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__Areas of Parallelograms and Triangles__

__Areas of Parallelograms and Triangles__

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**Areas of Parallelograms**

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

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 ΔAFD and ΔBEC

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)

∴ CD = EF --------- (1)

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

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

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

CE = DF

AFD ≅ BEC (SSS congruence rule)

ar(ΔAFD) = ar(ΔBEC)

ar(||gm ABCD) = ar(trap ABED) + ar(ΔBEC)

= ar(trap ABED) + ar(ΔAFD)

= ar(||gm ABEF)

⇒ 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(ΔABP) = 12 ar( ABCD)

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**Areas of Triangles**

Two triangles on the same base or equal bases and between the same parallels are equal in area.

Given : Triangles ΔABC and ΔABD are on a common base AB and between same parallels AB and CD.

To Prove : ar(ΔABC) = ar(ΔABD)

Construction : Draw BL||AC , BM||AD

Proof : ABLC is a parallelogram

(∴ AC||BL and AB||CL)

ABMD is a parallelogram

(∴ AD||BM and AB||DM)

ar(||gm ABLC) = ar(||gm ABMD) ..........(1)

In (||gm ABLC) , ar(ΔABC) = ar (ΔCLB)

∴ar(||gm ABLC) = 2ar(ΔABC) ..........(2)

In (||gm ABMD) , ar(ΔABD) = ar (ΔBDM)

∴ar(||gm ABMD) = 2ar(ΔABD) ..........(3)

2ar(ΔABC) = 2 ar(ΔABD)..........(4)

⇒ ar(ΔABC) = ar(ΔABD)

Area of a triangle is half the product of its base and the corresponding altitude.

Median of a triangle divides it into two triangles of equal area.

Two triangles with same base (or equal bases) and equal areas will have equal corresponding altitudes.

Triangles on the same base and having equal areas lie between the same parallels.

If triangles ABC and ABD are on a common base AB and ar( ABC) = ar( ABD), then AB || CD.

Diagonals of a parallelogram divide it into four triangles of equal area.

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