# Math Help - Lateral Area of the Frustum and Pyramid

1. ## Lateral Area of the Frustum and Pyramid

A lighthouse consists of a tall tapering tower constructed of brick and has a circular cross section. The tower is surmounted by a conical top consisting of a tin roof with an overhang such that the eave is 1 ft. The overall height of the lighthouse is 40 ft. and the height of the tower 30 ft. Find the area of the brick surface if the lower base of the tower is 20 ft. in diameter and the upper base 12 ft. in diameter. Also find the amount of roofing material used in its construction.

I already got the lateral area of the brick. - 1521.3 sq. ft.

What i can't get is the lateral area of the conical roof.

The answer at the book is 259.15 sq. ft.
The closest answer i can get was 262.12 sq. ft. by adding the eave to the slant height and square root of 2 over 2 to the radius.
I already consumed 3 papers back to back for this problem. :|

2. Originally Posted by Dug
A lighthouse consists of a tall tapering tower constructed of brick and has a circular cross section. The tower is surmounted by a conical top consisting of a tin roof with an overhang such that the eave is 1 ft. The overall height of the lighthouse is 40 ft. and the height of the tower 30 ft. Find the area of the brick surface if the lower base of the tower is 20 ft. in diameter and the upper base 12 ft. in diameter. Also find the amount of roofing material used in its construction.

I already got the lateral area of the brick. - 1521.3 sq. ft.

What i can't get is the lateral area of the conical roof.

The answer at the book is 259.15 sq. ft.
The closest answer i can get was 262.12 sq. ft. by adding the eave to the slant height and square root of 2 over 2 to the radius.
I already consumed 3 papers back to back for this problem. :|
1. Draw a sketch!

2. The length of the slanted distance is $s = \sqrt{10^2+6^2}=\sqrt{136}$
Then the length of the complete slanted distance is $1+\sqrt{136}$

3. Use proportions in similar triangles:

$\dfrac r6=\dfrac{1+\sqrt{136}}{\sqrt{136}}~\implies\boxed { r=\dfrac{6(1+\sqrt{136})}{\sqrt{136}}}$

4. The curved surface area of a cone is calculated by

$a = \pi \cdot r \cdot s$

s meaning the complete slanted distance of the roof.

5. Plug in all values:

$a = \pi \cdot \dfrac{6(1+\sqrt{136})}{\sqrt{136}} \cdot (1+\sqrt{136}) = 6 \pi \cdot \dfrac{(1+\sqrt{136})^2}{\sqrt{136}} \approx 259.137$