Area of isosceles triangle

We know that

  Area of triangle =   1/2 × Base × Height

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

Find area of triangle ABC

We know that

  Area of triangle =   1/2 × Base × Height

Here,

  Base = BC = b = 4 cm

  Height = h = AD = ?

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

So,

  Height = h

             = AD

             = √5 cm

Now,

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

= 1/2 × 4 × √5

= 2 × √5

= 2√5 cm ²

Find Area of triangle Δ ABC

We know that

  Area of triangle =   1/2 × Base × Height

Here,

  Base = b

   = BC

   = 2

      Height = h

 = AD

 = ?

Finding height,

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

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 − 1 = 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