I WILL GIVE BRAINLIEST!!! In a certain isosceles right triangle, the altitude to the hypotenuse has length 4sqrt2. What is the area of the triangle?
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Step-by-step explanation:
[tex] {x}^{2} + {x}^{2} = {4}^{2} \\ 2 {x}^{2} = {4}^{2} \\ {x}^{2} = 8 \\ then \: the \: area \: of \: a \: triangle \: is \\ \frac{1}{2} \times {x}^{2} \\ \frac{1}{2} \times 8 \\ 4 \: square \: unit[/tex]