## Maharashtra State Board Class 10 Maths Solutions Chapter 7 Mensuration Practice Set 7.2

Question 1.

The radii of two circular ends of frustum shaped bucket are 14 cm and 7 cm. Height of the bucket is 30 cm. How many litres of water it can hold? (1 litre = 1000 cm^{3})

Given: Radii (r_{1}) = 14 cm, and (r_{2}) = 7 cm,

height (h) = 30 cm

To find: Amount of water the bucket can hold.

Solution:

Volume of frustum = \(\frac { 1 }{ 3 } \) πh (r_{1}^{2 }+ r_{2}^{2} + r_{1} × r_{2})

∴ The bucket can hold 10.78 litres of water.

Question 2.

The radii of ends of a frustum are 14 cm and 6 cm respectively and its height is 6 cm. Find its

i. curved surface area,

ii. total surface area,

iii. volume, (π = 3.14)

Given: Radii (r_{1}) = 14 cm, and (r_{2}) = 6 cm,

height (h) = 6 cm

Solution:

i. Curved surface area of frustum

= πl (r_{1} + r_{2})

= 3.14 × 10(14 + 6)

= 3.14 × 10 × 20 = 628 cm^{2}

∴ The curved surface area of the frustum is 628 cm^{2}.

ii. Total surface area of frustum

= πl (r_{1}+ r_{2}) + πr_{1}^{2} + πr_{2}^{2}

= 628 + 3.14 × (14)^{2} + 3.14 × (6)^{2}

= 628 + 3.14 × 196 + 3.14 × 36

= 628 + 3.14(196 + 36)

= 628 + 3.14 × 232

= 628 + 728.48

= 1356.48 cm^{2}

∴ The total surface area of the frustum is 1356.48 cm^{2}.

iii. Volume of frustum

= \(\frac { 1 }{ 3 } \) πth(r_{1}^{2} +r_{2}^{2} + r_{1} × r_{2})

= \(\frac { 1 }{ 3 } \) × 3.14 × 6(14^{2} + 6^{2} + 14 × 6)

= 3.14 × 2(196 + 36 + 84)

= 3.14 × 2 × 316

= 1984.48 cm^{3}

∴ The volume of the frustum is 1984.48 cm^{3}.

Question 3.

The circumferences of circular faces of a frustum are 132 cm and 88 cm and its height is 24 cm. To find the curved surface area of frustum, complete the following activity. (π = \(\frac { 22 }{ 7 } \))

Solution:

Circumference1 = 27πr_{1} = 132 cm

Curved surface area of frustum = π (r_{1} + r_{2}) l

= π (21 + 14) × 25

=π × 35 × 35

= \(\frac { 22 }{ 7 } \) × 35 × 25

= 2750 cm^{2}