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

Given : Two sides of a triangle are 18 cm and 10 cm and the perimeter is 42 cm.

Let the given sides be a = 18 cm, b = 10 cm

& Let the third side be = c cm

Perimeter of a triangle = a + b + c

42 = 18 + 10 + c

42 = 28 + c

c = 42 - 28  

c = 14

Third side (c) = 14 cm

Semi Perimeter of the ∆,s = (a + b + c) /2

s = (18 + 10 + 14) / 2

s = 42/2

s = 21 cm

Semi Perimeter of the ∆ = 21 cm

Using Heron’s formula :  

Area of the wall , A = √s (s - a) (s - b) (s - c)

A = √21(21 - 18) (21 - 10) (21 - 14)

A = √21 × 3 × 11 × 7

A = √(3 × 7) × (3) × (11) × (7)

A = √(3 × 3) × (7 × 7) × (11)  

A = 3 × 7√ 11

A = 21√11 cm²

Hence, the  area of a triangle is 21√11 cm².

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Answer 2

Third side = Perimeter - (Sum of other two sides)

=> 42 - (18 + 10)

=> 42 - 28

=> 14 cm

Formula to be used:

Heron's Formula

Heron's Formula: $\tt{\sqrt{s(s-a)(s-b)(s-c)}}$

Where:

s = semi perimeter

a, b, c = sides of triangle

a, b, c = 18 cm, 10 cm, 14 cm respectively

s = (a + b + c)/2 = (18 + 10 + 14)/2 = 42/2 = 21

Area of triangle = $\tt{\sqrt{s(s-a)(s-b)(s-c)}}$

=> $\tt{\sqrt{21(21-18)(21-10)(21-14)}}$

=> $\tt{\sqrt{21(3)(11)(7)}}$

=> $\tt{21\sqrt{11}}$ sq cm.