Let us take triangle ABC

8.jpg

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

9.jpg

 

10.jpg

 

If we join both these triangles

11.jpg

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

 

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

12.jpg

Height = h = AB

  Base = b = BC

 

So,

Area of ∆ABC = 1/2 × b × h

= 1/2 × AB × BC

13.jpg

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

 

14.jpg

 

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

15.jpg

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

16.jpg

 

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

17.jpg

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

  1. Chapter 11 Class 7 Perimeter and Area
  2. Concept wise

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.