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

The area of ABCD is 36 cm².

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.

`(BC)/(AD)=(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* 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* 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* 18=36`

Therefore the area of ABCD is 36 cm².