Question

Parallelogram A B C D is shown. Line segment X Y goes from point X on side A B to point Y on side C D to form 2 trapezoids. The length of X Y is 11 and the length of Y C is 4. Sides X B and C Y are congruent.
Figure ABCD is a parallelogram. Points X and Y are placed so that BX ≅ DY and CD ⊥ XY.

The area of BXYC is 71.5 square units. What is the area of ABCD?

99 square units
110 square units
126 square units
143 square units

Answer 1

Place the parallelogram in coordinates: A(0,0), B(w,0), C(w+u,11), D(u,11). (XY ⟂ CD so XY is vertical, hence the height h = XY = 11.)

Y is 4 units from C along CD, so Y = (w+u−4,11). X has the same x-coordinate, so X = (w+u−4,0).

Given BX ≅ DY:
BX = Bx − Xx = w − (w+u−4) = 4 − u,
DY = Yx − Dx = (w+u−4) − u = w − 4,
so 4 − u = w − 4 ⇒ w + u = 8.

Compute area of B–X–Y–C by shoelace (or as a trapezoid): with these coordinates
Area(BXYC) = |u − 8|·11/2 = 71.5.
So |u − 8| = 143/11 = 13 ⇒ u − 8 = ±13.

If u − 8 = 13 then u = 21 and w = 8 − u = −13 (impossible). Thus u − 8 = −13 ⇒ u = −5 and w = 8 − u = 13.

Area of parallelogram = base·height = w·11 = 13·11 = 143.

Answer: 143 square units.