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Geometry Difficulty 1.4 Junior Find the answer Canada

In the diagram, $\$\triangle$
ABC $is right-angled at$ B$
and $BCDE$ is a square.

If $AB=8$ and $AC=17$ , the area of square $BCDE$ is

Pick one

Solution

In $\triangle ABC$ , $AC^2=AB^2+BC^2$ by the Pythagorean
Theorem.

Substituting, we get $17^2=8^2+BC^2$
or $BC^2=289-64=225$ . Since $BCDE$ is a square, then its area is equal
to $BC^2=225$ .

(Alternately, we could have determined that $BC=\sqrt{225}=15$ and so the area of the
square is $15\times15=225$ .)

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.