Area Theorems

# Area Theorems

## Area

Area of a plane closed figure is the amount of surface (region) enclosed by it.
Area is measured in square units.
Two congruent figures have equal area but two figures which have equal area are not always congruent.

## Theorems on Area

Theorem 1: A diagonal of a parallelogram divides it into two congruent triangles having equal area.

Proof: In the given parallelogram ABCD, BD divides it into two triangles ABD and CBD.
In ABD and CBD, AB = CD and AD = BC (opposites sides of a parallelogram are equal.)
BD is common.

By SSS congruency, ABD CBD.
Since, congruent triangles have equal area.
Thus, ar(ABD) = ar(CBD).

Theorem 2: Parallelograms on the same base and between the same pair of parallel lines are equal in area.
In the given figure, ABCD and ABFE are two parallelograms between the two parallel lines AB and EC and on the same base AB.
Therefore, ar (||gm ABCD) = ar (||gm ABFE)

Corollary 1: A rectangle and a parallelogram on the same base and between the same pair of parallel lines have equal area.

Since a rectangle is also a parallelogram and ||gm ABCD and rectangle ABFE are on the same base, their areas will be same.
Therefore, ar (||gm ABCD) = ar(rect ABFE)

Corollary 2: Parallelograms with equal bases and between the same parallel lines are equal in area.

Proof: In the given figure, ABCD and EFGH are two parallelograms with bases AB = EF and both of them lie between two parallel lines m and n.
Here, ar (||gm ABCD) = AB × DY and ar (||gm EFGH) = EF × GX
But AB = EF (Given) and DY = GX (Perpendicular distance between two parallel lines are equal)
Therefore, ar (||gm ABCD) = ar (||gm EFGH)

Theorem 3: If a triangle and a parallelogram are on the same base and between the same parallel lines, then the area of the triangle is equal to half the area of the parallelogram.

In the given figures, ABX and ||gm ABCD are on the same base AB and between two parallel lines AB and CD.
Therefore, ar (ABX) = 1/2 ar (||gm ABCD)

Corollary: Area of a triangle is equal to half of the base multiplied by the height. In the given figure, ABE and ||gm ABCD are on the same base AB and height of ABE and ||gm ABCD is EF.

From the above theorem, ar(ABE) = 1/2 ar (||gm ABCD)
= 1/2 × AB × EF = 1/2 × base × height

Theorem 4: Triangles on the same base and between the same pair of parallel lines have equal area.
In the given figure, ABC and ABD have same base AB and are between the same pair of parallel lines AB and CD.
Therefore, ar(ABC) = ar(ABD)

Corollary: Triangles with equal base and between the same pair of parallel lines are equal.
In the given figure, AE||CF and AB = DE.
Therefore, ar(ABC) = ar(DEF)

Example 1: In the given figure, area of ABX = 32 cm2 . Find the area of ||gm ABCD.
Solution: Ar (||gm ABCD) = 2 × ar(ABX)
= 2 × 32 cm2 = 64 cm2

Example 2: In the given figure, ABCD is a parallelogram.