# Thread: Rectangle in a square

1. ## 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?

2. 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,
$x=10\sqrt{2}-1$

3. 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.

4. 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, $\frac{1}{2}d\cdot d=(1/2)d^2$

5. Hello, Natasha!

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: $x\;=\;\sqrt{2}(DB) \;=\;\sqrt{2}\left(10 -\frac{\sqrt{2}}{2}\right) \;= \;10\sqrt{2} - 1$

6. Got it! Thanks. It works my way too