Question

what will be the lateral surface area of frustum of a cone if the area of base is 64 ∏ m ² and radius of upper one is 6m and slant height is 10 m

Answer 1

The lateral surface area of a cone is

$\pi l(r_{1} +r_{2} )$

Here

$l=slant height, r_{1} and r_{2}$

are radii of top & bottom

Area of base = 64π

$\pi r^2=64\pi$

Dividing by π

$r^2=64$

Taking square root

r= 8

So now the two radii are 8 & 6

The slant height of the cone is 10

Lateral area of the frustum

$\pi (10)(8+6)$

=440 m²

Lateral surface area of frustum of a cone if the area of base is 64 ∏ m ² and radius of upper one is 6m and slant height is 10 m is 440 m².

Answer 2

HELLO DEAR,

WE KNOW:-

The lateral surface area of a frustum of cone is

∏l{radius of base(R) + radius of top(r)}m²

where, l = slant height,

GIVEN:-

Area of base = 64∏

so, ∏R² = 64 ∏

R² = 64

R = 8m

now the radii of base = 8m, and radii of top = 6m

The slant height of the cone is 10(GIVEN)

Lateral area of the frustum of cone = ∏ * 10 * (8 + 6)m²

=22/7 * 10 * 14m²

= 22 * 10 * 2m²

= 440m²

Lateral surface area of frustum of a cone if the area of base is 64 ∏ m ² and radius of upper one is 6m and slant height is 10m is 440m²

I HOPE ITS HELP YOU DEAR,

THANKS