Question

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.

Answer 1

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,

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