We know that

  Area of triangle =   1/2 × Base × Height

49.jpg

Here,

  Base = BC = b

  Height = AD

 

Finding height

Now,

  In an isosceles triangle,

  Median & Altitude are the same

 

So, D is mid-point of BC

∴ BD = DC = b/2

51.jpg

52.jpg

Find area of triangle ABC

53.jpg

We know that

  Area of triangle =   1/2 × Base × Height

 

Here,

  Base = BC = b = 4 cm

  Height = h = AD = ?

54.jpg

Now,

In an isosceles triangle,

Median and altitude are the same

 

So, D is mid-point of BC

∴ BD = DC = 4/2

   = 2cm

 

Now,

In ∆ADC, right angled at D

By Pythagoras theorem,

AC 2 = AD 2 + DC 2

(3) 2 = AD 2 + (2) 2

   9 = AD 2 + 4

9 − 4 = AD 2

5 = AD 2

AD 2 = 5

AD = √5 cm

54.jpg

So,

  Height = h

             = AD

             = √5 cm

 

Now,

  Area of a triangle = 1/2 × Base × Height

= 1/2 × 4 × √5

= 2 × √5

= 2√5 cm2

 

Find Area of triangle Δ ABC

55.jpg

We know that

  Area of triangle =   1/2 × Base × Height

 

Here,

  Base = b

   = BC

   = 2

      Height = h

 = AD

 = ?

Finding height,

56.jpg

Now,

  In an isosceles triangle,

  Median and altitude on base are the same.

 

So, D is the mid-point of BC

∴ BD = DC = 2/2

   = 1 cm

    56.jpg

Now,

In ∆ADC, right angled at D

By Pythagoras theorem.

 AC 2 = AD 2 + CD 2

(4) 2 = AD 2 + (1) 2

16 = AD 2 + 1

16 − 4 = AD 2

15 = AD 2

AD 2 = 15

AD = √15 cm

 

Thus,

Height = AD = √15 cm

 

Now,

  Area of a triangle = 1/2 × Base × Height

= 1/2 × 2 × √15

= √15 cm 2

 

  1. Chapter 6 Class 7 Triangle and its Properties
  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.