Question

А 16 In the adjoining figure, E is mid-point of the side AB of a triangle ABC and EBCF is a parallelogram. If the area of AABC is 25 sq. units, find the area of || gm EBCF. Hint. Let EF meet AC at G. As E is mid-point of AB and EF || BC, G is mid-point of AC i.e. AG = GC. Also ZEAG = ZGCF and ZEGA= ZCGF G E F = B AAEGACFG.
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Answer 1

Answer:

The area of the  || gram is 25 square units

Step-by-step explanation:

These sort of questions look hard but are very easy once you observe and practice them :D

So, I'll explain how to do it!

Given,

AG = GC

Now, Triangles AGE and CGF are to be proven congruent :)

In Δ AGE and Δ CGF,

AG = GC [Side]

∠AGE = ∠CGF [angle]

∠EAG = ∠CFG [angle]

∴ΔAGE ≅ Δ CGF

Now, since they have been proven congruent, their sides, angles and areas are also congruent.

So this can be put down as follows:

Area of || gram = Ar. Of BCEG + Ar. of Δ AEG                        

                          = Ar. of Δ ABC

We know that,

Area of Δ ABC = 25 sq.units

∴ Area of || Gram = 25 sq. units

Hope this helps!