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Area of rectangle. Kap PS Problem

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Area of rectangle. Kap PS Problem [#permalink]

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New post 26 Jun 2009, 21:34
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Steps to find area of BEC?
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Re: Area of rectangle. Kap PS Problem [#permalink]

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New post 18 Jul 2009, 14:00
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mbaMission, That point with the two right angles I will call point M (Main point), to make the explanation easier.

Lets call
MD = b (base)
MC = h (height)

The area of the shadow triangle plus the area of the triangle CMD is the half of the area of the rectangle.

Area of the shadow triangle = \((25 - b)*h/2\)
Area of the triangle \(CMD = b*h/2\)
Half of the area of the rectangle is \(15*20/2 = 150\)

Thus
\((25 - b)*h/2 + b*h/2 = 150\)
Resolving for h you will find \(h = 12\)

If the hypotenuse is 15, and one of its legs is 12, the other leg is 9. So \(b = 9\);

To calculate the area of the shadow triangle
\((25 - b)*h/2 = (25 - 9)*12/2 = 96\) :wink:

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Re: Area of rectangle. Kap PS Problem [#permalink]

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New post 22 Jul 2009, 01:06
Well explained coelholds. +1 Kudo

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Re: Area of rectangle. Kap PS Problem [#permalink]

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New post 22 Jul 2009, 05:36
mbaMission wrote:
Steps to find area of BEC?

The diagonal is 25 (15^2 + 20^2).
Now, let the two segment be x and y respectively.
x+y=25
and 20^2-x^2=15^2-y^2
it will give you x^2-y^2=35*5
which will give x-y=7
x=16 from 1 and 2
so area is 1/2*16*12=96

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Re: Area of rectangle. Kap PS Problem [#permalink]

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New post 22 Jul 2009, 05:57
Thanks irajeevsingh.

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Re: Area of rectangle. Kap PS Problem   [#permalink] 22 Jul 2009, 05:57
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