Question

prove that parallelogram on the same base and betwwen the same parallels are equal in area.​

Answer 1

Answer:

We have to prove that area of parallelograms ABCD and EFCD are equal.

In ΔADE and ΔBCF

∠DAE=∠CBF

Since, AD is parallel to BC and AB is a transversal. So these are corresponding angles.

∠AED=∠BFC

Since, DE is parallel to CF and EF is a transversal. So these are corresponding angles.

AD=BC

Since, ABCD is a parallelogram and these are opposite sides of a parallelogram.

Hence, ΔADE≅ΔBCF, by ASA congruence rule i.e., Angle – Side – Angle congruence rule. We know that congruent figures have the same area.

⇒area(ΔADE)=area(ΔBCF),,,,,,,(1)

Now, we have:

area(ABCD)=area(ADE)⊥area(EDCB)

Since, area(ΔADE)=area(ΔBCF) from (1)

⇒area(ABCD)=area(ΔBCF)+area(EDCB)=area(EDCF)

Thus, the area of the two parallelograms ABCD and EFCD are the same.

Hence, it is proved that parallelograms on the same base and between the same parallels are equal in area.

Step-by-step explanation:

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