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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.

 Dec 28, 2014

Best Answer 

 #1
avatar+99238 
+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.

 Dec 28, 2014
 #1
avatar+99238 
+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

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