detailed solution

Correct option is A

We know that, 

The Heron's formula for calculating the area of the triangle,

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

where, s is the semi-perimeter = half of the perimeter = 21 cm

and  a, b and c are the sides of the triangle 

Triangular sides of walls are, a = 18 cm, b = 10 cm, c = ?

Thus, Semi-perimeter 

s = (a $+$ b $+$ c)/2

21 = (18 $+$ 10 $+$ c)/2

21 $\times$ 2 = 28$+$ c

42 = 28 $+$c

c =  14 cm

Area of a triangular wall = $\sqrt{s (s - a) (s - b) (s -c)}$

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

= $\sqrt{21 \times 3 \times 11 \times 7}$

= 21$\sqrt{11}$  cm²

Hence, Area of the wall of the park to be painted in colour = 21 $\sqrt{11}$cm².

detailed solution

Correct answer is 1

We know that, 

The Heron's formula for calculating the area of the triangle,

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

where, s is the semi-perimeter = half of the perimeter = 21 cm

and  a, b and c are the sides of the triangle 

Triangular sides of walls are, a = 18 cm, b = 10 cm, c = ?

Thus, Semi-perimeter 

s = (a $+$ b $+$ c)/2

21 = (18 $+$ 10 $+$ c)/2

21 $\times$ 2 = 28$+$ c

42 = 28 $+$c

c =  14 cm

Area of a triangular wall = $\sqrt{s (s - a) (s - b) (s -c)}$

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

= $\sqrt{21 \times 3 \times 11 \times 7}$

= 21$\sqrt{11}$  cm²

Hence, Area of the wall of the park to be painted in colour = 21 $\sqrt{11}$cm².