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Math Help - Rectangle Within Semicircle

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    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.
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    Quote Originally Posted by blueridge View Post
    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
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    Quote Originally Posted by blueridge View Post
    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}
    Attached Thumbnails Attached Thumbnails Rectangle Within Semicircle-rechteck_halbkreis.gif  
    Last edited by earboth; June 14th 2007 at 03:40 AM.
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