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
}

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