1. ## 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.

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

3. ## ok

Great but tell me:

how did you get x/sqrt{2}?

Shouldn't we RATIONALIZE the denominator?

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

EB

5. Originally Posted by earboth
Hello,

an isosceles right triangle is the half of a square.

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

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

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

6. ## ok

What a great reply! I thank you.

Earboth,

Tell me, how do you make lovely geometric pictures?

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### area of isosceles right triangle

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