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Math Help - Isosceles Right Triangle

  1. #1
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    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.
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  2. #2
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    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
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  3. #3
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    ok

    Great but tell me:

    how did you get x/sqrt{2}?

    Shouldn't we RATIONALIZE the denominator?
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  4. #4
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    Quote Originally Posted by symmetry View Post
    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
    Attached Thumbnails Attached Thumbnails Isosceles Right Triangle-halbquadrt.gif  
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  5. #5
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    Quote Originally Posted by earboth View Post
    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
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  6. #6
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    ok

    What a great reply! I thank you.

    Earboth,

    Tell me, how do you make lovely geometric pictures?
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