× × × ×= =

×

= 17.5 cm

Area of circle 222

17.5 17.5 cm7

= × ×

= 962.5 cm2.

49. Three cubes of silver with edges 3 cm, 4 cm and 5 cm are melted and

recast into a single cube. Find the cost of coating the surface of the

new cube with gold at the rate of Rs. 3.50 per square centimetre?

Ans. Let a cm be the edge of new cube

According to the given condition

a3 = 33 + 43 + 53 = 27 + 64 + 125 = 216 cm3

3 216a = ⇒ a = 6 cm.

Suface area of new cube = 6 × (side)2 = 6 × (6)2 = 216 cm2

Cost of coating the surface of new cube = Rs 3.50 × 216 = Rs 756.

50. In the adjoining figure, the area enclosed between the concentric circles

is 770 cm2. If the radius of the outer circle is 21 cm, calculate the

radius of the inner circle.

Ans. Radius of outer circle (R) = 21 cm.

Radius of inner circle (r) = r cm.

Area of shaded protion = 770 cm2

⇒ 2 2( ) 770π − =R r

⇒2 222

(21 ) 7707

− =r

⇒2 7

441 77022

r− = × = 35 × 7 = 245

⇒ 2441 245r = − ⇒ r2 = 196

⇒ r2 = 196 ⇒ 196r =

⇒ r = 14 cm.

51. Find the area of a flat circular ring formed by two concentric circles

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(circles with same centre whose radii are 9 cm and 5 cm

Ans.

r =9 cm

1r =5 cm2

External radius, r1 = 9 cm

internal radius, r2 = 5 cm

∴ Area of ring = 2 21 2π − πr r 2 2

1 2( )= π −r r 2 2(9 5 )= π −

22 22[81 25] 56

7 7= − = × = 176 cm2.

52. Find the area of the shaded portion in each the following diagrams :

7 cm

( )i( )ii

4.5

2.5 m

Ans. (i) Radius of circle (r) = 7 cm

Side of square = 7 + 7 = 14 cm

Area of circle 2

r= π

22

7 77

= × × = 154 cm2

Area of square = 14 × 14

Area of shaded portion

= 196 – 154 = 42 cm2

(ii) Radii of concentric circles are

21 cm

7 cm

7 cm

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r1 = 4.5 m, r

2 = 2.5 m

Hence, area of shaded portion = 2 21 2r rπ − π

2 21 2[ ]r r= π −

2 222 2220.25 6.25 (4.5) (2.5)

7 7 = × − = −

22214 44 cm

7= × =

53. The shape of a park is a rectangle bounded by semicircles at the ends,

each of radius 17.5 m, as shown in the adjoining figure. Find the

area and the perimeter of the park.3

5 m

35 m

40 m

Ans. Diameter of the part = 35 cm

i.e. radius of the circular part = 17.5 m

∴ Area of two semi-circular parts

2 2 2 21 222 (17.5) m

2 7r r= × π = π = ×

2 22217.5 17.5 m 962.5 m

7= × × =

Now, area of the rectangular part = (40 × 35) m2 = 1400 m2

∴ Area of the park = (962.5 + 1400) m2 = 2362.5 m2.

Now, Kcircumference of a semi-circle

1 222 17.5 m 55 m

2 7r r= × π = π = × =

Hence, perimeter of the park = 40 + 55 + 40 + 55 = 190 m

54. Find the area of the space enclosed by two concentric circles of radii

r =2.5cm2

r=4

.5cm

1

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17 cm and 11cm.

Ans. Here, R = 17 cm and r = 11 cm

Required area 2 2 2[(17) (11) ] cm= π −

222(17 11) (17 11) cm

7= × + −

2 22228 6 cm 528 cm

7= × × =

Hence, area of the space = 528 cm2

55. The diameter of a circular park is 52 m. On

its outside, there is path 4 m wide, running

around it. Find the cost of turfing the path at

Rs. 2.50 per square metre.

