# Rectangle Within Semicircle

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• June 13th 2007, 01:45 PM
blueridge
Rectangle Within Semicircle
A rectangle is inscribed in a semicircle of radius 2. Let P = (x,y) be the point in quadrant 1 that is a vertex of the rectangle and is on the circle.

(a) Express the area A of the rectangle as a function of x.

(b) Express the perimeter p of the rectangle as a function of x.
• June 13th 2007, 01:58 PM
topsquark
Quote:

Originally Posted by blueridge
A rectangle is inscribed in a semicircle of radius 2. Let P = (x,y) be the point in quadrant 1 that is a vertex of the rectangle and is on the circle.

(a) Express the area A of the rectangle as a function of x.

(b) Express the perimeter p of the rectangle as a function of x.

a) The width (or height or whatever you wish to call it) of the rectangle is x, the length is 2y. So the area A = x(2y) = 2xy.

b) P = 2l + 2w = 2(2y) + 2(x) = 2x + 4y

-Dan
• June 13th 2007, 09:56 PM
earboth
1 Attachment(s)
Quote:

Originally Posted by blueridge
A rectangle is inscribed in a semicircle of radius 2. Let P = (x,y) be the point in quadrant 1 that is a vertex of the rectangle and is on the circle.

(a) Express the area A of the rectangle as a function of x.

(b) Express the perimeter p of the rectangle as a function of x.

Hello,

the circle line of the semicircle is described by:

$y = \sqrt{4-x^2}$ according to Pythagorean theorem.

If the height of the rectangle isy then the length of the rectangle is 2x.

The area of a rectangle can be calculated by:

$A_{\text{rectangle}} = length \cdot height$ Plug in the terms of length and height:

$A(x) = 2x \cdot \sqrt{4-x^2}\ , 0 \leq x \leq 2$

.................................................. ......

EDIT: I forgot to show you how to calculate the perimeter of the rectangle:

In general the perimeter is: P = 2*length + 2*height

Using the terms for length and height you get:

$p = 2 \cdot 2x + 2 \cdot \sqrt{4-x^2} = 4x + 2 \cdot \sqrt{4-x^2}$
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