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Find the area of a triangle with side lengths 13, 17, and \(12\sqrt2\) .

tertre  Mar 30, 2018
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5+0 Answers

 #1
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Semi-perimeter = 1/2[13+17+12sqrt(2)]

S = 23.485......

Area = Sqrt[23.485 x (23.485 - 17) x (23.485 - 12sqrt(2)) x (23.485 - 13)]

Area = 102 square units.

Guest Mar 30, 2018
 #3
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Thanks so much, guest!

tertre  Mar 31, 2018
 #2
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Melody  Mar 31, 2018
 #4
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Yep, that's correct, Melody!

tertre  Mar 31, 2018
 #5
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Sloppy maths though, which misses part of the structure of the question.

The semi-perimeter,

\(\displaystyle s=\frac{1}{2}(13+17+12\sqrt{2})=15+6\sqrt{2},\)

so applying Heron's formula, the area will equal

\(\displaystyle \sqrt\{ (15+6\sqrt{2})(6\sqrt{2}-2)(6\sqrt{2}+2)(15-6\sqrt{2})\}\, \\=\text{(difference of two squares twice,)}\\\sqrt{(15^{2}-(6\sqrt{2})^{2})((6\sqrt{2})^{2}-2^{2})}\\=\sqrt{153\times68}=\sqrt{9\times17\times4\times17}=3\times2\times17=102.\)

Guest Mar 31, 2018

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