# Rectangle in a square

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• May 30th 2006, 03:59 PM
Natasha1
Rectangle in a square
The diagram shows a 1 by x rectangular plank which fits neatly inside a 10 ×10 square frame. What is the value of x?
• May 30th 2006, 04:10 PM
ThePerfectHacker
Note there are 2 big 45-45-90triangles, 2 small triangles 45-45-90 triangles and a rectangle in the middle.
Thus, the sum of all these areas is the area of big square which is 10x10=100.

The area of the rectange in the middle is 1xX=X
The area of the small triangles can easily be determined because they add up to a perfect square with diagnol of 1, thus the area is 1/2
The area of the big triangles can easily be determined because they add up to a perfect square with diagnol of x thus thee area is (1/2)x^2

Thus we have,
$\frac{1}{2}x^2+x+\frac{1}{2}=100$
Thus,
$x^2+2x+1=200$
Thus,
$(x+1)^2=200$
Thus,
$x=10\sqrt{2}-1$
• May 30th 2006, 04:28 PM
Natasha1
Thanks ;). I had a question though, do you find those little triangles by going through this reasoning.

We know the diagonals of these triangles is of 1 unit long then their sides are of lengths 1 / SQRT 2.

Therefore one little triangle on its own has area 1/4 but both have area 1/2.
• May 30th 2006, 07:25 PM
ThePerfectHacker
Quote:

Originally Posted by Natasha1
Thanks ;). I had a question though, do you find those little triangles by going through this reasoning.

We know the diagonals of these triangles is of 1 unit long then their sides are of lengths 1 / SQRT 2.

Therefore one little triangle on its own has area 1/4 but both have area 1/2.

You can do that.

I actually used a different reasoning. Since as I said the little triangles combine to give you a square. With a diagnol of 1 I used the following theorem. If the diagnols of a rhombus are d1 and d2 then is area is (1/2)d1d2 . Since a square is a rhombus and both its diagnols are equal we have its area as, $\frac{1}{2}d\cdot d=(1/2)d^2$
• May 30th 2006, 07:49 PM
Soroban
Hello, Natasha!

Quote:

1) The diagram shows a 1 by x rectangular plank which fits neatly inside a 10 ×10 square frame.
What is the value of x?
Code:

      A  D        B       * - * - - - - *       | *1  *      |     G *      * x  |       | *      *  |       |  *      * |       |    *      * E       |      *  * |       * - - - - * - *       D        F  C
We see that $\Delta DAG$ is a isosceles right triangle with hypotenuse $DG = 1$

. . Then: $AD = \frac{\sqrt{2}}{2}\quad\Rightarrow\quad DB = 10 - \frac{\sqrt{2}}{2}$

Since $\Delta DBE$ is an isosceles right triangle with legs $10 - \frac{\sqrt{2}}{2}$
. . then: $x\;=\;\sqrt{2}(DB) \;=\;\sqrt{2}\left(10 -\frac{\sqrt{2}}{2}\right) \;= \;10\sqrt{2} - 1$
• May 31st 2006, 12:25 AM
Natasha1
Got it! Thanks. It works my way too :)