Question

plz anyone give answer of this question​

Answer 1

Step-by-step explanation:

two parallelograms ABCD and EFCD are given on the same base DC and between the same parallels.

We need to prove that,

Area (ABCD) = Area (EFCD)

In ∆ ADE and ∆ BCF,

∠ DAE = ∠ CBF … (1)

These are corresponding angles from AD parallel to BC and transversal to AF.

∠ AED = ∠ BFC … (2)

These are corresponding angles from ED parallel to FC and transversal to AF.

Therefore, using the angle sum property of triangles,

∠ ADE = ∠ BCF … (3)

Also, being the opposite sides of the parallelogram ABCD, AD = BC … (4)

So, by using the Angle-Side-Angle (ASA) rule of congruence and (1), (3) and (4), we have

∆ ADE ≅ ∆ BCF

We know that congruent figures have equal areas.

Hence,

Area (ADE) = Area (BCF) … (5)

Now, Area (ABCD) = Area (ADE) + Area (EDCB)

From the equation (5),

Therefore

Area (ABCD) = Area (BCF) + Area (EDCB) … [From (5)]= Area (EFCD)

So, the area of parallelogram ABCD and EFCD is equal.