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Simora [160]
3 years ago
9

May I please have some assistance on this? Geometry.

Mathematics
1 answer:
NARA [144]3 years ago
8 0
The vertex of the right triangle will be in one of 2 places. It will be where lines extend straight out from both points A and B and meet. The two possible points are (-1,1) and (4,-2).
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Quadrilateral PQRS is a square whos side length is 10. Let X and Y be points outside the square so that XQ = YS = 6 and XP = YR
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392

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Triangles XQP and YRS are right triangles because triples 6, 8, 10 are Pythagorean triples.

Extend lines XQ, YR, YS and XP and mark their intersection as A and B.

Quadrilateral XAYB is a square because all right triangles PXQ, QAR, RYS and SBP are congruent (by ASA postulate) and therefore

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Segment XY is the diagonal of the square XAYB, by Pythagorean theorem,

XY^2=XA^2+AY^2\\ \\XY^2=14^2+14^2\\ \\XY^2=196+196\\ \\XY^2=392  

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