# Math Help - length of the bigger square..

1. ## length of the bigger square..

The ratio of the sides of 2 squares is 4:5. If the sum of their areas is 180 sq.cm, what is the length of the bigger square?

please do me a favor on this...
as i thought , can this be 4x + 5x = 180?

can anybody let me understand on this...
thanks a lot

2. ## Re: length of the bigger square..

Let $x$ be a side of the smaller square and let $y$ be a side of the bigger square.
The area of the smaller square is: $x^2$
The area of the bigger square is: $y^2$
So: $x^2+y^2=180$ (1)

You know the ratio of the sides of 2 square is: $\frac{4}{5}$, try to express this in function of $x$ and $y$ and then you can substitute this given into equation (1).

3. ## Re: length of the bigger square..

4/5 as x how about for y? we have only 1 equation... i cannot perform elimination in here.

thanks..

4. ## Re: length of the bigger square..

No, he did not say x was 4/5. Your problem says that the ratio of lengths is 4 to 5: x=(4/5)y.
$x^2+ y^2= 180$ becomes $\frac{16}{25}y^2+ y^2= 180$.

5. ## Re: length of the bigger square..

Hello, rcs!

The ratio of the sides of two squares is 4:5.
If the sum of their areas is 180 sq.cm,
what is the length of the bigger square?

You had a good start . . .

Let $4x$ = side of smaller square.
Let $5x$ = side of larger square.

Their areas are: . $\begin{Bmatrix}(4x)^2 &=& 16x^2 \\ (5x)^2 &=& 25x^2 \end{Bmatrix}$

The sum of their areas is 180: . $16x^2 + 25x^2 \:=\:180$

Go for it!

6. ## Re: length of the bigger square..

Sir Soroban, HallsofIvy you guys are an angel !