Question

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

We have a parallelogram ABCD and points X and Y that are placed such that BX = DY and CD ⊥ XY.Area ( BXYC ) = 71.5 sq. unitsThen the area of AXYD = BXYCArea ( ABCD ) = 2 * 71.5 = 143Answer:The area of ABCD is D ) 143 square units.

Answer 2

Answer-

The area of ABCD is 143 sq. units.

Solution-

$Area \ of \ parallelogram \ ABCD= Base \times Height = AB \times XY \ or \ CD \times XY$

(We can consider XY as height as it is perpendicular to CD)

$Area \ of \ trapzium \ BXYC=\frac{1}{2}(sum \ of \ parallel \ sides)(Height)$

and its area is given 71.5 sq. units.

$\Rightarrow \frac{1}{2}(BX+YC)(XY)=71.5$

We can replace BX as DY as BX ≅ DY is given,

$\Rightarrow \frac{1}{2}(DY+YC)(XY)=71.5$

And also CD= DY+YC,

$\Rightarrow \frac{1}{2}(CD)(XY)=71.5$

$\Rightarrow (CD)(XY)=143$

$\Rightarrow Area \ of \ parallelogram \ ABCD=143$