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= the radius of the upper base, r2= the radius of the lower base, and L= the slant height. For any Frustum, the volume is
V=31(A1+A2+A1A2)h
The volume of the frustrum of the cone
Vc=31(πr12+πr22+πr12πr22)h
=3π(r12+r22+r1r2)h
Given
r2=3r1,h=3cm,V=52πcm3
V=3π(r12+(3r1)2+r1(3r1))(3)=52π
13r12=52
r1=2cm
6. From the right triangle by the Pythagorean Theorem
L2=(r2−r1)2+h2
L=(3r1−r1)2+h2
L=4r12+h2
L=4(2)2+(3)2=5(cm)
7. The lateral area of the frustum of a right circular cone is
AL=21(2πr2+2πr1)L
AL=π(3r1+r1)L
AL=4πr1L
AL=4π(2)(5)=40π(cm2)≈125.66(cm2)
The lateral surface area of the opening is 40πcm2≈125.66cm2.
Comments