The diagram shows a 1 by x rectangular plank which fits neatly inside a 10 ×10 square frame. What is the value of x?
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,
$\displaystyle x=10\sqrt{2}-1$
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.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, $\displaystyle \frac{1}{2}d\cdot d=(1/2)d^2$
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?We see that $\displaystyle \Delta DAG$ is a isosceles right triangle with hypotenuse $\displaystyle DG = 1$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: $\displaystyle x\;=\;\sqrt{2}(DB) \;=\;\sqrt{2}\left(10 -\frac{\sqrt{2}}{2}\right) \;= \;10\sqrt{2} - 1$