Find the area of the triangle of sides 9 cm, 12 cm and 15 cm

Option 1 : 54

**Given: **

Sides of the triangle are 9cm, 12cm, and 15cm

**Concept Used:**

Area of any triangle with side given

\({\rm{\Delta }} = {\rm{\;}}\sqrt {{\rm{s}} \times \left( {{\rm{s}} - {\rm{a}}} \right) \times \left( {{\rm{s}} - {\rm{b}}} \right) \times \left( {{\rm{s}} - {\rm{c}}} \right)} {\rm{\;Unit}}{{\rm{s}}^2}\)

Where, s = (a + b + c)/2 and a, b, c are sides of the triangle

Area of a right triangle = ½ × base × height

**Calculation:**

Here,

a = 9 cm

b =12 cm

And c = 15 cm

S = (9 + 12 + 15)/2

⇒ 18

\(\sqrt {{\rm{s}} \times \left( {{\rm{s}} - {\rm{a}}} \right) \times \left( {{\rm{s}} - {\rm{b}}} \right) \times \left( {{\rm{s}} - {\rm{c}}} \right)} \)

⇒ √ 18 × (18 – 9) × (18 – 12) × (18 – 15)

⇒ √ 18 × 9 × 6 × 3

⇒ 54

**∴**** Area of the triangle is 54 cm ^{2}**

9^{2} + 12^{2} = 15^{2}

The given triangle is a right triangle and 9 cm and 12 cm are its base and height

Area of the triangle = ½ × 12 × 9

⇒ 54 cm^{2}

**∴ The area of the triangle is 54 cm ^{2}**