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Mensuration

1. The perimeter of a right - angled triangle is 60

c.m. Its hypotenuse is 28 c.m. The area of the

triangle is :-

(1) 120 c.m.2 (2) 240 c.m.2

(3) 75 c.m.2 (4) 60 c.m.2

2. The cost of tiling a rectangular room is Rs.

12852 at the rate of Rs. 68 per square ft. The

length of the room is 8 ft. less than the side of a

square whose area is 841 square ft. What is the

breadth of the rectangular room?

(1) 8 ft. (2) 12 ft.

(3) 7 ft. (4) None of these

3. The area of a rectangle is equal to the area of a

circle with circumference equal to 52.8 meters.

What is the length of the rectangle if its breadth

is 5.5 meters?

(1) 40.32 m (2) 42.56 m

(3) 36.82 m (4) None of these

4. A rectangular carpet has an area of 176 sq.

meters and a perimeter of 52 meters. The length

of its diagonal is?

(1) 19 m (2) 22 m

(3) 16 m (4) None of these

5. The length of a rectangle is three times of its

breadth If the length of its diagonal is 7 c.m,

then the perimeter of the rectangle is?

(1) 48 cm. (2) 52 cm.

(3) 56 cm. (4) 64 cm.

6. A cuboid of dimension 27 cm. × 16 cm. × 15

cm. is melted and smaller cubes are of side 6

cm. is formed. Find how many such cubes can

be formed?

(1) 15 (2) 30

(3) 45 (4) 40

7. A circular wire of diameter 84 cm. is bent in the

form of a rectangle whose sides are in the ratio

13:9. The area of the rectangle is?

(1) 4096 cm2 (2) 4224 cm2

(3) 4212 cm2 (4) 4190 cm2

8. The area of a circle is 440 square cm. The area

of square inscribed in this circle will be?

(1) 320 cm.2 (2) 260 cm.2

(3) 280cm.2 (4) 240 cm2

9. A copper sphere of diameter 36 c.m is drown

into a wire of diameter 8 mm. Find the length of

the wire?

(1) 512 m (2) 486 m

(3) 408 m (4) 398 m

10. Find the difference between the volume of a

cylinder and volume of a cone having same

radius and same height. Radius is equal to length

of a rectangle having area 1120 cm2 and breadth

32 cm and height is half of the radius.

(1) 40001.01 cm3 (2) 44916.67 cm3

(3) 43333.33 cm3 (4) 51111.99 cm3

11. A solid cone of radius 21 cm and height 10/7 of

its radius is melted to form four solid cylinders

of radius 7 cm. Find the difference between the

height of the cone and height of each of the

solid cylinders.

(1) 9 cm (2) 6 cm

(3) 7.5 cm (4) 5.5 cm

12. A path of width 4m is running around a

rectangular field of length 24 m and breadth 18

m. Find the cost of flooring the path at the cost

of Rs.300/m2.

(1) Rs.60000 (2) Rs.90000

(3) Rs.120000 (4) Rs.100000

10



13. The sum of areas of two rectangles (R1 and R2)

of same breadth is 720 cm2, and the ratio of

length of rectangle R1 and length of rectangle

R2 is 5:4 respectively. If the area of square

having length of side equal to the breadth of

rectangle is 256 cm2, then find the area of

rectangle R1.

(1) 320 cm2 (2) 400 cm2

(3) 240 cm2 (4) 480 cm2

14. If the length and breadth of a rectangular

seminar hall are each increased by 2m, then the

area of floor is increased by 42 sq.m. If the

length is increased by 2m and breadth is

decreased by 2m then the area is decreased by

10sq.m. The perimeter of the floor is

(1) 36m (2) 38m

(3) 34m (4) 32m

15. The difference between two parallel sides of

trapezium is 8cm. The perpendicular distance

between them is 38cm. if the area of the

trapezium is 950 cm^2, find the length of the

parallel sides.

