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 ∆DEF = b × h
Since both triangles are same,
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
