ABCD is a parallelogram with a base of BC. BFE is an straight line cut through the AD in equal 2 part at point F. CDE is another extended

straight line that touches the point E. If the area of the parallelogram ABCD is 80 cm^2, DE = CD, and AF = FD, find the area of a) triangle BCE, b) triangle ABF, c) triangle EFC.

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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**

First we have to drawthe parallelogram ABCD.

Let h1 be the distance between the parallel lines CD and AB. Then the area of the parallelogram ABCD = CD*h1 = 80 sq cm.

a)Triangle BCE with base BD has the same height h1 as the distance between the parallel lines AB and CD . Also CE = 2CD by data. So the area of the triangle BCE = (1/2)CE*h1 = (1/2)*2CD*h1 = CD*h1 = **80 sq cm.**

b) AF = FD = (1/2)AD. So the area of ABCD= AD*h2, where h2 is the distance between the parallel lines AD and BC. So the area of the triangle ABF = (1/2)AF*h2 = (1/2){(1/2)AD*h2)} = (1/4)AD*h2 = (1/4)area of ABCD = (1/4)(80) sq cm = **20 sq cm.**

c)Consider the triangles ABF and EDF: sides ED = CD by data. So ED = AB, as CD = AB in the parallelogram ABCD.

Sides FD = AF by data.

Angles EDF = angle FAB alternate angles , as AB || CD. AD is the intersecting the parallel lines AB and CD

Therefore by side angle side postulate triangles ABF and CDF are congruent. So CDF = ABF = 2o sq cm.

So area EFC = area CDF, as DE = DC and the height of F is same from EC to F.

So area EFC = 2 area CDF = 2*20 sq cm = **40 sq cm.**

**So area of BCE = 80 sq cm, area of ABF = 20 sq cm and area of EFC = 40 sq cm.**

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