(1) 27, 35 (2) 31, 40

(3) 29, 21 (4) 41, 49

16. The diameter of a rod is 2 cm and its length is

32 cm. This rod is drawn into a wire of length

128 cm and uniform thickness. What is the

thickness of the new wire?

(1) 2 cm (2) √2 cm

(3) 1 cm (4) 1.5 cm

17. Rajan is trying to make a cube using three old

cubes having edges 6 cm, 8 cm and 10 cm. What

will be the edge of new cube?

(1) 3 cm (2) 5 √2 cm

(3) 12 cm (4) 6 cm

18. The area of a rectangular parking space is 144

m2. If the length had been 6 meters more, the

area would have been 54 m2 more. The original

length of the parking space is

(1) 22 metres (2) 18 metres

(3) 16 metres (4) 24 metres

19. What is the cost of ploughing a land which

length and breadth in the ratio of 12:6 and its

perimeter is 840 feet. If the cost of needed to

plough a land is Rs.11 per 100 sq. feet.

(1) Rs.5648 (2) Rs.4312

(3) Rs. 6,660 (4) Rs.7,550

20. The barrel of a Ink pen is cylindrical in shape

which radius of base as 0.12 cm and is 7 cm

long. One such barrel in the pen can be used to

write 450 words. A barrel full of ink which has a

capacity of 17 cu. cm can be used to write how

many words approximately?

(1) 15423 (2) 21342

(3) 17645 (4) 24147

21. PQR is a right angle triangle with right angle at

Q. if the semi-circle on PQ with PQ as diameter

encloses on area of 243 sq.cm and the semicircle

on QR with QR as diameter encloses an area of

108 sq.cm then the area of the semicircle on PR

with PR as diameter will be.

(1) 305 sq.cm (2) 325 sq.cm

(3) 351 sq.cm (4) 362 sq.cm

22. The length and breadth of the hall are 60 feet

and 30 feet respectively. Square tiles of a 4 feet

length of different colors are to be laid on the

floor. Red tiles are laid in the 1st row on all

sides. If green tails are laid in the one – two time

of the remaining and yellow in the rest,

approximately how many yellow tiles will be

there?

(1) 52 (2) 63

(3) 72 (4) 35

23. A rectangular sheet, when folded into 2

corresponding parts had a perimeter 102cm for

each part folded along one set of sides and the

same is 114cm when folded along the other set

of sides. What is the area of the sheet?

(1) 1520 sq.cm (2) 1402 sq.cm

(3) 1310 sq.cm (4) 1260 sq.cm

24. Smallest angle of a triangle is equal to two-third

the smallest angle of a quadrilateral. The ratio

between the angles of the quadrilateral is

3:4:5:6. Largest angle of the triangle is twice its

smallest angle. What is the sum of second

largest angle of the triangle and largest angle of

the quadrilateral?

(1) 160º (2) 180º

(3) 190º (4) 170º

25. The volume ratio of Cylinder 1 to Cylinder 2 is

5:7 and its height is in the ratio of 2:3.Find the

Radius ratio of both the cylinders?

(1) 10:21 (2) 10 1/2 : 21 ½

(3) 15 1/2 : 14 ½ (4) 13:14

26. Amrita goes walking at morning. She walks

along the boundary of the walking park which is

in rectangular shape and she walks at a speed of



8kmph completes her one round in 15 minutes.

Find the area of park, if the length and breadth is

in the ratio of 3:2?

(1) 240000 (2) 126500

(3) 365200 (4) 250000

27. The radius of the base of a cylindrical vessel is

21 cm and its height is 12 cm. The vessel is full

of milk. How many cylindrical vessels each

having diameter 14 cm and height 4 cm can be

filled with the milk in the big cylindrical vessel?

