PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3

PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3

PSEB 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3

Note: Assume π = 22/7, unless stated otherwise.

Question 1.
Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.
Answer:

Question 5.
What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm.
(Use π = 3.14)
Answer:
For the conical tent to be made, radius r = 6 m and height h = 8 m.
l² = h² + r² = 8² + 6² = 64 + 36 = 100
∴ l = √100 = 10 m
Area of the tarpaulin used in making tent
= Curved surface area of conical tent
= πrl
= 3.14 × 6 × 10 m²
= 188.4 m²
Now, the width of the tarpaulin is 3 m.
∴ Length of tarpaulin required = 188.4/3 m
= 62.8 m
But, 20 cm, i.e., 0.2 m of tarpaulin is required more for margins and wastage.
∴ Total length of the tarpaulin required = 62.8 + 0.2 m = 63 m
Thus, total length of tarpaulin required is 63 m.

Question 6.
The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of whitewashing its curved surface at the rate of ₹ 210 per 100 m².
Answer:

Question 7.
A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
Answer:
For the conical cap, radius r = 7 cm and height h = 24 cm.

Follow on Facebook page – Click Here

Google News join in – Click Here

Read More Asia News – Click Here

Read More Sports News – Click Here

Read More Crypto News – Click Here