Answer: [tex]36 cm^2[/tex]
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 [tex]cm^2[/tex].
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,
[tex]\triangle BEF \sim \triangle ACD[/tex],
Thus, [tex]\frac{area of \triangle ADF }{area of \triangle BFE} =(\frac{AD}{BE})^2[/tex]
But, AD = 2 BE,
Therefore, [tex]\frac{area of \triangle ADF }{area of \triangle BFE} =(\frac{2BE}{BE})^2[/tex] = 4/1
Thus, area of Δ ADF = [tex]12 cm^2[/tex]
Similarly, [tex]\triangle ADB \sim \triangle BAE[/tex]
Now, let the area of the triangle AFB = x [tex]cm^2[/tex]
Thus, x + 12 = 2(x+3) ( because the area of the ΔADB = 2(area of ΔBAE) )
⇒ x = 6 [tex]cm^2[/tex]
Therefore, area of ΔAFB= 6 [tex]cm^2[/tex]
⇒ area of ΔABD = 12 + 6 = 18 [tex]cm^2[/tex]
By the definition of diagonal of parallelogram,
Area of parallelogram ABCD = 2× area of Δ ABD= 2 × 18 = 36 [tex]cm^2[/tex]
