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how to find the area of the 45-45-90 triangle with a hypotenuse of 24?

Sagot :

AL2006
The area of a triangle is 1/2(base x height) .

The two legs of this triangle are equal.  If you position it properly,
one of them is the base, and the other one is the height.

Since it's a right triangle,    A² + B² = C²

But  A = B  and  C = 24 .

2A² = (24)²

A² = (24)² / 2

But since  A=B,  A² is also (base x height) of the triangle.
Area is just 1/2 of it.

Area = 1/2(24²/2) = (24)²/4 = (24/2)² = (12)² = 144 . 


[tex]45^o-45^o-90^o\ triangle\ is\ a\ half\ of\ the\ square\ (look\ at\ he\ picture).\\\\The\ square\ is\ a\ rhombus\\then\ area\ equal\ the\ half\ of\ multiply\ diameters:\\\\A_{\fbox{}}=\frac{d^2}{2}\\\\A_\Delta=\frac{1}{2}A_{\fbox{}}\Rightarrow A_\Delta=\frac{1}{2}\cdot\frac{d^2}{2}=\frac{d^2}{4}\Rightarrow A_\Delta=\frac{24^2}{4}=\frac{576}{4}=144.[/tex]
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