Question

1. A lampshade is in the form of a frustum of a cone of which the top and the bottom radius is thrice the top radius, and the height is 14cm.

The volume of the lampshade is 763cm³

Find:

a. The area of material required to cover the lampshade

b. Bottom Radius ?

Answer 1

Solution 1. a

Let the volume of the frustum be V, the top radius be r and bottom radius be R and vertical height be h.

To calculate the volume V, we use the formula

$V=πh/3(R^2+Rr+r^2)$

We are given that V=763cm³ , h=14cm, R=3r

We then substitute the figures in the formula as follows;

763=14π/3[ (3r)² + (3r)(r)+r²]

763=14π/3[9r²+3r²+r²], we sum the terms in the brackets.

763=14π/3[13r²] then we multiply 14 by 13

763=182πr²/3 When we multiply by 3/182π both sides, we get the following:

763(3/182π)= 182πr²/3 (3/182π), we are left with r2 on the right hand side giving

2289/182π=r² To find r alone, we need to take the square root of both sides

$\sqrt {(2289/182π)}= \sqrt{r^2}$

Working out the square root of both sides we get r=2.00cm

Therefore, radius r=2.00cm and R= 3r=3(2)=6.00cm

We want to find the area of material required to cover the lampshade. This is basically the surface area and it is given by;

$Surface Area=πl(R+r)$ But $l$ is the slant height and is given by

$l= \sqrt{h^2+R^2}$

h=14, R=6

$l=\sqrt{(14^2+6^2 )}$

$l=\sqrt{(196+36)}$

$l=\sqrt{232}$

l=15.23cm

Hence Surface Area=15.23π(6+3)

=15.23π(9)

430.62cm2

Solution 1. b

Bottom radius R =3r

= 3(2.00)

=6.00cm