Hello, Sane!

Here's the first one . . . it's very messy!

1) A sheet of thick plastic is used to make a traffic cone.

What area of plastic must be cut from the sheet to make

a cone of vertical height 40cm and diameter 30cm? Code:

*
/|\
/ | \
/ | \ R
/ |40 \
/ | \
*-----+-----*
15

We have a right triangle with legs 15 and 40.

. . The hypotenuse is: .$\displaystyle R \:=\:\sqrt{15^2 + 40^2} \:=\:\sqrt{1875} \:=\:5\sqrt{73}$

The cone will be cut from a circle of radius $\displaystyle R$. Code:

* * *
* *
* *
* *
* O *
* * *
* *:::* R *
*:::::::*
* *:::::::::::* *
A*:::::::::::::::*B
*:::::::::::*
* * *

The area of the entire circle is: .$\displaystyle A \:=\:\pi(5\sqrt{73})^2 \:=\:1825\pi$ cm²

The cone takes up some fraction of the circle . . . What fraction?

The arc AB is some fraction of the circumference of the circle.

. . And *that* is the fraction we seek.

The circle has circumference: .$\displaystyle C_1 \:=\:2\pi(5\sqrt{73}) \:=\:10\sqrt{73}\pi$ cm

Arc AB is the circumference of the base-circle of the cone.

. . The base-circle has: .$\displaystyle C_2 \:=\:\pi D \:=\:30\pi$ cm

There is our fraction . . . $\displaystyle \frac{C_2}{C_1}\;=\;\frac{30\pi}{10\sqrt{73}} \;=\;\frac{3}{\sqrt{73}} $

The area of the cone is: .$\displaystyle \frac{3}{\sqrt{73}} \times 1825\pi \;=\;75\sqrt{73}\pi$ cm²