Question

Point E is the midpoint of side BC of parallelogram ABCD (labeled counterclockwise) and AE ∩ BD =F. Find the area of ABCD if the area of △BEF is 3 cm^2

Answer 1

Answer:

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Step-by-step explanation:

Answer 2

Answer:

The area of parallelogram ABCD is 36 cm².

Step-by-step explanation:

Since here ABCD is the parallelogram.

Where E is the mid point of the line segment BC.

And, F is the intersection point of the segments AE and BD,

Also,  Area of △BEF is 3 .

We have to find Area of parallelogram ABCD.

Since,In  ΔBEF  and ΔACD,

$\angle ADF=\angle EBF$              ( because AD ║ BE )

$\angle DFA=\angle BFE$               (vertically opposite angle)

Thus, By AA similarity postulate,

$\triangle BEF\sim \triangle ACD$,

Thus, $\frac{\text{Area of }\triangle ADF}{\text{Area of }\triangle BFE}=(\frac{AD}{BE})^2$

But, $AD=2BE$,

$\frac{\text{Area of }\triangle ADF}{\text{Area of }\triangle BFE}=(\frac{2BE}{BE})^2=\frac{4}{1}$

$\text{Area of }\triangle ADF=12$

Similarly$\triangle ADB\sim \triangle BAE$,

Now, let the area of the triangle AFB be x.

Thus, x + 12 = 2(x+3) ( because the area of the ΔADB = 2(area of ΔBAE) )

⇒ x = 6

⇒ $\text{Area of }\triangle AFB=6$

⇒ $\text{Area of }\triangle ABD=12+6=18$

By the definition of diagonal of parallelogram,

$\text{Area of parallelogram ABCD}=2\times \text{Area of }\triangle ABD=2\times 18=36$

Therefore, the area of parallelogram ABCD is 36 cm².