Question #147377
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.
1
Expert's answer
2020-12-01T02:51:55-0500

5. Consider frustrum of a right circular cone

Let r1=r_1= the radius of the upper base, r2=r_2= the radius of the lower base, and L=L= the slant height. For any Frustum, the volume is


V=13(A1+A2+A1A2)hV=\dfrac{1}{3}(A_1+A_2+\sqrt{A_1A_2})h

The volume of the frustrum of the cone


Vc=13(πr12+πr22+πr12πr22)hV_c=\dfrac{1}{3}(\pi r_1^2+\pi r_2^2+\sqrt{\pi r_1^2\pi r_2^2})h

=π3(r12+r22+r1r2)h=\dfrac{\pi}{3}( r_1^2+ r_2^2+r_1r_2)h

Given

r2=3r1,h=3 cm,V=52π cm3r_2=3r_1, h=3\ cm,V=52 \pi \ cm^3

V=π3(r12+(3r1)2+r1(3r1))(3)=52πV=\dfrac{\pi}{3}( r_1^2+ (3r_1)^2+r_1(3r_1))(3)=52\pi

13r12=5213r_1^2=52

r1=2 cmr_1=2 \ cm

6. From the right triangle by the Pythagorean Theorem


L2=(r2r1)2+h2L^2=(r_2-r_1)^2+h^2

L=(3r1r1)2+h2L=\sqrt{(3r_1-r_1)^2+h^2}


L=4r12+h2L=\sqrt{4r_1^2+h^2}

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

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


AL=12(2πr2+2πr1)LA_L=\dfrac{1}{2}(2\pi r_2+2\pi r_1)L

AL=π(3r1+r1)LA_L=\pi(3 r_1+ r_1)L

AL=4πr1LA_L=4\pi r_1L

AL=4π(2)(5)=40π(cm2)125.66(cm2)A_L=4\pi (2)(5)=40\pi(cm^2)\approx125.66(cm^2)

The lateral surface area of the opening is 40π cm2125.66 cm2.40\pi\ cm^2\approx125.66\ cm^2.



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