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Area of an isosceles right triangle
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02 Sep 2021, 23:41
02 Sep 2021, 23:41
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Question Stats:
86% (00:59) correct
13% (00:53) wrong based on 15 sessions
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Joined: 10 Apr 2015
Posts: 6218
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Re: Area of an isosceles right triangle
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03 Sep 2021, 07:31
03 Sep 2021, 07:31
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RA911 wrote:
Our isosceles right triangle looks like this:
From here we can apply the Pythagorean theorem to write: \(x^2 + x^2 = 6^2\)
Simplify to get: \(2x^2 + x^2 = 36\)
Divide both sides of the equation by 2 to get: \(x^2 = 18\)
Solve: \(x = \sqrt{18}\)
Area of triangle \(= \frac{(base)(height)}{2}\)
For this triangle, the base and height both equal x.
So we get: area \(= \frac{(\sqrt{18})(\sqrt{18})}{2}\)
\(= \frac{18}{2}\)
\(= 9\)
Answer: 9
Re: Area of an isosceles right triangle
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22 May 2024, 11:40
22 May 2024, 11:40
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