1. ## sum of areas

A piece of wire 18 m long is cut into two pieces, the length of the first piece being x m. The first piece is bent into a circle, and the other is bent into a rectangle with length twice the width. Give an expression for the total area A enclosed in the two shapes in terms of x.

2. start with the circle its perimetre is X then X=2pi*R then R=X/2pi

Area of the circle is = pi*R^2 = (pi * X^2 )/(4pi^2) = X^2/(4pi)

now the rectangle : we have 18-X perimetre of the rectangle its perimetre = 2L+2W and we have L=2W => 18-X=6W => W=3-X/6 and L = 2W = 6-X/3

=> Area of the rectangle is = W*L = ( 3-X/6 ) ( 6-X/3 )

Total Area = ( 3-X/6 ) ( 6-X/3 ) + X^2/(4pi)

3. T.A = 18 - X -X +X^2/18 + X^2/4pi
T.A = (1/18+1/4pi)X^2 - 2X + 18

4. Originally Posted by samtheman17
A piece of wire 18 m long is cut into two pieces, the length of the first piece being x m. The first piece is bent into a circle, and the other is bent into a rectangle with length twice the width. Give an expression for the total area A enclosed in the two shapes in terms of x.
For circle:
circumference x = 2 $\pi$ r
$r = \frac{x}{2\pi}$
area of circle = $(\pi)(\frac{x}{2\pi})^2 = \frac{x^2}{4\pi}$
For rectangle:
perimeter = 18 - x
sum of a length and a width = $\frac{18 - x}{2}$
length = $(\frac{2}{3})(\frac{18 - x}{2})$
width = $(\frac{1}{3})(\frac{18 - x}{2})$
area = $(\frac{2}{3})(\frac{18 - x}{2})(\frac{1}{3})(\frac{18 - x}{2})$
= $\frac{(18 - x)^2}{18}$
= $\frac{324 - 36x + x^2}{18}$
= $\frac{x^2}{18} -2x + 18$