(1) 24 (2) 20

(3) 22 (4) 27

28. Length, breadth and height of a cuboidal box is

6 cm, 4 cm and 3 cm, respectively. A person

keeps 18 such box one above the other and then

paints it, and amount incurred was Rs. x. If all

boxes are painted separately then amount

incurred would be Rs. y. Find the difference

between ‘x’ and ‘y’ if the cost of painting the

box per cm2 is Rs. 2.

(1) Rs. 1520 (2) Rs. 1632

(3) Rs. 1266 (4) Rs. 1484

29. Length of a cuboid is 3 cm, and the ratio of its

breadth and height is 1:3. If the length of

diagonal is 13 cm2, then find the volume of the

cuboid.

(1) 184 cm3 (2) 164 cm3

(3) 254 cm3 (4) 144 cm3

30. A solid cone is cut vertically into two equal

parts. If radius and volume of the whole solid

cone before it is cut is 6 cm and 216√3 cm3,

respectively, then find the total surface area of

one of the cut portions.

[Use π = 3]

(1) 224.3 cm2 (2) 252.2 cm2

(3) 286.7 cm2 (4) 184.9 cm2

31. A cone is inscribed in a hollow cylinder. The

height and the base of cone and cylinder is the

same. The cost of painting the curved surface

area of the cylinder is Rs. 9072 at Rs. 14 per

cm2, and the curved surface area of cone is 405

cm2. Find the space remaining in the cylinder

after inscribing cone. [Take π = 3]

(1) 1902 cm3 (2) 1856 cm3

(3) 1920 cm3 (4) 1944 cm3

32. A piece of iron size, 14 inches in length, 11

inches in breadth and 3 inches in thick was

melted and remodified in the form of a rod of 14

inches diameter. Then find the length of the rod.

(1) 8 inches (2) 2 inches

(3) 7 inches (4) 3 inches

33. Radius of curvature at two points (A and B) of

spiral is given. What should be the

circumference of the spiral?

(1) 65. 84 cm (2) 62.80 cm

(3) 57.76 cm (4) 50.24 cm

34. Outside a rectangular garden a path of uniform

width of 3 meter is made. If the area of

rectangular garden is 640m2 and area of garden

including path is 916m2, then find the perimeter

of the garden.

(1) 30 m (2) 50 m

(3) 20 m (4) 80 m

35. The perimeter of a square is equal to the

perimeter of a rectangle of length 14 cm and

breadth 20 cm. Find the circumference of a

semicircle (approx.) whose diameter is equal to

the side of the square.

(1) 30.23cm (2) 43.71 cm

(3) 17.5cm (4) 50.47cm

36. Area of a rectangular plot is 75√3 cm2. If the

sum of its diagonals is four times the shorter

side, then the longer side would be?

(1) 12 cm (2) 15 cm

(3) 18 cm (4) 21 cm

37. A circle of maximum possible area is cut out

from a rectangular cardboard sheet of length 10

cm and breadth 6 cm. What is the remaining

area of the cardboard?

(1) 3(20 - 6π) cm2 (2) 3(20 - 3π) cm2

(3) 3(20 - 5π) cm2 (4) 3(20 - 2π) cm2

38. The ratio between sides of a square and an

equilateral triangle is 11:8. If the area of triangle

is 144√3 m2, find the perimeter of square

(1) 124 meter (2) 96 meter

(3) 132 meter (4) None of these



39. A rectangle having length as 21 cm and breadth

as 10 cm is rotated about its longer edge as axis.

Find the volume of the solid generated.

(1) 72.07 cm3 (2) 167.05 cm3

(3) 214.06 cm3 (4) 102.51 cm3

40. A solid spherical ball of volume 38808 cm3 is

broken into two equal halves. Find the increase

in the total surface area of both hemispheres

with respect to the sphere (in cm2).

(1) 2772cm2 (2) 2277cm2

(3) 2727cm2 (4) 2722cm2

41. The sum of areas of two rectangles (R1 and R2)

of same length is 594 cm2, and the ratio of

breadth of rectangle R1 to breadth of rectangle

R2 is 5:4. If the area of square having length of

side equal to the length of rectangle is 484 cm2,

then find the area of rectangle R1.

