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

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,

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

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

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

Answer 2

If the area of  BXYC  is 71.5, the area of the parallelogram is going to be calculated as  143 square units.

What is a parallelogram?

This is a four sided shape that is known to have two sides that are opposite but parallel to each other.

In the question, we know that  BX ≅ DY

Therefore Area of ABCD = 2(71.5)

=  143 square units

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