+42 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.
+1610 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=32A1
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=32A1, we can find A1:
72=32A1
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.
