Question

Approximate the change in the lateral surface area (excluding the area of the base) of a right circular cone of fixed height of h=6 m when its radius decreases from r= 10m to r= 9.9 m

Answer 1

Answer:

The change in the lateral surface area is approximate $6.32\ m^{2}$

Step-by-step explanation:

we know that

The lateral surface area of the cone is equal to

$LA=\pi r l$

where

r is the radius of the base

l is the slant height

Part 1

we have

$r=10\ m, h=6\ m$

Calculate the slant height l (applying the Pythagoras Theorem)

$l^{2}=r^{2}+h^{2}$

substitute the values

$l^{2}=10^{2}+6^{2}$

$l^{2}=136$

$l=\sqrt{136}\ m$

Find the lateral area of the cone

$LA=\pi (10)(\sqrt{136})$

$LA=10\pi \sqrt{136}\ m^{2}$

Part 2

we have

$r=9.9\ m, h=6\ m$

Calculate the slant height l (applying the Pythagoras Theorem)

$l^{2}=r^{2}+h^{2}$

substitute the values

$l^{2}=9.9^{2}+6^{2}$

$l^{2}=134.01$

$l=\sqrt{134.01}\ m$

Find the lateral area of the cone

$LA=\pi (9.9)(\sqrt{134.01})$

$LA=9.9\pi \sqrt{134.01}\ m^{2}$

Part 3

Find  the change in the lateral surface area

$10\pi \sqrt{136}-9.9\pi \sqrt{134.01}$

assume $\pi =3.14$

$10(3.14)\sqrt{136}-9.9(3.14)\sqrt{134.01}=6.32\ m^{2}$