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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 2 years 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 \(9^2=81\).

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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.

Yes you are right thanks.

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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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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 \(9^2=81\).

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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.

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Yes you are right thanks.

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