Question

ABCD is a parallelogram and e is the midpoint of BC show that area triangle DEC is equals to 1/4 area(ABCD)

Answer 1

Given that

⇒ ABCD is a parallelogram.

Kindly see the attachment also.

The diagonal BD divides the parallelogram ABCD in two equal triangles

Δ ABD and Δ BCD

Area of Δ ABD + Area of Δ BCD = Area of Prallelogram ABCD

(∵ Area of Δ ABD = Area of Δ BCD ) , so we have

Area of Δ BCD + Area of Δ BCD = Area of Prallelogram ABCD

2 (Area of Δ BCD ) = Area of Prallelogram ABCD   .....(i)  

Now E is the midpoint of BC , so DE divides Δ BCD into two equal triangles Δ BED & Δ DEC. So

Area of Δ BCD = Area of Δ BED + Area of Δ DEC

( ∵ Area of Δ BED = Area of Δ DEC ) , so

Area of Δ BCD = Area of Δ DEC + Area of Δ DEC

Area of Δ BCD = 2 (Area of Δ DEC) , putting this value in (i) , we get

2 [2(Area of Δ DEC )] = Area of Prallelogram ABCD

4(Area of Δ DEC ) = Area of Prallelogram ABCD

⇒ Area of Δ DEC  = 1/4 (Area of Prallelogram ABCD)

Which is required.

I hope it will help you.

Answer 2

Step-by-step explanation:

Area of ∆DEC=1/4(Area of parallelogram ABCD).

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