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.
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.
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.