Hello, Dug!

You're wrong about the eaves.

A cylindrical tower 30 ft in diameter has a conical roof.

The length of whose eaves is 2 ft.

An element of the roof is inclined 45 degrees to the horizontal.

Find the weather surface.

Answer at the back of the book: **61.061 square yards.** Code:

A
o
* | * _
* | * 15√2
* |15 *
* | *
* | * B
C o - - - - - + - - - - - o 2
* | 15 M 15 | *
* | E * - o D
| |
| |

The diameter of the tower is 30 feet: .$\displaystyle CB = 30,\;CM = MB = 15$

$\displaystyle \text{Angle }ABC = 45^o \quad\Rightarrow\quad AB = 15\sqrt{2}$

The eaves are 2 feet: .$\displaystyle B\!D = 2$

Note that: $\displaystyle ED = \sqrt{2}$

$\displaystyle \text{The radius of the circular base of the cone is: }\:MB + ED \:=\:15 + \sqrt{2}$

$\displaystyle \text{The circumference of the circular base is: }\:2\pi(15 + \sqrt{2})$

Flatten the conical roof and we have a major sector of a circle.

Code:

* * *
P * * Q
o o
* * * *
* * R
* * O * *
* * *
* @ *
* *
* *
* *
* * *
_
s = 2π(15+√2)

$\displaystyle \text{We have the formula: }\:s \:=\:R\theta$

. . $\displaystyle \text{where }\:s \:=\:2\pi(15 + \sqrt{2}),\;\;R \:=\:15\sqrt{2} + 2 \:=\:\sqrt{2}(15 + \sqrt{2})$

$\displaystyle \text{Then: }\:2\pi(15+\sqrt{2})\:=\:\sqrt{2}(15+\sqrt{2})$$\displaystyle \theta$

. . $\displaystyle \theta \:=\:\frac{2\pi(15+\sqrt{2})}{\sqrt{2}(15+\sqrt{2} )} \;=\;\pi\sqrt{2}$

$\displaystyle \text{The area of the sector is: }\:A \:=\:\tfrac{1}{2}R^2\theta$

. . $\displaystyle \text{where: }\,R \:=\:\sqrt{2}(15+ \sqrt{2}),\;\theta \:=\:\pi\sqrt{2}$

$\displaystyle \text{Hence: }\;A \;=\; \tfrac{1}{2}\bigg[\sqrt{2}(15 + \sqrt{2})\bigg]^2(\pi\sqrt{2}) \;=\;1197.029986\text{ ft}^2 $

. . $\displaystyle \text{Therefore: }\;A \;=\;\frac{1197.029986}{9} \;\approx\;133\text{ yd}^2$