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Thread: [SOLVED] Incribed Rectangle and Triangle

  1. #1
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    Question [SOLVED] Incribed Rectangle and Triangle

    A rectangle of base b and height h and an isosceles triangle of base b are inscribed in a circle of radius one as shown. For what values of b and h will the rectangle and triangle have the same area?


    How would you solve this? I got an expression for the height of the triangle, but it led to h=0, or b=0. Thanks in advance.
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  2. #2
    MHF Contributor red_dog's Avatar
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    The height of the triangle is $\displaystyle 1-\frac{h}{2}$.

    The area of the rectangle is $\displaystyle S_1=bh$

    The area of the triangle is $\displaystyle S_2=\frac{b\left(1-\frac{h}{2}\right)}{2}=\frac{b(2-h)}{4}$

    $\displaystyle S_1=S_2\Rightarrow bh=\frac{b(2-h)}{4}\Rightarrow h=\frac{2}{5}$

    We have $\displaystyle b^2+h^2=4$ Replacing h we get $\displaystyle b=\frac{4\sqrt{6}}{5}$
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  3. #3
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    Thanks for replying, but I how did you know the center of the rectangle was on the center of the circle (i.e. why is the height of the triangle $\displaystyle 1-h/2$)? Also, where did you get $\displaystyle b^2+h^2=4$?

    Thanks again.
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  4. #4
    MHF Contributor red_dog's Avatar
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    Quote Originally Posted by amla View Post
    how did you know the center of the rectangle was on the center of the circle
    The angles of the rectangle are of $\displaystyle 90^{\circ}$, so each diagonal is diameter of the circle.

    Also, where did you get $\displaystyle b^2+h^2=4$?
    In a triangle formed by two sides of the rectangle and a diameter apply Pitagora.
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    Thanks!
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