# Semi-Circle Rectangle Area, perimeter

• Oct 28th 2009, 11:10 PM
goliath
Semi-Circle Rectangle Area, perimeter
A window is in the shape of a semi-circle on top of a rectangle. The height of the rectangle is twice its width.

a) Give an expression for A, the area of the window, in terms of w, its width.
b) Give an expression for P, the perimeter of the window, in terms of w, its width.
c) Give an expression for A, in terms of P only.
• Oct 28th 2009, 11:31 PM
earboth
Quote:

Originally Posted by goliath
A window is in the shape of a semi-circle on top of a rectangle. The height of the rectangle is twice its width.

a) Give an expression for A, the area of the window, in terms of w, its width.
b) Give an expression for P, the perimeter of the window, in terms of w, its width.
c) Give an expression for A, in terms of P only.

I'll show you how you can do a):

The area of a rectangle is calculated by $\displaystyle A_{rect}=length \cdot width$

The area of a semi-circle is calcvulated by $\displaystyle A_{sem-circ}=\frac12 \cdot \pi \cdot (radius)^2$

... and now it's your turn. :D

$\displaystyle length = 2 \cdot width$
$\displaystyle radius = \frac12 \cdot width$

Therefore the area of the window is:

$\displaystyle A = 2 \cdot width \cdot width + \frac12 \cdot \pi \cdot \left(\frac12 width\right)^2 = (width)^2 \left(2 + \frac18 \pi\right)$
• Oct 29th 2009, 04:50 AM
Wilmer
Quote:

Originally Posted by goliath
b) Give an expression for P, the perimeter of the window, in terms of w, its width.

Do you "see" that P = height*2 + width + half-circle-circumference,
which is 5w + pi(w)/2 ?
• Nov 2nd 2009, 10:38 PM
goliath
K guys, I got those 2, I need help with part C
• Nov 2nd 2009, 11:07 PM
earboth
Quote:

Originally Posted by goliath
...

c) Give an expression for A, in terms of P only.

1. If you need further information or assistance ask publicly at MHF.

2. It would be much easier for us to help you if you could show some work so we can see where you have difficulties. Then we have a chance to offer adaquate help.
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3. From Wilmer's post you learned that:

$\displaystyle P=w\left(5+\frac12 \pi\right) ~\implies~w=\dfrac{P}{5+\frac12 \pi}$

Plug the term for w into the equation of $\displaystyle A=w^2\left(2+\frac18 \pi\right)$

You'll get:

$\displaystyle A=\left(\dfrac{P}{5+\frac12 \pi} \right)^2\left(2+\frac18 \pi\right)$

After moving around some stuff you'll get:

$\displaystyle A=\dfrac{\pi + 16}{2(\pi +10)^2} \cdot P^2$
• Nov 2nd 2009, 11:10 PM
nikhil
area of window A=2w^2+(pi/2)(w/2)^2..........eq 1
perimeter.........p=5w+pi(w/2)
or w(5+pi/2)=p
or w=p/(5+pi/2)
put this value in eq 1
A interm of P will be obtained
• Nov 2nd 2009, 11:13 PM
goliath
K got it! Awesome guys, thanks!