Question

The volume of a frustum of a right circular cone is 52π cm3. Its altitude is 3cm and the measure of its lower radius is three times the measure of its

upper radius.


5. What is the radius of the upper base?

6. What is the slant height of the frustum?

7. Find the lateral area of the frustum.

Answer 1

5. Consider frustrum of a right circular cone

Let $r_1=$ the radius of the upper base, $r_2=$ the radius of the lower base, and $L=$ the slant height. For any Frustum, the volume is

$$V=\dfrac{1}{3}(A_1+A_2+\sqrt{A_1A_2})h$$

The volume of the frustrum of the cone

$$V_c=\dfrac{1}{3}(\pi r_1^2+\pi r_2^2+\sqrt{\pi r_1^2\pi r_2^2})h$$

$$=\dfrac{\pi}{3}( r_1^2+ r_2^2+r_1r_2)h$$

Given

$r_2=3r_1, h=3\ cm,V=52 \pi \ cm^3$

$$V=\dfrac{\pi}{3}( r_1^2+ (3r_1)^2+r_1(3r_1))(3)=52\pi$$

$$13r_1^2=52$$

$$r_1=2 \ cm$$

6. From the right triangle by the Pythagorean Theorem

$$L^2=(r_2-r_1)^2+h^2$$

$$L=\sqrt{(3r_1-r_1)^2+h^2}$$

$$L=\sqrt{4r_1^2+h^2}$$

$$L=\sqrt{4(2)^2+(3)^2}=5 (cm)$$

7. The lateral area of the frustum of a right circular cone is

$$A_L=\dfrac{1}{2}(2\pi r_2+2\pi r_1)L$$

$$A_L=\pi(3 r_1+ r_1)L$$

$$A_L=4\pi r_1L$$

$$A_L=4\pi (2)(5)=40\pi(cm^2)\approx125.66(cm^2)$$

The lateral surface area of the opening is $40\pi\ cm^2\approx125.66\ cm^2.$