Ans. Diameter of the park = 52 m. So, Radius

of the park 52

26 m2

= =

Width of path = 4 m

Here, r = 26 m and R = (26 + 4) m = 30 m

Area of the path 2 2[ ]R r= π − 2 2 2[(30) (26) ] m= π −

222(30 26) (30 26) m

7= + −

22256 4 m

7= × × = 704 m2

Now, cost of turfing 1m2 = Rs 2.50

Cost of turfing 704 m2 = Rs 2.50 × 704 = Rs 1760.

56. The sum of diameters of two circles is 2.8 m and the difference of

their circumferences is 0.88 m. Find the radii of two circles.

Ans. Let diameter of two circles be d1 and d

2

∴ d1 + d

2 = 2.8 m

or 2r1 + 2r

2 = 2.8 (∵ d = 2r)

r1 + r

2 = 1.4 ...(i)

Also, 1 22 2 0.88 mr rπ − π =

17 cm

11 cmO

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1 22 ( ) 0.88r rπ − = m

1 21 7

0.88 0.142 22

r r− = × × = ...(ii)

(i) and (ii), we have

2r1 = 1.54. r

1 = 0.77 m or 77 cm.

r2 = 1.4 – r

2 = 1.4 – 0.77

= 0.63 m or 63 cm.

57. Calculate the length of the boundary and the area of the shaded

region in the following diagrams. All measurements are in

centimetres.

10

7

28

(i) Unshaded part is a semi-circle

(ii) Four semi-circles on a square

Ans. (i) Length of boundary

1AB BC CD (2 )

2r= + + + π

22 710 7 10

7 2= + + + ×

= 10 + 7 + 10 + 11 = 38 cm.

Area of shaded portion = area of rectangle ABCD – area of

semi-circle

2110 7 ( )

2= × − πr

1 22 7 7 7770 70

2 7 2 2 4= − × × × = −

1010

7

A

B

D

C

+7/2 7/2

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= 70 – 19.25 = 50.75 cm2

(ii) Length of one semi-circle = rπ

227 22 cm.

7= × =

Length of 4 semi-circles= 22 × 4 = 88 cm.

Area of four semi-circles = Area of two circles = 22πr

22 227 7 308 cm .

7

×= × × =

Area of square = 14 × 14 = 196 cm2

Area of shaded portion = 308 + 196 = 504 cm2.

58. In the adjoining figure, ABCD is a square of side 14 cm. A, B, C and

D are centres of circular arcs of radius 7 cm. Find the perimeter and

the area of the shaded region.

D

A

C

B

Ans. Perimeter of shaded portion

� � � �EF FG GH EH= + + + = Perimeter of circle = 2 rπ

222 7 44 cm.

7= × × =

14 1414

7

7

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D

A

CE7

7

7

7

7

7

7

7

F H

BG

Area of square = 14 × 14 = 196 cm2

Area of four quadrants or area of on circle

2 222

7 7 154 cm7

r= π = × × =

Area of shaded portion = (196 – 154) cm2 = 42 cm2.

59. The boundary of the shaded region in the given figure consists of

three semi-circles, the smaller being equal. If the diameter of the

larger one is 28 cm, find

(i) the length of the boundary

(ii) the area of the shaded region

Ans. Length of boundary

length of � �2(length of )AB AQ+

1 28 1 282 2 2

2 2 2 4

= × π + × π

14 14= π× + π×

22(14 14) 28

7π × = × = 22 × 2 + 22 × 2 = 88 cm.

Area of the shaded region = 21

( )2

rπ

21 22

14 14 308 cm2 7

= × × × =

A

Q

O PB

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60.Find the volume, the total surface area and the lateral surface area of

the cuboid having

(i) length = 24 cm, breadth = 16 cm and height = 7.5 cm

(ii) length = 10 m, breadth = 35 cm and height = 1.2 m

Ans. (i) Here l = 24 cm, b = 16 cm, h = 7.5 cm

∴ Volume of the cuboid

= 2 (lb + bh + hl)

