1. In the examination figure the sides of the triangle b=501, c=392.6, d=512.8 find the area to the nearest 10 square feet

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- Apr 4th 2008, 10:25 AM #1

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- Apr 4th 2008, 10:35 AM #2

- Apr 4th 2008, 10:49 AM #3
There are two methods. I've showed both.

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1st method, Heron's Formula:

$\displaystyle s = \frac{(a+b+c)}{2} = \frac{(501+392.6+512.8}{2} = 703.2$

$\displaystyle a = \sqrt{(703.2)(703.2 - 501)(703.2 - 392.6)(703.2 - 512.8)} = 91700$ $\displaystyle \mathrm {units} ^2$

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2nd method, Geometry:

**EDIT:**I have attached a diagram so you can see how it is geometrically.

Find the height by using Pythagoras and then use the standard triangle area formula.

$\displaystyle x^2 + h^2 = 501^2$ ---------(1)

$\displaystyle (392.6 - x)^2 + h^2 = 512.8^2$ --------------(2)

(2) - (1)

$\displaystyle (392.6 - x)^2 - x^2 = 512.8^2 - 501^2 $

Solve for $\displaystyle x$:

$\displaystyle x = 181$

Substitute $\displaystyle x$ value into formula (1) or (2) to get $\displaystyle h$.

$\displaystyle h = 467$

$\displaystyle \mathrm {Area} = \frac{1}{2}(\mathrm {Base})(\mathrm {Height}) = \frac{1}{2} (392.6)(467) = 91700$ $\displaystyle \mathrm {units} ^2$