Inscribed rectangle, please help.
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Inscribed rectangle, please help.

[From: ] [author: ] [Date: 12-06-08] [Hit: ]
This equations is for the area of a rectangle inscribed in a semicircle with the radius of 3. I simply cant fathom as to why i need to subtract 2x to find the height of the rectangle.Sketch the semicircle with the radius of 3, (diameter 6) to see the situation.x equals the half the base of a rectangle, (so base is 2x).......
I have the equation a=2x(square root (9-2x)) where x equals the half the base of a rectangle. This equations is for the area of a rectangle inscribed in a semicircle with the radius of 3. I simply can't fathom as to why i need to subtract 2x to find the height of the rectangle.

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I have the equation a=2x(square root (9-2x)) where x equals the half the base of a rectangle. This equations is for the area of a rectangle inscribed in a semicircle with the radius of 3. I simply can't fathom as to why i need to subtract 2x to find the height of the rectangle.

Sketch the semicircle with the radius of 3, (diameter 6) to see the situation.

x equals the half the base of a rectangle, (so base is 2x). OK so far.
What is the height when x coordinate is x ?

Well, the equation of the circle is x^2 + y^2 = 3^2 = 9
So, y^2 = 9 - x^2 and
y = sqrt[9 - x^2]
The area A = base 2x times height sqrt[9 - x^2]
You should have got the area as
A =2xsqrt[9 - x^2]

It was a slight miscopy to write a=2x(square root (9-2x))

Regards - Ian
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