In an isosceles triangle ABC with AB = AC and AD is perpendicular to BC, if AD = 6 cm and the perimeter of ΔABC is 36 cm, then the area of ΔABC is: 

Given 

Δ ABC is an isosceles triangle , AB = AC 

Ad = 6 cm 

Formula used 

Area of triangle = 1/2 × base × height 

Figure:

Calculation:

Let, Base = b, and two equal sides = a 

According to the given data,

⇒ 2a + b = 36     ---(1)

⇒ 2a = 36 - b

⇒ a = (36 - b)/2     ---(2)

Now, AD perpendicular to BC ,so ΔADB is a right angle triangle.

So, AB² = AD² + BD²

⇒ a² = 6² + (b/2)²

⇒ a2 = 36 + b²/4

⇒ a2 - b2/4 = 36 

⇒ 4a2 - b2 = 36 × 4

⇒ (2a + b) (2a - b) = 144

Putting the value of (2a + b) from equation (1),

⇒ 36 × (2a - b) = 144

⇒ 2a - b = 4

Putting the value of b from equation (2),

⇒ 2 × (36 - b)/2 - b = 4

⇒ (36 - b) - b = 4

⇒ 36 - 2b = 4

⇒ 2b = 36 - 4 

⇒ b = 32/2 = 16 cm

So, the area of the Δ ABC = 1/2 × 16 × 6 = 48 cm²

∴ Area of triangle ABC is 48 cm2

Shortcut Trick

AD perpendicular to BC ,so ΔADB is a right angle triangle

We know the triplet  (6,8,10) 

so AB= 10 , AD = 6 cm and BD = 8 cm 

BC = 2 × BD      (In isosceles triangle, altitude and median are same)

BC = 16 cm 

Area of triangle ABC  = 1/2 × 16 × 6 = 48 cm²

∴ Area of triangle ABC is 48 cm2