Sir please help me to find out the chapter surface areas and volumes of ncert class 10 .Find the best and easiest way to solve the tough question of exercise 13.4 question number 2 . Give me the easiest and simplest question of this chapter.The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the surface area of the frustum.
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The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the surface area of the frustum.Q.2
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Given,
Slant height (l) = 4 cm
Circumference of upper circular end of the frustum = 18 cm
∴ 2πr1 = 18
Or, r1 = 9/π
Similarly, circumference of lower end of the frustum = 6 cm
∴ 2πr2 = 6
Or, r2 = 6/π
Now, CSA of frustum = π(r1+r2) × l
= π(9/π+6/π) × 4
= 12×4 = 48 cm2