Answer:
27 square units
Step-by-step explanation:
Given the described geometry where ∆QPO has 3 times the base length and 3/4 the height of parallelogram UVWX, you want to know its area when the parallelogram area is 24 square units.
Dimensions
∆VNP ≅ ∆WNX, so VP = WX
∆UMQ ≅ ∆XMW, so UQ = XW
This makes segment PQ be 3 times the length of segment UV.
Since M and N are midpoints of UX and VW, respectively, parallelogram MNWX has half the height of parallelogram UVWX. The diagonals NX and MW cross at their respective midpoints, point O. If AB is the height line of the parallelogram through point O, then AO is the altitude of triangle QPO, and it is 3/4 of AB.
Area formulas
The result of these observations is that ∆QPO has 3 times the base and 3/4 the height of parallelogram UVWX. Then the area of the triangle is ...
A = 1/2bh = 1/2(QP)(AO)
= 1/2(3UV)(3/4AB) = 9/8·UV·AB
Of course, the area of the parallelogram is 24 = UV·AB, so the area of the triangle is ...
Area ∆QPO = 9/8(24) = 27 . . . . square units
