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Problem 261

Geometry Difficulty 1.4 Multiple choice CEMC Pascal · Canada · 2025

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

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Official 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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