Practice Set 7.2 Geometry 10th Standard Maths Part 2 Chapter 7 Mensuration Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 7.2 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 7 Mensuration.

10th Standard Maths 2 Practice Set 7.2 Chapter 7 Mensuration Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 7.2 Chapter 7 Mensuration Questions With Answers Maharashtra Board

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