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# geometry

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In triangle PQR, M is the midpoint of $$\overline{PQ}$$. Let X be the point on $$\overline{QR}$$ such that $$\overline{PX}$$ bisects $$\angle{QPR}$$ and let the perpendicular bisector of $$\overline{PQ}$$ intersect $$\overline{PX}$$ at Y. If PQ = 36, PR = 22, and MY = 8, then find the area of triangle PYR.

Apr 29, 2024

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Since M is the midpoint of PQ​, we know PM=MQ=2PQ​=18.

Since PX bisects ∠QPR, triangles QPX and RPX have equal area.

Let the area of triangle QPX (and therefore also the area of triangle RPX) be A1​.

Now, let's consider triangle PYM. Triangle PYM shares height with triangle QPX (the altitude from Y to PQ​ ), and the base of triangle PYM (segment PM with length 18) is 32​ the length of the base of triangle QPX (segment PQ with length 36).

Therefore, the area of triangle PYM (denoted by A2​) is 32​ the area of triangle QPX (denoted by A1​):

A2​=32​A1​

We are also given that MY=8. Since MY is an altitude of triangle PYM, we can find A2​ using the formula for the area of a triangle:

A2​=21​⋅base⋅height

A2​=21​⋅PM⋅MY=21​⋅18⋅8=72

We now know that A2​=72. Substituting this back into the equation A2​=32​A1​, we can find A1​:

72=32​A1​

A1​=72⋅23​=108

Finally, triangle PYR is the sum of triangles PYM and RPX. Since we found that A1​=A2​=108, the total area of triangle PYR is:

Area of PYR = A1​+A2 ​= 108 + 108 = 216​.

Apr 29, 2024