A conical frustum has one base with a radius of 9, and the other base has a radius of 12. The height of the frustum is 4. Let A be the total surface area of the frustum, in square units.

Question

Let V be the volume of the frustum, in cubic units. Find A+V.

Answer 1

The total surface area of the frustum of cone  = π l₁ (R+r) +πR² +πr²

The slant height (l₁)   = √[H² +(R-r)²]

here

R=12

r=9

H=4

l1 = √ (4² + (12-9)²) = √ (16 +9)= 5

total surface area = 5π ( 12+9) + π (12)² + π (9)²

=105 π + 144π +81π

= 330 π

The Volume of Frustum of Cone = 1/3 π H [(R³-r³)/ r

=1/3 π 4(12³ -9³) /9

=1/3 π(999)/9

=1/3 π 111

=111 π /3

A+V = 330π sq unit+ 111π/3 cu unit