Question

The slant height of a frustum of a cone is 4 cm and the perimeters (circumferences) of its circular ends are 18 cm and 6cm. Find the curved surface area and the volume of the frustum numerically equal ? Discuss its truthfulness.​

Answer 1

Answer:

L = 4 cm, C1 = 18 cm and C2 = 6 cm.

CSA of a frustum of a cone = πl (r1 + r2)

= l/2 ( 2πr1 + 2πr2 )

= 4/2 ( C1 + C2 )

= 2 ( 18 + 6 )

= 48 cm^2

Volume of a frustum of a cone = 1/3 πh ( r1^2 + r2^2 + r1r2 )

= h/3 (πr1^2 + πr2^2 + √πr1^2 • πr2^2)

= h/3 (A1 + A2 + √A1A2)

with the help of C1 & C2 r1 = 2.86 & r2 = 0.95

and with the help of formula l^2 = h^2 + r^2 we will get h = 3.52 cm.

Volume of frustum of cone= πh [r1^2 + r2^2 + r1r2]

=22/7*3.52/3[2.86^2 +0.95^2 + 2.717]

=22/7 * 3.52/3 [ 8.1796 + 0.9025 + 2.717 ]

= 43.511 cm^3

HOPE IT HELPS...