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Math Help - The rectangle

  1. #1
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    The rectangle

    The rectangle below has a base with length 26 and a height of 12. Inside the rectangle, there is an inscribed triangle whose base is the base of the rectangle, and whose third vertex lies on the top of the rectangle at a distance X from the corner. What should X be for the triangle to be a right triangle?
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  2. #2
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    Quote Originally Posted by perash View Post
    The rectangle below has a base with length 26 and a height of 12. Inside the rectangle, there is an inscribed triangle whose base is the base of the rectangle, and whose third vertex lies on the top of the rectangle at a distance X from the corner. What should X be for the triangle to be a right triangle?
    I don't know, but whenever I hear "right triangle", the Pythagorean Theorem always comes to my mind.

    We will use here the Pythagorean Theorem.

    let the short leg of the inscribed triangle be u, and the long leg be v.

    In the triangle containing x, and whose hypotenuse is u,
    u^2 = 12^2 +x^2

    In the triangle whose hypotenuse is v,
    v^2 = 12^2 +(26 -x)^2

    In the inscribed triangle in question,
    26^2 = u^2 +v^2
    So,
    26^2 = [12^2 +x^2] +[12^2 +(26 -x)^2]
    26^2 = 12^2 +x^2 +12^2 +26^2 -52x +x^2
    0 = 2(12^2) -52x +2x^2
    0 = 12^2 -26x +x^2
    0 = 144 -26x +x^2
    By Quadratic formula,
    x = {26 +,-sqrt[(26)^2 -4(1)(144)]} / 2(1)
    x = 8 or 18 ------------------answer.
    Last edited by ticbol; August 4th 2008 at 09:06 PM.
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  3. #3
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    Hello, perash!

    Another approach . . .


    The rectangle below has a base with length 26 and a height of 12.
    Inside the rectangle, there is an inscribed triangle whose base is the base of the rectangle,
    and whose third vertex lies on the top of the rectangle at a distance x from the corner.
    What should x be for the triangle to be a right triangle?
    Code:
          A  x  E      26-x       B
          * - - * - - - - - - - - *
          |    *   *              |
          |   *       *           |
       12 |  *           *        | 12
          | *               *     |
          |*                   *  |
          * - - - - - - - - - - - *
          D         26            C

    If \angle DEC = 90^o, then: . \Delta DAE \sim \Delta EBC

    Hence: . \frac{x}{12}\:=\:\frac{12}{26-x} \quad\Rightarrow\quad x^2 - 26x + 144 \:=\:0 \quad\Rightarrow\quad (x - 8)(x - 18) \:=\:0


    . . Therefore: . x \:=\:8,\:18

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  4. #4
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    The triangle will also be a right triangle if either of the other 2 angles are 90 degrees. So x can also be 0 or 26.
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