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square \(ABCD\) divided into 18 smaller square.
17 squares of the sides of length 1.
\(ABCD\) square area Find.

Note by Mahla Salarmohammadi
1 year, 5 months ago

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Let 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 \(9^2=81\). Brilliant Member · 1 year, 5 months ago

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@Brilliant Member 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. Sal Gard · 1 year, 3 months ago

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@Sal Gard Yes you are right thanks. Brilliant Member · 1 year, 3 months ago

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