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The magnitude or measure of the part of the planar region enclosed by a simple closed figure is called its area.

Two figures are congruent if they have same shape and same size. Congruent figures have equal area but the converse need not be true.

Two figures are said to be on the same base and between the same parallels, if they have a common base and the vertices (or vertex) of the which are opposite to the common base of each figure lie on a line parallel to the base.

Parallelograms on the same base or equal bases and lie between the same parallels are equal in area. A parallelogram and a rectangle on the same base and between the same parallels are equal in equal area.

Given : Parallelograms ABCD and ABEF are on the same base AB, and between the same parallels AB and CF.

To Prove: The area of parallelogram ABCD is equal to the area of parallelogram ABEF.

Proof : In Î”AFD and Î”BEC

AF = BE (Opposite sides of parallelogram ABEF)

AD = BC (Opposite sides of parallelogram ABCD)

AB = CD (Opposite sides of parallelogram ABCD)

AB = EF (Opposite sides of parallelogram ABEF)

âˆ´ CD = EF --------- (1)

CD = CE + ED --------- (2)

EF = ED + DF --------- (3)

CE+~~ED~~ = ~~ED~~+DF --------- (4)

CE = DF

AFD â‰… BEC (SSS congruence rule)

ar(Î”AFD) = ar(Î”BEC)

ar(||gm ABCD) = ar(trap ABED) + ar(Î”BEC)

= ar(trap ABED) + ar(Î”AFD)

= ar(||gm ABEF)

â‡’ ar(||gm ABCD) = ar(||gm ABEF)

All figures that are on the same base and between same parallels need not have equal areas. Area of a parallelogram is equal to the product of any of its side and the corresponding height.

Parallelograms on the same base or equal bases that have equal areas lie between the same parallels.

If parallelograms ABCD and ABEF are on the same base AB and ar(ABCD) = ar(ABEF) then, AB || FC.

If a parallelogram and a triangle are on the same base and between the same parallels then the area of the triangle is equal to half the area of the parallelogram.

If triangle ABP and parallelogram ABCD are on the same base AB and between the same parallels AB and PC then, ar(Î”ABP) = $\frac{\text{1}}{\text{2}}$ ar( ABCD)

The magnitude or measure of the part of the planar region enclosed by a simple closed figure is called its area.

Two figures are congruent if they have same shape and same size. Congruent figures have equal area but the converse need not be true.

Two figures are said to be on the same base and between the same parallels, if they have a common base and the vertices (or vertex) of the which are opposite to the common base of each figure lie on a line parallel to the base.

Parallelograms on the same base or equal bases and lie between the same parallels are equal in area. A parallelogram and a rectangle on the same base and between the same parallels are equal in equal area.

Given : Parallelograms ABCD and ABEF are on the same base AB, and between the same parallels AB and CF.

To Prove: The area of parallelogram ABCD is equal to the area of parallelogram ABEF.

Proof : In Î”AFD and Î”BEC

AF = BE (Opposite sides of parallelogram ABEF)

AD = BC (Opposite sides of parallelogram ABCD)

AB = CD (Opposite sides of parallelogram ABCD)

AB = EF (Opposite sides of parallelogram ABEF)

âˆ´ CD = EF --------- (1)

CD = CE + ED --------- (2)

EF = ED + DF --------- (3)

CE+~~ED~~ = ~~ED~~+DF --------- (4)

CE = DF

AFD â‰… BEC (SSS congruence rule)

ar(Î”AFD) = ar(Î”BEC)

ar(||gm ABCD) = ar(trap ABED) + ar(Î”BEC)

= ar(trap ABED) + ar(Î”AFD)

= ar(||gm ABEF)

â‡’ ar(||gm ABCD) = ar(||gm ABEF)

All figures that are on the same base and between same parallels need not have equal areas. Area of a parallelogram is equal to the product of any of its side and the corresponding height.

Parallelograms on the same base or equal bases that have equal areas lie between the same parallels.

If parallelograms ABCD and ABEF are on the same base AB and ar(ABCD) = ar(ABEF) then, AB || FC.

If a parallelogram and a triangle are on the same base and between the same parallels then the area of the triangle is equal to half the area of the parallelogram.

If triangle ABP and parallelogram ABCD are on the same base AB and between the same parallels AB and PC then, ar(Î”ABP) = $\frac{\text{1}}{\text{2}}$ ar( ABCD)