Let us take triangle ABC

Let there be another triangle which is exactly same as ∆ABC

If we join both these triangles

We get a parallelogram ABCD

So, two equal triangles form a parallelogram

Now,

Area of parallelogram = Base × Height

Area of ∆ABC + Area of ∆DCA = b × h

Sincebothtriangles aresame,

their area will be equal

Area of ∆ABC + Area of ∆ ABC = b × h

2 Area of ∆ABC = b × h

Area of ∆ABC = b × h

Area of ∆ABC = 1/2 × b × h

So, our formula is

Area of triangle = 1/2 × b × h

#### For right angled triangle ∆ABC

**
**

Height = h = AB

Base = b = BC

So,

Area of ∆ABC = 1/2 × b × h

= 1/2 × AB × BC

Here,

base is any side of triangle

& height is perpendicular from opposite vertex to the base

Let’s take some examples

#### Find area of Δ ABC

In ∆ABC,

Height = h = AD

= 4 cm

Base = b = BC

= 5 cm

So,

Area of ∆ABC = 1/2 × b × h

= 1/2 × 5 × 4

= 5 × 2

= 10 cm
^{
2
}

∴ Area of ∆ABC is 10 cm
^{
2
}

Note:Here, height is outside the triangle

#### Find area

In ∆ABC,

Height = h = AD

= 3 cm

Base = b = BC

= 4 cm

So,

Area of ∆ABC = 1/2 × b × h

= 1/2 × 4 × 3

= 2 × 3

= 6 cm
^{
2
}

∴ Area of ∆ABC is 6 cm
^{
2
}

#### Find area

In ∆ABC,

Height = h = BE

= 4 m

Base = b = AC

= 8 m

So,

Area of ∆ABC = 1/2 × b × h

= 1/2 × 8 × 4

= 4 × 4

= 16 m
^{
2
}

∴ Area of ∆ABC is 16 m
^{
2
}

#### Find area

In ∆ABC,

Height = h = BE

= 3 m

Base = b = AC

= 3 m

So,

Area of ∆ABC = 1/2 × b × h

= 1/2 × 3 × 3

= 1/2 × 9

= 9/2

= 4.5 m
^{
2
}

∴ Area of ∆ABC is 4.5 m
^{
2
}