Answer to Question #191653 in Algebra for Angelo

Question

A semi-circle of radius 14 cm is formed from a piece of wire. If it is bent into a rectangle whose length is 2 cm more than its width, find the area of the rectangle.

Answer 1

$\text{ perimeter of semi circle}=(2+\pi)r=(2+\pi)×14=71.9823 cm\newline\text{ Given, length of rectangle(l) is 2cm more than its width(b).}\newline l=2+b\newline\text{ perimeter of rectangle}=2(l+b)\newline\text{ Given, perimeter of semi circle=perimeter of rectangle}\newline (2+\pi)×14=2(l+2+l)\newline (2+\pi)×14=4(l+1)\newline l+1=\frac{(2+\pi)×14}{4}\newline l=\frac{(2+\pi)×14}{4}-1\newline l=16.9956 cm\newline\text{ Therefore, width of rectangle is }18.9956 cm.\newline \text{Area of the rectangle}=lb=16.9956×18.9956=322.8416 cm^2$