Question

prove that parallelograms on the same base and between same parallels lines are equal in area

Answer 1

for this you should know that the area of a parallelogram is equal to base*height where height is the perpendicular distance between the 2 parallels of the parallelogram. in the question it is given that the bases are equal and the parallelograms are between the same parallels. therefore both base and height are equal and hence their areas are also equal.

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 !