Geonetry

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

#Geometry

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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

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 $9^2=81$ .

A Former Brilliant Member - 5 years, 2 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.

Sal Gard - 5 years ago

Yes you are right thanks.

A Former Brilliant Member - 5 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$