# Isosceles Right Triangle

• Jan 18th 2007, 02:00 PM
symmetry
Isosceles Right Triangle
Express the area A of an isosceles right triangle as a function of the length x of one of the two equal sides.
• Jan 18th 2007, 02:06 PM
ThePerfectHacker
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,
$\frac{1}{2} \cdot \frac{x}{\sqrt{2}}\cdot \frac{x}{\sqrt{2}}=\frac{1}{4}x^2$
• Jan 18th 2007, 02:18 PM
symmetry
ok
Great but tell me:

how did you get x/sqrt{2}?

Shouldn't we RATIONALIZE the denominator?
• Jan 22nd 2007, 05:48 AM
earboth
Quote:

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
• Jan 22nd 2007, 05:51 AM
Quick
Quote:

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: $A=\frac{1}{2}x^2$
• Jan 22nd 2007, 02:21 PM
symmetry
ok
What a great reply! I thank you.

Earboth,

Tell me, how do you make lovely geometric pictures?