Question 234640
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Ordinarily you would use Heron's Formula to find the area of a triangle with sides measuring *[tex \Large a], *[tex \Large b], and *[tex \Large c].



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ \sqrt{s(s\ -\ a)(s\ -\ b)(s\ -\ c)}]


where *[tex \Large s] is the perimeter divided by 2, that is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ s\ =\ \frac{a\ +\ b\ +\ c}{2}]


Step 1:  Substitute *[tex \Large a], *[tex \Large b], and *[tex \Large c] to calculate *[tex \Large s]


Step 2:  Substitute *[tex \Large s], *[tex \Large a], *[tex \Large b], and *[tex \Large c] to calculate the area.


However, the first statement in your problem is false.  That is to say it is not true that a triangle has sides 18, 29, and 47.  That is because a three-sided figure is not a triangle unless the sum of the measures of the two short sides is GREATER than the measure of the longest side.  Not having a triangle, the question "What is the area..." is absurd.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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