length of a side in a rectangle

**sanee66** · June 25th 2007, 08:45 PM

Could someone please look at this? I think the answer is 26 for the length of KL because KL=2(13). Is this right?

**Jhevon** · June 25th 2007, 09:02 PM

Originally Posted by sanee66

Could someone please look at this? I think the answer is 26 for the length of KL because KL=2(13). Is this right?

Why would you assume KL = 2(13)?

See the slightly modified diagram below.

MJKN is a square, so the diagonal bisects the angles at the corners. so we have 45 degree angles in the positions shown

Remember, $MK = \sqrt {50}$ and $KI = 13$

Now, $\sin 45^{ \circ} = \frac {JK}{MK} = \frac {JK}{ \sqrt {50}}$

$\Rightarrow JK = \sqrt {50} \sin 45^{ \circ} = 5$

Now note that JK = MN = IL

By Pythagoras' Theorem:

$IJ = LK = \sqrt { (IK)^2 + (IL)^2} = \sqrt {13^2 + 5^2} = 12$

**Soroban** · June 25th 2007, 09:17 PM

Hello, sanee66!

Sorry, your answer makes no sense . . .

In square $JMNK$ , diagonal $MK$ measures $\sqrt{50}$
and in rectangle $IJKL$ , diagonal $KI$ measures $13$ .
What is the length of IJ (or KL)?

Code:

      I                       J           M
      * - - - - - - - - - - - * - - - - - *
      |   *                   |    __   * |
      |       *  13           |   √50 *   |
    x |           *           |     *     | x
      |               *       |   *       |
      |                   *   | *         |
      * - - - - - - - - - - - * - - - - - *
      L         y             K     x     N

The sides of the square have length $x$ .
From Pythagorus, we have: . $x^2 + x^2 \:=\:(\sqrt{50})^2$

Hence: . $2x^2 = 50\quad\Rightarrow\quad x^2 = 25\quad\Rightarrow\quad x = 5$

In right triangle $ILK$ , we have: . $IL = 5,\:IK = 13$
From Pythagorus, we have: . $y^2 + 5^2\:=\:13^2\quad\Rightarrow\quad y^2 = 144\quad\Rightarrow\quad y = 12$

Therefore: . $KL \,=\,IJ \,=\,12$

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