= 2 (24 × 16 + 16 × 7.5 + 7.5 × 24) cm2

= 2 (384 + 120 + 180) cm2

= 2 × 684 cm2 = 1368 cm2

Lateral surface area of the cuboid = 2 (l + b) × h

= 2 [24 + 16] × 7.5 cm2

= 2 × 40 × 7.5 cm2 = 600 cm2

(ii) Here, l = 10 m, b = 35 cm and h = 1.2 m

∴ Volume of the cuboid = l × b × h

= 10 × 0.35 × 1.2 m2 = 4.2 m3

Total surface area of the cuboid

= 2 [lb + bh + hl]

= 2 [10 × 0.35 + 0.35 × 1.2 + 1.2 × 10] m2

= 2 [3.5 + 0.42 + 12] m2 = 2 × 15.92 m2 = 31.84 m2

Lateral suface area of the cuboid

= 2 (l + b) × h = 2 [10 + 0.35] × 1.2 m2

= 2 × 10.35 × 1.2 m2 = 24.84 m2

61. A wall of length 13.5 m, width 60 cm and height 1.6 m is to be

constructed by using bricks, each of dimensions 22.5 cm by 12

cm 8 cm. How many bricks will be needed ?

Ans.Length of the wall (l) = 13.5 m

= 13.5 × 100 cm = 1350 cm

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Breadth of the wall (b) = 60 cm

Height of the wall (h) = 1.6 m

= 1.6 × 100 cm = 160 cm

Volume of the wall = l × b × h

= (1350 × 60 × 160) cm3

Volume of a brick = (22.5 × 12 × 8) cm3

Hence, required number of bricks used

1350 60 160

22.5 12 8

× ×=

× ×

1350 60 160 10

225 12 8

× × ×=

× ×

= 6 × 5 × 20 × 10 = 6000

62. Find the length of each edge of a cube, if its volume is :

(i) 216 cm3 (ii) 1.728 m3

Ans. (i) (Edge)3 = Volume of a cube

(Edge)3 = 216 cm3

⇒ 1/ 3Edge (3 3 3 2 2 2)= × × × × ×

⇒ Edge = 3 × 2 ⇒ Edge = 6 cm

(ii) (Edge)3 = Volume of a cube

∴ (Edge)3 = 1.728 m3

⇒ 3 1.728 1728(Edge)

1000 1000= = ⇒

1/ 31728

Edge1000

=

⇒1/ 3

2 2 2 2 2 2 3 3 3Edge

10 10 10

× × × × × × × × =

× ×

⇒2 2 3

Edge10

× ×= ⇒

12Edge m

10=

⇒ Edge = 1.2 m

63. The total surface area of a cube is 216 cm2. Find its volume.

Ans. 6(Edge)2 = Total surface area of a cube.

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6 (Edge)2 = 216 cm2

⇒2 216

(Edge)6

= ⇒ (Edge)2 = 36

⇒ Edge 36= ⇒ Edge = 6 cm

Volume of the given cube = (Edge)3

= (6)3 = 6 × 6 × 6 = 216 cm3.

64. A wall 9 m long, 6 m high and 20 cm thick, is to be constructed ;

using bricks of dimensions 30 cm, 15 cm and 10 cm. How many

bricks will be required.

Ans. Length of the wall = 9 m = 9 × 100 cm = 900 cm

Height of the wall = 6 m = 6 × 100 cm = 600 cm

Breadth of the wall = 20 cm

Volume of the wall = (900 × 600 × 20) cm3 = 10800000 cm3

Volume of the one brick = 30 × 15 × 10 cm3 = 4500 cm3

Number of bricks required to construct the wall

Volume of wall

Volume of one brick=

10800000

4500= = 2400.

65. The dining-hall of a hotel is 75 m long; 60 m broad and 16 m high.

It has five-doors 4 m by 3 m each and four windows 3 m by 1.6 m

each. Find the cost of :

(i) papering its walls at the rate of Rs. 12 per m2;

(ii) carpetting its floor at the rate of Rs. 25 per m2.

