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: 36 cm2
Answers
Step-by-step explanation:
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∠ADF=∠EBF ( because AD ║ BE )
\angle DFA=\angle BFE∠DFA=∠BFE (vertically opposite angle)
Thus, By AA similarity postulate,
\triangle BEF\sim \triangle ACD△BEF∼△ACD ,
Thus, \frac{\text{Area of }\triangle ADF}{\text{Area of }\triangle BFE}=(\frac{AD}{BE})^2Area of △BFEArea of △ADF=(BEAD)2
But, AD=2BEAD=2BE ,
\frac{\text{Area of }\triangle ADF}{\text{Area of }\triangle BFE}=(\frac{2BE}{BE})^2=\frac{4}{1}Area of △BFEArea of △ADF=(BE2BE)2=14
\text{Area of }\triangle ADF=12Area of △ADF=12
Similarly\triangle ADB\sim \triangle BAE△ADB∼△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=6Area of △AFB=6
⇒ \text{Area of }\triangle ABD=12+6=18Area of △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=36Area of parallelogram ABCD=2×Area of △ABD=2×18=36
Therefore, the area of parallelogram ABCD is 36 cm².
