# Thread: need help, please!

1. ## need help, please!

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

2. Originally Posted by hamadouousmane
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
Do a search for Heron's formula.

-Dan

3. Originally Posted by hamadouousmane
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
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 $\displaystyle h$.
$\displaystyle h = 467$
$\displaystyle \mathrm {Area} = \frac{1}{2}(\mathrm {Base})(\mathrm {Height}) = \frac{1}{2} (392.6)(467) = 91700$ $\displaystyle \mathrm {units} ^2$