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+7π4x+2=14+7π4x=12+7πx=12+7π4
so the area of the rectangle is
Area=x(x+1)=12+7π4(12+7π4+1)
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+7π4x+2=14+7π4x=12+7πx=12+7π4
so the area of the rectangle is
Area=x(x+1)=12+7π4(12+7π4+1)
and you can do that on the calc on the home page. It will be in cm squared.