Chapter 7 Mensuration Class 10 Maharashtra Board TextBook Solution
Practice set 7.2 | Q 2| Page 148
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
Answer:-
Radius of one circular end, r1 = 14 cm
Radius of other circular end, r2 = 7 cm
Height of the bucket, h = 30 cm
Therefore, Volume of water in the bucket = Volume of frustum of cone = (1/3) * π * h * (r1^2 + r1r2 + r2^2)
= (1/3) * (22/7) * 30 * (14^2 + 147 + 7^2)
= (1/3) * (22/7) * 30 * 343
= 10780 cm^3
= 10780/1000
= 10.780 L
Thus, the bucket can hold 10.780 litres of water.
Solution
Here, r1 = 14 cm, r2 = 6 cm and h = 6 cm.
Slant height of the frustum, l
[= sqrt{h^2 + left( r_2 – r_1 right)^2} ]
[ =sqrt{6^2 + left( 14 – 6 right)^2}]
[= sqrt{6^2 + 8^2} ]
[ =sqrt{36 + 64} ]
[= sqrt{100}= 10 cm]
[ Curved surface area of frustrum ]
[= pileft( r_1 + r_2 right)l]
[= 3 . 14 times left( 14 + 6 right) times 10]
[= 3 . 14 times 20 times 10]
[= 628 {cm}^2]
[∴ the curved surface area of frustum is 628 sq. cm ]
[Here, r1
= 14 cm, r2
= 6 cm and h ]
[= 6 cm.
Slant height of the frustum, l ]
[= sqrt{h^2 + left( r_2 – r_1 right)^2} ]
[= sqrt{6^2 + left( 14 – 6 right)^2} ]
[= sqrt{6^2 + 8^2} ]
[= sqrt{36 + 64} ]
[= sqrt{100}]
[= 10 cm
Total surface area of frustrum
[= pileft( r_1 + r_2 right)l + pi r_1^2 + pi r_2^2 ]
[= 3 . 14 times left( 14 + 6 right) times 10 + 3 . 14 times {14}^2 + 3 . 14 times 6^2 ]
[= 3 . 14 times 20 times 10 + 3 . 14 times 196 + 3 . 14 times 36]
[= 628 + 615 . 44 + 113 . 04]
[ = 1356 . 48 { cm}^2]
Chapter 7 Mensuration Textbook Solution 7.2
Practice set 7.2
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