# Key with Explanations for Mensuration Practice Problems

Friends, here are the answers with explanations for our Mensuration Practice Problems. Please check the practice problems from here and try to solve on your own before checking these solutions. Happy Reading.

1. Width of the plate =  236.251/17.5 = 13.5 cm

Perimeter = 2 (length + width) = 2 (17.5 + 13.5) = 62 cm

2. Let the length of the plot be X m

So, Breadth = (75/100)x X  = (3X/4)m

But, here we know that the Perimeter = 1050

=> 1050 = 2 (X + (3X/4)

=> X = 300 m

So, the Area of the plot = 300 x ((3 x 300)/4)  = 67500 m2

3. Let the length be 1 and breadth also be 10.

So, the Area = 10 x 10 = 100

New length (10% excess) = 11,
New breadth (5% excess) = 9.5
New area = 11 x 9.5 = 104.5

So, Change in area = 104.5 - 100 = 4.5

Percentage change = ((4.5/100) x 100) = 4.5%

4. Area of the floor of the room = 800 x 500 = 400000 cm2

Area of the Tile = 20 x 10 cm2

So, number of Tiles required = (400000 / (20 x 10)) = 2000

5. Let the radius of larger circle be 'R' and small circle be 'r'.

So, 2πR = 132

=> 2 x (22/7) x R = 132

=> R = 21

2πr = 88

=> 2 x (22/7) x R= 88

=> r = 14

Difference in the areas = πR2- πr2
= π (R2- r2)

=> (22/7)(212 - 142) => (22/7) (21+14)(21-14)

=> (22/7)(35)(7) = 770 sq metres

6. Volume of sphere = (4/3)πr3

Volume of cone = (1/3)πr2h

Let the number of cones be 'X'

=>  (4/3)πr3  = ((1/3)πr2h )(X)

=> ((4/3) π 73 = (1/3) π (72)(2)(X)

=> X = 14

7. Length of the carpet = 10 - (0.5 + 0.5) = 9 m

Width of the carpet = 8 - (0.5+0.5) = 7 m

Area of carpet = 9 x 7 = 63

Required length of carpet = 63/0.25 = 252 m

So, Cost of carpet = 252 x Rs. 20 = Rs. 5,040.

8. Side of the square field = √ 56025 = 225 m

Length of the wire required = 4 x 225 = 900 m

So, Total cost = Rs. 15 x 900 = Rs. 13,500

9. Circumference of the circular plot

= (2640/12) = 220 m

2πr = 220 => 2 . (22/7) . r = 220

=> r = 35 m

Area of the plot

= πr2 = (22/7) . 35 . 35 = 3850 sq.m

So, cost of carpeting = 3850 x Rs. 85 = Rs. 3,27,250

10. Perimeter of rectangle = 2 (18+26) = 88 cm

Circumference of circle => 2πr = 88

=> 2 x (22/7) x r = 88

=> r = 14 cm

Area of circle = πr2
= (22/7) . 14 . 14 = 616 cm2

11. Radius of outer circle = 21 + 7 = 28 m

Road area = Area of outside circle - Area of inside circle

Road area =  π.(28)2- π.(21) 2 = 1078 sq.m

So, the cost = 1078 x Rs. 5 = 5,390

12. Lateral surface area of cylinder = 2πrh

Base area of cylinder = πr2

So, 2πrh = 3πr2

=> h:r = 3:2

13. Let the sides of two cubes be a1 and a2

a13-a23 =  8:27 => a1:a2 = 2:3

So, surface areas ratio = 6.a12-6. a22 = (2)2-(3)2 = 4:9

14. Length of the rope will be the radius of the circle = 14 m

Area of the circle = πr2  = (22/7) . 14 . 14 = 616 sq.m

So, Area of the ground that can be grazed = 616 sq.m

15. Length of the stick = length of the diagonal of the room

√(122+92+82) = √289 = 17 m

16. Quantity of the earth taken from the tank = 30 x 20 x 12 cu.m

Area of the field for the earth to be spreaded

= (375 x 40) - (30 x 20) = 14400

So, The rise in the level of the field = ((30 x 20 x 12) / 14400) x 100 = 50 cm

17. Area of the road = Area of outside rectangle - Area of inside rectangle

= 70 x 50 - 60 x 40

= 3500 - 2400 = 1100 m2

So, the cost of making 1100 m2  = 500 x 1100 = Rs. 5,50,000

18. Area of four walls = 2 x 5 (7.5 + 5.5) = 130 m2

Area of required paper =  130 m2

Breadth of the paper = 40 cm = 0.4 m

So, length of the paper = 130/0.4 = 325 m

So, cost of paper at 75 paise per meter = 325 x 0.75 = Rs. 243.75

19. Let the length and breadth be 10 and 10

Area = 10 x 10 = 100

New length = 12 (20% increase)

New breadth = 11 (10% increase)

New area = 12 x 11 = 132

So, Increment in area = 132 - 100 = 32 %

20. Let the number of spherical balls be "X"

Volume of cylinder = (X). (Volume of sphere)

π. 72. 28 = X . (4/3). π. 72

X = 3