Thread: 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?

Originally Posted by perash

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.

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 $\displaystyle x$ from the corner.
What should $\displaystyle x$ be for the triangle to be a right triangle?

Code:

      A  x  E      26-x       B
      * - - * - - - - - - - - *
      |    *   *              |
      |   *       *           |
   12 |  *           *        | 12
      | *               *     |
      |*                   *  |
      * - - - - - - - - - - - *
      D         26            C

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

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


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

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.