**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 cm

^{2}= 64 cm^{2}^{}

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

Prove that
ar(∆ADO) = ar(∆BCO).

**Solution:**ar(∆ABD) = ar(∆BAC)

ar(∆ADO) + ar(∆AOB) = ar(∆BCO) + ar(∆AOB)

ar(∆ADO) = ar(∆BCO)

**Example 3:**In the given figure, prove that area (∆BAE) = area (∆CAD) if BC|| DE.

**Solution:**Area (∆BDE) = area (∆CDE) [Since, BC||DE]

Also, ∆ADE is common to both ∆BAE and ∆CAD.

Therefore,
area (∆BAE) = area (∆CAD).