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:

The area of ABCD is 36 cm².

Step-by-step explanation:

Given information: ABCD is a parallelogram, E is the midpoint of BC and AE ∩ BD=F.

In triangle EBF and ADF

$\angle FBE=\angle FDA$          (Alternate interior angles)

$\angle FEB=\angle FAD$          (Alternate interior angles)

$\angle BFE=\angle DFA$          (Vertically opposite angles)

Therefore triangle EBF and ADF are similar triangles by AA rule.

The sides BC and AD are opposite sides of parallelogram, therefore their lengths are equal. E is midpoint of BC.

$\frac{BC}{AD}=\frac{1}{2}$

If a point divides the side of a triangle in m:n, then the line segment between the point and the opposite vertex, divides the area of triangle is m:n.

Therefore the sides of triangle are in proportion of 1:2 and we can say that the point F divides the line EA and BD in 1:2.

In triangle ABE, the line BF divides the area of triangle is 1:2.

$\text{ Area of }\triangle BEF:\text{ Area of }\triangle ABF=1:2$

$\text{ Area of }\triangle ABF=2\text{ Area of }\triangle BEF$

$A_{\triangle ABF}=2A_{\triangle BEF}$

$A_{\triangle ABF}=2\times 3=6$

In triangle ABD, the line FA divides the area of triangle is 1:2.

$\text{ Area of }\triangle ABF:\text{ Area of }\triangle DFA=1:2$

$\text{ Area of }\triangle DFA=2\text{ Area of }\triangle ABF$

$A_{\triangle DFA}=2A_{\triangle ABF}$

$A_{\triangle DFA}=2\times 6=12$

$A_{\triangle ABD}=A_{\triangle ABF}+A_{\triangle ADF}$

$A_{\triangle ABD}=6+12=18$

Since AD is a diagonal of the parallelogram, therefore AD divides the area of parallelogram in two equal parts.

$A_{\triangle ABCD}=2A_{\triangle ABD}$

$A_{\triangle ABCD}=2\times 18=36$

Therefore the area of ABCD is 36 cm².