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Math Help - [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 1-\frac{h}{2}.

    The area of the rectangle is S_1=bh

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

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

    We have b^2+h^2=4 Replacing h we get 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 1-h/2)? Also, where did you get 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 90^{\circ}, so each diagonal is diameter of the circle.

    Also, where did you get b^2+h^2=4?
    In a triangle formed by two sides of the rectangle and a diameter apply Pitagora.
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  5. #5
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    Thanks!
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