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There is a point X inside a square PQRS such that PX = 1, QX = 2 and triangles PXQ and PXS are congruent. What is the area,
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-> SOLUTION: Hi tutors, can u help me answer this question? thanks
There is a point X inside a square PQRS such that PX = 1, QX = 2 and triangles PXQ and PXS are congruent. What is the area,
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Question 977682: Hi tutors, can u help me answer this question? thanks
There is a point X inside a square PQRS such that PX = 1, QX = 2 and triangles PXQ and PXS are congruent. What is the area, in square units, of the square?
A.
B. 4
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D.
E. 5 Answer by solver91311(24713) (Show Source):
In order for triangles PXQ and PXS to be congruent, angles SPX and QPX must be congruent, which is to say they bisect the right angle vertex P. Construct a segment XY that is parallel to side PS and is perpendicular to PQ, intersecting PQ at Y.
Triangle PXY is then an isosceles right triangle with hypotenuse 1. And therefore segments PY and XY both measure
Now consider triangle QXY which is a right triangle with hypotenuse 2 and leg that measures . From this information and the Pythagorean Theorem, you can calculate the measure of QY.
The measure of the side of the square is PY + QY. Square this value to find the area of your square.
John
My calculator said it, I believe it, that settles it
