Question

Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.​

Answer 1

Step-by-step explanation:

Given :-

Two sides of a triangle are 18 cm and

10 cm

Perimeter of the triangle is 42 cm

To find :-

Area of the given triangle.

Solution :-

Given that

Two sides of a triangle are 18 cm and

10 cm

Let a = 18 cm

Let b = 10 cm

Let the third side be c cm

We know that

Perimeter of a triangle is equal to the sum of the lengths of the three sides.

Perimeter of the given triangle

= (a+b+c) cm

=> P = (18+10+c) cm

=> P = (28+c) cm

According to the given problem

Perimeter of the triangle = 42 cm

=> 28+c = 42

=> c = 42-28

=> c = 14 cm

The thrid side of the triangle = 14 cm

Now

We know that

Area of a triangle by Heron's formula

∆= √[S(S-a)(S-b)(S-c)] sq.units

Where, S = (a+b+c)/2 units

Now,

S = Perimeter/2 = 42/2 = 21 cm

The value of S = 21 cm

Now,

Area of the triangle

=> ∆ =√[21(21-18)(21-10)(21-14)] cm²

=> ∆ = √[21(3)(11)(7)] cm²

=> ∆ =√[21×21×11]

=> ∆ = 21√11 cm²

Or

We know that

√11 = 3.32(approximately)

Therefore, ∆ = 21×3.32 = 69.72 cm² (approximately)

Answer :-

Area of the given triangle is 21√11 cm² (or) 69.72 cm²

Used formulae:-

→ Area of a triangle by Heron's formula

∆= √[S(S-a)(S-b)(S-c)] sq.units

• Where, S = (a+b+c)/2 units
• a,b,c are the three sides

→ Perimeter of a triangle is equal to the sum of the lengths of the three sides.

→ √11 = 3.316... = 3.32 (approximately)