square \(ABCD\) divided into 18 smaller square.

17 squares of the sides of length 1.

\(ABCD\) square area Find.

17 squares of the sides of length 1.

\(ABCD\) square area Find.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestLet area of ABCD be \(x^2\) and area of the last of the 18 squares be \(y^2\).

We have \(x^2=17+y^2\) or \((x+y)(x-y)=17\). We must have \(x+y=17\) and \(x-y=1\).

Solving we get \(x=9\) and \(y=8\).

Therefore area of ABCD is \(9^2=81\). – Svatejas Shivakumar · 8 months ago

Log in to reply

– Sal Gard · 6 months ago

Actually, x^2=17+y^2 if I'm not mistaken because there are seventeen squares of side length one. Therefore the area is 81. At least I know this case works because it can be drawn out. Maybe the wording messed one of us up.Log in to reply

– Svatejas Shivakumar · 6 months ago

Yes you are right thanks.Log in to reply