We have a parallelogram ABCD, with the point F dividing AD in two equal parts.

Let us take the distance between the sides BC and AD as d.

Now, BC*d = AD*d = 80 cm^2.

We have AF = AD/2

The area of the triangle ABF is (1/2)*(AD/2)*d

= 80/4 = 20 cm^2

Similarly if we take the triangle DFC, the area is (1/2)*(AD/2)*h

= 80/4 = 20 cm^2

The area of the triangle BFC is 80 - 20 - 20 = 40 cm^2.

Now we draw a line EX perpendicular to BC from the point E.

If we consider angle ECX,

sin ECX = EX / EC = d / DC

we have DC = (EC /2)

So we get EX = EC*d/DC

=> EC*d*/(EC/2)

=> d* 2

The area of triangle BCE = (1/2)*(BC)*EX

=> (1/2)*BC*d*2

=> BC*d

=> 80 cm^2

The area of the triangle EFC = area of triangle BCE - area of triangle BFC.

We have derived the area of the triangle BFC as 40, so the area of EFC = 80 - 40 = 40 cm^2,

**Therefore the area of the triangle BCE is 80 cm^2, of triangle ABF is 20 cm^2 and of triangle EFC is 40 cm^2**

Posted on

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now