Isosceles Right Triangle

**symmetry** · January 18th 2007, 01:00 PM

Express the area A of an isosceles right triangle as a function of the length x of one of the two equal sides.

**ThePerfectHacker** · January 18th 2007, 01:06 PM

The hypotenuse is $x$ and this is a 45-45-90 Triangle.
Thus, the equal sides are $\frac{x}{\sqrt{2}}$
That means the area is,
$\frac{1}{2} \cdot \frac{x}{\sqrt{2}}\cdot \frac{x}{\sqrt{2}}=\frac{1}{4}x^2$

**symmetry** · January 18th 2007, 01:18 PM

Great but tell me:

how did you get x/sqrt{2}?

Shouldn't we RATIONALIZE the denominator?

**earboth** · January 22nd 2007, 04:48 AM

Originally Posted by symmetry

Express the area A of an isosceles right triangle as a function of the length x of one of the two equal sides.

Hello,

an isosceles right triangle is the half of a square.

Area of the square: $A_{square}=x^2$ . Therefore the area of the isosceles right triangle:
$A_{\text{isosceles right trinagle}}=\frac{1}{2} \cdot x^2$

EB

**Quick** · January 22nd 2007, 04:51 AM

Originally Posted by earboth

Hello,

an isosceles right triangle is the half of a square.

Area of the square: $A_{square}=x^2$ . Therefore the area of the isosceles right triangle:
$A_{\text{isosceles right trinagle}}=\frac{1}{2} \cdot x^2$

EB

of course you could just say that the area of a triangle is $A=\frac{1}{2}bh$

So: $A=\frac{1}{2}x^2$

**symmetry** · January 22nd 2007, 01:21 PM

What a great reply! I thank you.

Earboth,

Tell me, how do you make lovely geometric pictures?

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