2 squares. need to find a side

[Guest]

So basically. I have a math problem that really annoys me

Can any one help me solve it?

[nathi123]

Let the side on the larger square is x and side on the small square is y.Then [tex]x^{2}-y ^{2}=17; 4(x+y)=68 \Rightarrow x+y=17 \Rightarrow (x+y)(x-y)=17 \Rightarrow x-y=1[/tex]
[tex]\Rightarrow x=1+y \Rightarrow (1+y) ^{2} -y ^{2} =17 \Leftrightarrow y=8 \Rightarrow x=1+8=9[/tex]

[Guest]

It took me a moment to understand math123's method so here is my explanation.

Taking x as the length of a side of the larger square and y as the length of the smaller square. The area of the larger square is $x^2$ and the area of the smaller square is $y^2$ so the area of the shaded part is $x^2- y^2= 17$. Now, the vertex of the smaller square on each sides divide the side into two parts. Let the length of the longer part be p and the length of the shorter part be q. Then each of the four shaded parts is a right triangle with legs of length p and q and hypotenuse y: $p^2+ q^2= y^2$.

So we have $x^2- y^2= 17$, $p+ q= x$, $p^2+ q^2= y^2$ and $4(p^2+ q^2)= 17$.

From the last two equations, $4(p^2+ y^2)= 4y^2= 17$ so $y^2= \frac{17}{4}$ and $y= \frac{\sqrt{17}}{2}$.
Then $x^2- y^2= x^2+ \frac{17}{4}= 17$. $x^2= 17- \frac{17}{4}= \frac{51}{4}$
$x= \frac{\sqrt{51}}{2}$.

A side of the larger square has length $\frac{\sqrt{51}}{2}$ and a side of the smaller square has length $\frac{\sqrt{17}}{2}$.