Ans. Length of the dining hall of a hotel = 75 m

Breadth of the dining hall of a hotel = 60 m

Height of the dining hall of a hotel = 16 m

(i) Area of four walls of the dining hall

= 2 (L + B) × H = 2(75 + 60) × 16

= 2 (135) × 16 = 270 × 16 = 4320 m2

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Area of one door = 4 × 3 m2 = 12 m2

Area of 5 doors = 12 × 5 = 60 m2

Area of one window = 3 × 1.6 = 4.8 m2

Area of 4 windows = 4.8 × 4 = 19.2 m2

Area of the walls to be papered

= 4320 – (60 + 19.2) = 4320 – 79.2 = 4240.8 m2

Cost of papering the walls at the rate of Rs 12 per m2

= 4240.8 × 12 = Rs 50889.60

(ii) Area of floor = L × B = 75 × 60 = 4500 m2

Cost of carpetting the floor at the rate Rs 25 per m2

= 4500 × 25 = Rs 112500

66. Find the volume of wood required to make a closed box of exter-

nal dimensions 80 cm, 75 cm and 60 cm, the thickness of wall of

the box being 2 cm throughout.

Ans.External length of the closed box = 80 cm

External breadth of the closed box = 75 cm

External Height of the closed box = 60 cm

External volume of the closed box = 80 × 75 × 60 = 360000 cm3

Internal length of the closed box = 80 – 4 = 76 cm

Internal Breadth of the closed box = 75 – 4 = 71 cm

Internal Height of the closed box = 60 – 4 = 56 cm

Internal volume of the closed box = 76 × 71 × 56 cm = 302176

cm3

Volume of wood required to make the closed box = External vol-

ume of the closed box – Internal volume of the closed box

= 360000 – 302176 = 57824 cm3.

67. Find the length of the longest pole that can be placed in a room 12

m long, 8 m broad and 9 m high.

Ans.Length of room (l) = 12 m

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Breadth (b) = 8 m

and height = (h) = 9 m

Longest pole 2 2 2l b h= + + 2 2 2(12) (8) (9)= + +

144 64 81 289= + + = = 17 m.

68. The volume of cuboid is 972 m3. If its length and breadth be 16 m

and 13.5 m respectvely, find its height.

Ans.Here, length of the cuboid (l) = 16 m

Breadth of the cuboid (b) = 13.5 m

Let height of the cuboid be h m

Volume of the cuboid = 972 m3

⇒ l × b × h = 972 ⇒ 16 × 13.5 × h = 972

⇒ 972

m16 13.5

h =×

⇒ 9

m2

h =

⇒ h = 4.5 m

Hence, height of the cuboid = 4.5 m

69. The volume of a cuboid is 1296 m3. Its length is 24 m and its

breadth and height are in the ratio 3 : 2. Find the breadth and

height of the cuboid.

Ans.Length of the cuboid = 24 m

Let breadth of the cuboid be 3x

and height of the cuboid be 2x

Volume of the cuboid = 1296 m2

⇒ l × b × h = 1296 ⇒ 24 × 3x × 2x = 1296⇒ 144x2 = 1296

⇒ 2 1296

144x = ⇒ x2 = 9 m ⇒ 9x = ⇒ x = 3

Hence, the breadth of the cuboid = 3 × 3 = 9 m

Height of the cuboid = 2 × 3 = 6 m.

70. The surface area of a cuboid is 468 cm2. Its length and breadth are

12 cm and 9 cm respectively. Find its height.

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Ans. Let height of the cuboid be h cm

Surface area of the cuboid = 468 cm2

⇒ 2 [lb + bh + hl] = 468

⇒ 2 [ 12 × 9 + 9 × h + h × 12] = 468

⇒ 2 [ 108 + 9h + 12h] = 468

⇒468

21 1082

h + =

⇒ 21h = 234 – 108

⇒ 21h = 126 ⇒ 126

21h = ⇒ h = 6 cm

Hence, height of the cuboid = 6 cm.

71. A room 9 m long, 6 m broad and 3.6 m high has one door 1.4 m by 2

m and two windows, each 1.6 m by 75 cm. Find :

(i) the area of 4 walls, excluding doors and windows.

(ii) the cost of white-washing the walls from inside at the rate of

Rs. 2.50 per m2;

(iii) the cost of painting its ceiling at Rs. 5 per m2.

