 # Areas of Parallelogram and Triangles

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## 09 Areas of Parallelogram and Triangles

#### AREAS OF PARALLELOGRAM AND TRIANGLES

Theorem 1:

Parallelograms on the same base and between the same parallels are equal in area. Given:

$\square \text{ABCD}$ and $\square \text{EFCD}$ are two parallelograms.

To prove:

Proof :

In $\Delta \text{ADE}$ and $\Delta \text{BCF}$

(Corresponding angles, $\text{AD}\parallel \text{BC}$,and AF transversal.)

(Corresponding angles, $\text{DE}\parallel \text{FC}$, and $\text{AF}$ transversal.)

So,

(Third angle of the triangles, by angle sum property)

(Opposite sides of parallelogram $\text{ABCD}$ )

$\Delta \text{ADE}\cong \Delta \text{BCF}$ (ASA rule)

So

(From the fig.)

.

Theorem 2: Area of triangle is half of parallelogram with common base and between same parallel.

Given:

Let $\Delta \text{ABC}$ and parallelogram $\text{BCED}$ be with common base $\text{BC}$.

$\mathrm{BC}\parallel ED$ To prove :

Construction: Draw $\text{AL}\perp \text{BC}$, $\text{EM}\perp \text{BC}$.

Extend $\text{BC}$ Up to $\text{M}$.

Proof:

AS $\text{BC}\parallel \text{DE}$,

(Distance between two parallel lines is equal)

Now,

.

Theorem 3: Two triangles on the same base (or equal bases) and between the same parallels are equal in area.

Given:

$\Delta \text{ABC}$ And $\Delta \text{PBC}$ be two triangles with common base $\text{BC}$ and between same

Parallel lines

$\text{BC}\parallel \text{AP}$ To prove:

Construction: Draw $\text{BD}\parallel \text{CA}$, and $\text{CQ}\parallel \text{BP}$,

Let line $\text{AP}$ meets $\text{BD}$ at $\text{D}$ and $\text{CQ}$ at $\text{Q}$.

Proof:

(Triangle and parallel gram with common base and

Between parallel lines) …(1)

And …(2)

$\square \text{BCAD}$ is parallelogram

$\square \text{BCQP}$ is parallelogram.

Both the parallelogram have same base and between two parallel lines,

….(3)

So, (from and $\text{3}$ )

Theorem 4 : Two triangles having the same base and equal areas lie between the same

Parallels.

Given:

$\Delta \text{ABC}$ and $\Delta PBC$ have equal base BC.

,

Let $\text{AN}\perp \text{BC}$, $\text{PM}\perp \text{BC}$ To prove:

Triangles lie between two parallel lines.

i.e.

Proof :

As

AS corresponding heights are equal, we say point $\text{A}$ and $\text{P}$ lie on the line parallel

to $\text{BC}$, through $\text{A}$.

Hence the result.

Change E’ to F