A semi-circle of radius 14 cm is bent to form a rectangle whose length is 1 cm more than its width. Find the area of the rectangle.

Guest Dec 28, 2014

#1**+10 **

Well

the perimeter of the semicircle is 2r+0.5(2pi*r) = 2r+pi*r

the perimeter of this particular semicircle is 14+7pi

Let the sides of the rectangle be x, x, x+1, and x+1

So

$$\\x+x+x+1+x+1=14+7\pi\\

4x+2=14+7\pi\\

4x=12+7\pi\\

x=\frac{12+7\pi}{4}$$

so the area of the rectangle is

$$\\Area=x(x+1)\\\\

=\frac{12+7\pi}{4}\left(\frac{12+7\pi}{4}+1\right)\\\\$$

and you can do that on the calc on the home page. It will be in cm squared.

Melody
Dec 28, 2014

#1**+10 **

Best Answer

Well

the perimeter of the semicircle is 2r+0.5(2pi*r) = 2r+pi*r

the perimeter of this particular semicircle is 14+7pi

Let the sides of the rectangle be x, x, x+1, and x+1

So

$$\\x+x+x+1+x+1=14+7\pi\\

4x+2=14+7\pi\\

4x=12+7\pi\\

x=\frac{12+7\pi}{4}$$

so the area of the rectangle is

$$\\Area=x(x+1)\\\\

=\frac{12+7\pi}{4}\left(\frac{12+7\pi}{4}+1\right)\\\\$$

and you can do that on the calc on the home page. It will be in cm squared.

Melody
Dec 28, 2014