(1) 240 cm2 (2) 280 cm2

(3) 330 cm2 (4) 360 cm2

42. Curved surface area of a cylindrical vessel is

1584 cm2. If the height of the vessel is 18 cm,

then find the cost of painting the total surface

area of the vessel at the rate of Rs. 2 per cm2.

(1) Rs. 5632 (2) Rs. 5864

(3) Rs. 5540 (4) Rs. 5780

43. The length and breadth of a rectangular park is

48 m and ‘x’ m, respectively. If the cost of

fencing the rectangular park at the rate of Rs.

14/m is Rs. 2352, then find the area of a square

field having side equal to ‘x’.

(1) 1296 m2 (2) 1024 m2

(3) 1600 m2 (4) 1225 m2

44. If the volume of the cylindrical tank of height 14

cm and radius ‘r’ cm is 6336 cm3, then find the

value of ‘r’.

(1) 9 cm (2) 10 cm

(3) 12 cm (4) 15 cm

45. Length and breadth of a rectangular park is 44 m

and 38 m, respectively. A path is made around

the park from outside of width 3.5 m and paved

with marbles. Find the cost of paving the path, if

it costs Rs. 2 per cm2.

(1) Rs. 1142 (2) Rs. 1870

(3) Rs. 1246 (4) Rs. 1484

46. A took 20 seconds to cross a rectangular field

diagonally walking at the rate of 68 m/min and

B took the same time to cross the same field

along its side walking at the rate of 50 m/min.

The area of the rectangular field is?

(1) 234 m² (2) 260 m²

(3) 216 cm² (4) None of these

47. If a is the area, b is the circumference and c is

the diameter of circle then the value of a/bc

(1) 4 : 1 (2) 1 : 4

(3) 1 : 2 (4) 2 : 1

48. The radius and height of a right circular cone are

in the ratio 5:12. If its volume is 800 π cm3,

what is its slant height?

(1) 9 cm (2) 13 cm

(3) 26 cm (4) 17

49. Area of the square is equal to the area of the

rectangle. The ratio of the length and breadth of

the rectangle is 4:1 and the perimeter of the

square is 64 cm. Find the perimeter of the

rectangle

(1) 64 cm (2) 48 cm

(3) 72 cm (4) 80 cm

50. A road of width 7m is running around a circular

field from outside. The radius of the circular

field is 28m. Find the cost of fencing the road at

the rate of Rs.16/m2.

(1) Rs.18564 (2) Rs.22456

(3) Rs.12986 (4) Rs.22176



Mensuration (Solution)

Ans.1(4) Let base = b cm. Height = h c.m.

b + h + 28 = 60

b + h = 32

(b+h)2 = (32)2..........(I)

Also, b2 + h2 = 282

2bh = 60×4

bh = 120

bh = 60

Area = 60 c.m.2

Ans.2(4) The area of rectangular room = Sq.ft

The side of the square room = = 29 ft

length = 29-8 = 21 ft

breadth = = 9 ft .

Ans.3(1) Circumference of the circle = 52.8 m

2r = 52.8

2× ×r = 52.8

r = 8.4 m

Area of the circle = r2

×8.4×8.4 = 221.76 m2

Area of rectangle

Length of the rectangle = = 40.32m

Ans.4(4) 2(l+b)=52

l+b = 26 , and lb = 176 m2

Diagonal =

=

= = = 18 m

Ans.5(3) Let breadth = x cm then, length = 3x cm

x2+(3x)2 =

10x2= 490

x = 7 cm.

then, length = 21 cm.

bredth = 7 cm.

Perimeter = 2 (l+b)

= 2 (7+21)

= 56 c.m

Ans.6(2) No. of cubes =

Ans.7(3) Length of wire = 2x =264 cm.

Perimeter of rectangle = 2(13x+9x)=264

cm.

