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 extendedstraight line that touches the point E. If the area of the...

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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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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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