Question

prove that parallelogram on the same base and between the same parallel lines are equal in area​

Answer 1

【 GIVEN】 :-

Two parallelograms ABCD and EFCD, on

the same base DC and between the same parallels

AF and DC are given

【TO PROVE THAT】 :-

We need to prove that ar (ABCD) = ar (EFCD).

【PROOF】 :-

In Δ ADE and Δ BCF,

∠ DAE = ∠ CBF (Corresponding angles from AD || BC and transversal AF) (1)

∠ AED = ∠ BFC (Corresponding angles from ED || FC and transversal AF) (2)

Therefore, ∠ ADE = ∠ BCF (Angle sum property of a triangle) (3)

Also, AD = BC (Opposite sides of the parallelogram ABCD) (4)

So, Δ ADE ≅ Δ BCF [By ASA rule, using (1), (3), and (4)]

Therefore, ar (ADE) = ar (BCF) (Congruent figures have equal areas) (5)

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

= ar (BCF) + ar (EDCB) [From(5)]

= ar (EFCD)

So, parallelograms ABCD and EFCD are equal in area.

Answer 2

Given:-

• There are two parallelograms ABCD and ABEF  on the same base AB and between equal parallel lines AB and FC.

To prove:-

• Area of prallelogram ABCD and area of parallelogram ABEF are equal.

Proof:-

• We know that, opposite sides of a parallelogram are equal in lengths.

Hence,

• In Parallelogram ABCD, AD = BC

• And In Parallelogram ABEF, AF = BE

• Again, WE know that, opposite sides of parallelogram are parallel to each other.

∴In Parallelogram ABCD , AD || BC

And, In Parallelogram ABEF, AF || BE

Then,

         ∠DAF = ∠CBE

Now,

In ΔADF and ΔBCE,

                               AD = BC

                           ∠DAF = ∠CBF

                               AF = BE

• ∴ From, Side-Angle-Angle  condition, ΔADF and ΔBCE are congruent.

∴Area of ΔADF = Area of ΔBCE    

Now,

• Area of Parallelogram ABCD = Area of ABED + Area of ΔBCE.

•       => Area of ABED + Area of ΔADF

•       => Area of parallelogram ABEF

Note:-

• Kindly check diagram is in attachment !