= 44x=264

x=6

Area of Rectangle = (13x6)x(9x6)

= 4212 cm2

Ans.8(3) R2 = 440 = R2=440× = 140

So, R = ×(Diagonal)

Diagonal = 2 R

Area of the square = ×(Diagonal)2

6= ×4R2 = 2R2

= 2×140

= 280 cm2

Ans.9(2) Volume of sphere = × × ×

= 7776 cm3

Volume of wire = ( x0. 4×0.4×h) cm2

7776 = × × × h

h = 48600 cm

h = m

h = 486 m

Ans.10(2) Area of rectangle = length x breadth

=> 1120 = length x 32

=> Length = 1120/32

=> Length = 35 cm

Radius = 35 cm

Height = 35/2 cm

Volume of cylinder – Volume of cone

= πr2h – 1/3 πr2h

= πr2h x (3 – 1)/3

= 2/3 x πr2h

= 2/3 x 22/7 x 35 x 35 x 35/2

= 44916.67 cm3

Ans.11(3) We know that volume of cone

= 1/3πr2h

And volume of cylinder = πr2h

1/3 x 22/7 x 21 x 21 x 10/7 x 21 = 4 x

22/7 x 7 x 7 x h

= h = 1/3 x 22/7 x 21 x 21 x 30 x ¼ x 7/22

x 1/7 x 1/7

= h = 22.5 cm

Required difference = 10/7 x 21 – 22.5

= 30 – 22.5

= 7.5 cm

Ans.12(3)

Area of the path = (32 x 26) – (24 x 18)

2 2 2 2 2( ) 32 28b h b h

1

2

12852

68

841

189

21

227

227

221.76

5.5

2 2 2l b (l b) 2lb

226 2x176

676 352 324

2

7 10

27x16x1530

6x6x6

22 84

7 2

π

7

22

1

2

1

2

4

3

36

2

36

2

36

2

4

10

4

10

48600

100



= 832 – 432

= 400 m2

Cost of flooring the path = 400 x 300

= Rs.120000

Ans.13(2) Let, length of rectangle R1 and rectangle

R2 be 5x cm and 4x cm respectively.

Breadth of rectangle = side of square

= √256 = 16 cm

According to question,

Sum of areas of two rectangles (R1 and

R2)

= 720

5x × 16 + 4x × 16 = 720

16(5x + 4x) = 720

16x = 720/9

x = 80/16, x = 5

So, the length of rectangle R1 = 5 × 5

= 25 cm

Therefore, area of rectangle R1 = 25 × 16

= 400 cm2

Ans.14(2) Let length = x metres

Breadth = y metres

Then (x+2) (y+2) – xy = 42

Xy+2x+2y+4 – xy = 42

2x +2y = 42 – 4

2x +2y = 38

X+y = 19 ------------->1

And xy – [(x+2)(y-2)] = 10

Xy – [ xy – 2x + 2y +4)] = 10

Xy – xy +2x – 2y + 4 = 10

2x – 2y =6

X – y = 3 ---------------->2

Adding (1) & (2)

2x = 22

X=11

Put x value in (1)

Y = 19 -11

=8

So length = 11m and breadth = 8m

Perimeter = 2(l+b)

=2(11+8)

=38m

Ans.15(3) Let the 2 parallel sides of the trapezium be

x cm and y cm

Then x – y = 8 ========>1

½ (x+y)× 38 = 950

19 (x+y) = 950

X+y = 950/19

X+ y = 50 ==========>2

Solving 1 and 2

2x = 58

x= 29, y = 21

So the parallel sides are 29cm and 21 cm.

Ans.16(3) πR2H = πr2h

12 × 32 = r2 × 128

r2 = ¼

r = ½

Diameter = 1 cm

Ans.17(3) It has been given that the new cube is

made of old cubes. Volume of new

cubes, 6 cm, 8 cm And 10 cm The sum of

the volume of the cube with the sides will

be equal to.

