A right isosceles triangle has an area of 18. What is the length of the longest side of the triangle?

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Answer 1 · 2017-02-03T10:51:39

$"The Desired Length"=6sqrt2" units."$

Consider a Right Isosceles $DeltaABC$ with, $AB=BC$

Clearly, $m/_B=90^@,$ so that, $AC$ being the Hypotenuse is

the longest side of $Delta ABC.$

Now, the Area of $Delta ABC=1/2xxABxxBC$

$=1/2AB^2, .........[because, AB=AC]$ .

$:. Area =18 rArr 1/2AB^2=18 rArr AB^2=36=BC^2.$

Using Pythagoras' Theorem, in $Delta ABC$ , we have,

$AB^2+BC^2=AC^2=36+36=72$

$:. AC=sqrt72=6sqrt2$ is the desired lemgth.