Ans. Length of the room (l) = 9 m

Breadth of the room (b) = 6 m

and height of the room (h) = 3.6 m

Area of 4 walls = 2 (l + b) × h

= 2 (9 + 6) × 3.6 = 2 × 15 × 3.6 m2 = 108 m2

Area of the door = 1.4 × 2 m2 = 2.8 m2.

Area of two windows 2752 1.6 m

100

= ×

2 23

2 1.6 m 2.4 m4

= × =

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(i) Area of 4 walls, excluding doors and windows

= (108 – 2.8 – 2.4) m2 = (108 – 5.2) m2 = 102.8 m2.

(ii) Cost of white-washing the walls of 1m2 = Rs 2.50

Cost of white-washing the walls of 102.8 m2 = Rs. 102.8

× 8 250 = Rs 257.

(iii) Area of the ceiling = l × b = (9 × 6) m2 = 54 m2

Cost of painting the ceiling of 1m2 = Rs 5

Cost of painting the ceiling of 54 m2

= Rs 54 × 5 = Rs 270.

72. The length, breadth and height of a cuboid are in the ratio 7 : 6 : 5.

If the surface area of the cuboid is 1926 cm2, find its dimensions.

and also, find the volume of the cuboid.

Ans. Surface area of cuboid = 1926 cm2

Ratio in the length, breadth and height of a cuboid = 7 : 6 : 5

Let length of the cuboid (l) = 7x

Preadth of the cuboid (b) = 6x

and height of the cuboid (h) = 5x

Surface area = 2 (l × b + b × h + h × l)

= 2(7x × 6x + 6x × 5x + 5x × 7x)

= 2(42x2 + 30x2 + 35x2) = 2 × 107 x2

= 241 x2

∴ 214x2 = 1926 ⇒ 2 1926

214x = = 9 = (3)2 ∴ x = 3

Hence, length (l) = 7x = 7 × 3 = 21 cm

Breadth (b) = 6x = 6 × 3 = 18 cm

and height (h) = 5x = 5 × 3 = 15 cm

Then the volume of cuboid = l × b × h = (21 × 18 × 15) cm3

= 5670 cm3

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73. If the areas of the three adjacent faces of a cuboidal box are 120 cm2,

72 cm2 and 60 cm2 respectively, then find the volume of the box.

Ans. Let l be the length, b be the breadth and h be the height of the cuboid

∴ l × b = 120 cm2 ...(i) b × h = 72 cm2 ...(ii)

h × l = 60 cm2 ...(iii)

Multiplying (i), (ii), (iii) we have

l2 b2 h2 = 120 × 72 × 60 = 518400 ∴ 518400 720lbh = =

Hence, volume of the cuboidal is box 20 cm3.

74. The external dimensions of wooden box, open at the top are 54 cm

by 30 cm by 16 cm. It is made of wood 2 cm thick. Calculate :

(i) the capacity of the box (ii) the volume of wood.

Ans. Internal length of the open box = 54 – 4 = 50 cm

Internal breadth of the open box = 30 – 4 = 26 cm

Internal height of the open box = 16 – 2 = 14 cm.

(i) Capacity of the open box = (50 × 26 × 14) cm3 = 18200 cm3.

(ii) Volume of the wood = (54 × 30 × 16 – 50 × 26 × 14) cm3

= (25920 – 18200) cm3 = 7720 cm3.

75. Three cubes of metal with edges 5 cm, 4 cm and 3 cm respectively

are melted to form a single cube. Find the lateral surface area of

the new cube formed.

Ans.Volume of three cubes

= [ (5)3 + (4)3 + (3)3] cm3 = [125 + 64 + 27] cm3 = 216 cm3

Now, volume of a big single cube = 216 cm3

i.e. edge of a big single cube 216 cm= = 6 cm

Hence, lateral surface area of the new big cube

= 4 (edge)2 = 4(6)2 cm2 = 4 × 36 cm2 = 144 cm2

perimeter and area of plane figures, volume and …©edulabz interna tional math class viii 1...

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