Volume of a cube ( =side)3

Thus, the volume of the new cube

= (6)3 + (8)3 + (10)3

⇒ Volume = 216 + 512 + 1000

= 1728cm3

∴ Side of 1728 cubic cm volume

=∛(volume) = √1728 =12 cm. Ans.18(3) l × b = 144

(l + 6) × b = 144+54 = 198

So, (l + 6)/l = 198/144

1 + 6/l = 11/8

6/l = 11/8 – 1 = 3/8

l = 16 m

Ans.19(2) Let length be l and breadth be b.

Then the perimeter of the floor = 2 (l+b)

2(l+b) = 840

l+b =420

l = 12/18 ×420 =280 feet

b= 6/18 × 420 = 140 feet

Area = 140×280 = 39200 sq.feet

Cost of laying = 392×11 = Rs.4312

Ans.20(4) Volume of the barrel of Ink pen = πr2h =

22/7 × 0.12×0.12 × 7 = 0.3168cu cm

A barrel which has capacity 0.3168 cu. cm

can write 450 words

So which has capacity 17 cu cm can write

= 450/0.3168×17= 24147 words

Ans.21(3) Required area = π/2 (PR/2)2

=π/2 (PR)2/4

=π/2 [(PQ2+QR2)/4]

=π/2 [PQ2/4 + QR2/4]

=π/2 (PQ2/4) + π/2 (QR2/4)

=243 +108

=351 sq.cm

Ans.22(4) Area left after laying red tiles = [(60-

8)×(30-8)]

= 52×22

= 1144 sq.ft

Area under green tiles = 1144 / 2 = 572

sq.ft

Area under yellow tiles = 1144-572=572

sq.ft

Number of yellow tiles = 572/ 16

= 35



Ans.23(4) When folded along breadth, we have 2

(l/2 + b) = 102

L + 2b = 102 ---------------->1

When folded along length, we have 2

(l+b/2) = 114

2l+b = 114 ------------------>2

Multiply (2) by 2, we get

4l + 2b =228 --------------->3

Subtract 1 from 3

3l = 126 => l=42

Put l value in (1) we get b = 30

Area of the sheet = (42×30) cm2

=1260 sq.cm

Ans.24(2) Let be ratio = x

Sum of angles of the quadrilateral

⇒ 3x + 4x + 5x + 6x = 360o

⇒ 18 x = 360o

⇒ x = 20o

Angles of quadrilateral = 60o, 80o, 100o,

120o

Smallest angle of triangle = 40o

Largest angle of triangle = 80o

Sum = 60o + 120o

=180o

Ans.25(3) πR2h1 / πr2h2 = 5/7

R2 / r2 3 =5/7

R/r =15 1/2 : 14 1/2

Ans.26(1) Circumference = (8000/60)×15

= 2000m

2(l+b) = 2000

l+b =1000

3x+2x =1000

5x= 1000

x=200

Area = 3x×2x= 600×400 =240000m2.

Ans.27(4) Radius of each of the small cylindrical

vessels

= 14/2 = 7 cm

Let, required number of cylindrical

vessels = n

We know that,

Volume of cylinder = πr2h

According to the question

22/7 x 21 x 21 x 12 = n x 22/7 x 7 x 7 x 4

=> n = 22/7 x 21 x 21 x 12 x 7/22 x 1/7 x

1/7 x ¼

=> n = 27

Ans.28(2) Total surface area of each box

= 2 × [(6 × 4) + (4 × 3) + (3 × 6)]

= 2 × (24 + 12 + 18) = 108 cm2

Total cost incurred for painting 18 boxes

separately = 2 × 18 × 108 = Rs. 3888

So, y = 3888

Total surface area ifall boxes are kept one

above the other = 2 × [(6 × 4) + (4 × 54) +

(54 × 6)] = 2 × (24 + 216 + 324)

