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 cm2.

Answer 1

Answer: $36 cm^2$

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 $cm^2$.

We have to find Area of parallelogram ABCD.

Since,In  ΔBEF  and ΔACD,

∠ADF=∠EBF ( because AD ║ BE )

∠DFA = ∠BFE (vertically opposite angle)

Thus, By AA similarity postulate,

$\triangle BEF \sim \triangle ACD$,

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

But, AD = 2 BE,

Therefore,  $\frac{area of \triangle ADF }{area of \triangle BFE} =(\frac{2BE}{BE})^2$ = 4/1

Thus, area of Δ ADF = $12 cm^2$

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

Now, let the area of the triangle AFB = x $cm^2$

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

⇒ x = 6 $cm^2$

Therefore, area of ΔAFB= 6 $cm^2$

⇒ area of ΔABD = 12 + 6 = 18 $cm^2$

By the definition of diagonal of parallelogram,

Area of parallelogram ABCD = 2× area of Δ ABD= 2 × 18 = 36 $cm^2$