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3 years ago

11

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

= 8. Find XY^2.

1 answer:

erik [133]3 years ago

3 0

Answer:

392

Step-by-step explanation:

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

all angles of the quadrilateral XAYB are right angles

all sides of XAYB are congruent and equal to 6 + 8 = 14 units.

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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Answer:

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Step-by-step explanation:

Part 1) we know that

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Part 2) we know that

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Part 4) we know that

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