= 1128 cm2

Total cost incurred for painting the boxes

that are kept one above the other

= 2 × 1128 = Rs. 2256

So, x = 2256

Required difference = 3888 – 2256

= Rs. 1632

Ans.29(4) Let breadth and height of the cuboid be x

cm and 3x cm, respectively.


x2 + (3x)2 + 32 = 132

x2 + 9x2 + 9 = 169

10x2 = 160

x2 = 16

x = 4 cm

Breadth of cuboid = 4 cm

Height of cuboid = 12 cm

Volume of cuboid = 3 × 4 × 12 = 144 cm3

Ans.30(1) Let, height of cone = ‘h’ cm

Given, volume of the solid cone

= 216√3 cm3

Radius of cone = 6 cm

So, 1/3 × π r2 h = 216√3

1/3 × 3 × 6 × 6 × h = 216√3

h = 6√3 cm

Diameter of base circular part of solid

cone

= 12 cm

Slant height of solid cone = √{(6√3)2 +

62}

= √144 = 12 cm

So, total curved surface area of the cut

portion

= (πrl/2) + (πr2/2) + (Area of the

equilateral triangle with side 12 cm)

= (3 × 6 × 12)/2 + (3 × 62)/ 2 + (√3/4 ×

122)

= 108 + 54 + 36√3

= (162 + 36√3) cm2

= 224.3 cm2

Ans.31(4) Let, radius and height of cone and

cylinder be r cm and h cm, respectively.

And, slant height of cone = l cm

Curved surface area of cylinder

= 9072/14 = 648 cm2

So, 2 × 3 × r × h = 648

rh = 108

r = 108/h --------- (i)

And, 3 × r × l = 405

r = 135/l --------- (ii)

From (i) and (ii)



108/h = 135/l

l/h = 135/108

l/h = 5/4

Let, slant height and height of cone be 5x

cm and 4x cm, respectively.

So, 25x2 = 16x2 + r2

9x2 = r2

r = 3x

So, 3x = 108/4x

12x2 = 108

x2= 9

x = 3

Height of cone = 4x = 12 cm

Required space = 3 × 9 × 9 × 12 – 1/3 × 3

× 9 × 9 × 12 = 1944 cm3

Ans.32(4) Given that diameter= 14/2= 7 inches

According to the question,

14×11×3= 22/7×7×7×l

462=154×l

l= 3 inches

Ans.33(3)

Minimum and maximum value of the

circumference of the spiral will be the

circumferences of the circles given in the

diagram

Circumference of smaller circle

= 2⨉3.14⨉8 = 50.24 cm

Circumference of bigger circle

= 2⨉3.14⨉10 = 62.80 cm

50.24 < Circumference of the spiral <

62.80

Thus option C is the valid answer

Ans.34(4) Let length and Width of the inner

boundary be x and y respectively

Given that area of the garden=640m2

(i.e) xy=640m2……………… (1)

Now length of outer boundary path=x+6

And width of outer boundary path=y+6

Hence Area of garden including path

=(x+6)(y+6)=916

(i.e) xy+6x+6y+36=916………………….

(2)

From (1) and (2), we have

640+6x+6y+36=916

x+y=40

Thus perimeter of the

garden=2(x+y)=80m

Ans.35(2)

Ans.36(2) 𝑙 × 𝑏 = 75√3 - - - - (1)

2√𝑙2 + 𝑏2 = 4𝑏

𝑙2 + 𝑏2= 4𝑏2

𝑙2 = 3𝑏2

𝑏 = 𝑙/√3 Putting this value in equation 1.

𝑙 × 𝑙/√3 = 75√3

𝑙2 =225

𝑙 = 15 Ans.37(2) A circle cut out from the rectangular

cardboard can attain a maximum diameter

of 6cm.

So, maximum possible area of the circle

= π × 32 = 9π cm2

Area of the rectangular cardboard

= 10×6 = 60 cm2

Therefore, remaining area of cardboard

= 60 - 9π cm2 = 3(20 - 3π) cm2

Hence, option 2.

