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

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- May 30th 2006, 02:59 PMNatasha1Rectangle 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, 03:10 PMThePerfectHacker
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,

$\displaystyle \frac{1}{2}x^2+x+\frac{1}{2}=100$

Thus,

$\displaystyle x^2+2x+1=200$

Thus,

$\displaystyle (x+1)^2=200$

Thus,

$\displaystyle x=10\sqrt{2}-1$ - May 30th 2006, 03:28 PMNatasha1
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, 06:25 PMThePerfectHackerQuote:

Originally Posted by**Natasha1**

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, $\displaystyle \frac{1}{2}d\cdot d=(1/2)d^2$ - May 30th 2006, 06:49 PMSoroban
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

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

Since $\displaystyle \Delta DBE$ is an isosceles right triangle with legs $\displaystyle 10 - \frac{\sqrt{2}}{2}$

. . then: $\displaystyle x\;=\;\sqrt{2}(DB) \;=\;\sqrt{2}\left(10 -\frac{\sqrt{2}}{2}\right) \;= \;10\sqrt{2} - 1$ - May 30th 2006, 11:25 PMNatasha1
Got it! Thanks. It works my way too :)