Ans.38(3) Let the side of a triangle = x meter

√3

4 x2 = 144√3

x2 = 144×4

x = 24 meter

perimeter of square = 33×4 = 132 meter

Ans.39(2) As the rectangle is rotated about its longer

edge, 21 cm becomes the height of the

cylinder while 10 cm becomes the

circumference of the base of the cylinder.

2 × π × r = 10

So, r = 35/22 cm

Thus, volume of cylinder = π × r2 × h

= (22/7) × (35/22) × (35/22) × 21

= 3675/22 cm3 = 167.05 cm3

Ans.40(1) Let the radius of the sphere be ‘r’ cm.

Given, Volume of sphere = 38808 cm3

(4/3) × π × r3 = 38808

r3 = 9261

r = 21 cm

Now, total surface area of sphere before

breaking into two halves = 4πr2

Total surface area of two hemispheres

= 2 × 3πr2 = 6πr2

Thus, increase in total surface area

= 6πr2– 4πr2 = 2πr2

= 2772 cm2

Ans.41(3) Let, breadth of rectangle R1 and rectangle

R2 be 5x cm and 4x cm respectively.

Length of rectangle = side of square

= √484 = 22 cm




Sum of areas of two rectangles (R1 and

R2)

= 594

22 × 5x + 22 × 4x = 594

22(5x + 4x) = 594

9x = 594/22

9x = 27, x = 27/9 = 3

So, the breadth of rectangle R1= 5 × 3

= 15 cm

Therefore, area of rectangle R1

= 22 × 15 = 330 cm2

Ans.42(1) Let, radius of vessel be ‘r’ cm.

So, 2 × 22/7 × r × 18 = 1584

r = 14 cm

Required cost = 2 × 22/7 × 14 × (14 + 18)

×2

= Rs. 5632

Ans.43(1) So, 2 × (48 + x) × 14 = 2352

48 + x = 84

x = 84 – 48 = 36

So, area of square field = 36 × 36

= 1296 m2

Ans.44(3) Volume of cylinder = π r2 h

6336 = 22/7 × r2 × 14

r2 = 144

r = 12 cm

So, the value of r = 12 cm

Ans.45(3) Area of path paved with marble

= (44 + 3.5 × 2) × (38 + 3.5 × 2) – 44 × 38

= 2295 – 1672 = 623 cm2

Required cost = 623 × 2 = Rs. 1246

Ans.46(4)

Ans.47(2) We know that,

Area of a circle = πd2 /4

Where, d = Diameter of the circle

⇒ a = πc2 /4

Also, Circumference of the circle = πd

⇒ b = πc

a/bc = (πc2 /4)/πc2 = 1 : 4

Ans.48(3) As per the given data

r/h = 5/12

⇒ h = 12r/5

Also given that volume of the cone is 800

π cm 3

⇒ 1/3 π r2 h = 800 π

⇒ 1/3 × r2x12r/5 = 800

⇒ 12r3 /15 = 800

⇒r3= 1000

⇒ r = 10 cm

∴height of the cone = 12r/5 = 12 (10)/5 =

24 cm

∴slant height of the cone = √(242 + 102) =

26 cm

Ans.49(4) Side of a square = 4a =64

=> a= 64/4

= 16 cm

Area of a rectangle = Area of the square

=16×16 = 256

(i.e) Area of a rectangle =256

(i.e) 4x×x = 256

=> x2 = 256/4

= 64

=> x= 8 cm

Perimeter of the rectangle = 2 (l+b) = 2

(4x+x)

= 2 (4×8+8)

= 2×40

= 80 cm

Ans.50(4) Area of the circular field = π(R2 – r2)

= [22/7 x (28 + 7)2 ]– [22/7 x (28)2]

= 22/7 x (35)2 – 22/7 x (28)2

= 3850 – 2464

= 1386 m2

Cost of fencing = 16 x 1386 = Rs.22176

Date post:	21-Feb-2022
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