NCERT Solutions for Class 9 Maths Chapter 12 Herons Formula

Ans : Side of traffic signal board = a 
Perimeter of traffic signal board  
\(2 s=3 a \Rightarrow s=\frac{3}{2} a\)
By Heron's formula, 
Area of given triangle \(=\sqrt{s(s-a)(s-b)(s-c)}\)
\(\begin{aligned} \text { Area of given triangle } &=\sqrt{\frac{3}{2} a\left(\frac{3}{2} a-a\right)\left(\frac{3}{2} a-a\right)\left(\frac{3}{2} a-a\right)} \\ &=\sqrt{\left(\frac{3}{2} a\right)\left(\frac{a}{2}\right)\left(\frac{a}{2}\right)\left(\frac{a}{2}\right)} \\ &=\frac{\sqrt{3}}{4} a^{2} \end{aligned}\)....(1)
Perimeter of traffic signal board = 180 cm
Side of traffic signal board \((a)=\left(\frac{180}{3}\right) \mathrm{cm}=60 \mathrm{cm}\) 
\(\begin{array}{l}{\text { Using equation }(1), \text { area of traffic signal board }=\frac{\sqrt{3}}{4}(60 \mathrm{cm})^{2}} \\ {=\left(\frac{3600}{4} \sqrt{3}\right) \mathrm{cm}^{2}=900 \sqrt{3} \mathrm{cm}^{2}}\end{array}\) 

Ans : The sides of the triangle (i.e., a, b, c) are of 122 m, 22 m, and 120 m respectively. 
Perimeter of triangle = (122 + 22 + 120) m 
2s = 264 m 
s = 132 m
By Heron's formula, 
Area of triangle \(=\sqrt{s(s-a)(s-b)(s-c)}\)
\(\begin{aligned} \text { Area of given triangle } &=[\sqrt{132(132-122)(132-22)(132-120)}] \mathrm{m}^{2} \\ &=[\sqrt{132(10)(110)(12)}] \mathrm{m}^{2}=1320 \mathrm{m}^{2} \end{aligned}\)
Rent of 1 \(m^{2}\) area per year = Rs 5000
Rent of 1 \(m^{2}\) area per month = Rs \(\frac{5000}{12}\)
Rent of 1320 \(m^{2}\) area per 3 month = \(\operatorname{Rs}\left(\frac{5000}{12} \times 3 \times 1320\right)\)
= Rs (5000 x 330) = Rs 1650000 
Therefore, the company had to pay Rs 1650000. 

Ans : Sides of the triangular wall are 15 m, 11m, an 6m.
Semi perimeter of triangular wall (s) = (15 + 11 + 6) / 2 m = 16 m
Using Heron’s Formula,
\(  \begin{aligned} \text { Area of the message }=& \sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})} \\ &=\sqrt{16(16-15)(16-11)(16-6)} \mathrm{m}^{2} \\ &=\sqrt{16 \times 1 \times 5 \times 10} \mathrm{m}^{2}=\sqrt{800} \mathrm{m}^{2} \\ &=20 \sqrt{2} \mathrm{m}^{2} \end{aligned} \) 

Ans : Let the third side of the triangle be x. 
Perimeter of the given triangle 42 cm 
18 cm + 10 cm + x = 42 
x = 14 cm 
\(s=\frac{\text { Perimeter }}{2}=\frac{42 \mathrm{cm}}{2}=21 \mathrm{cm}\)
By Heron's formula, 
Area of a triangle =\(=\sqrt{s(s-a)(s-b)(s-c)}\)
\(\begin{aligned} \text { Area of the given triangle } &=(\sqrt{21(21-18)(21-10)(21-14)}) \mathrm{cm}^{2} \\ &=(\sqrt{21(3)(11)(7)}) \mathrm{cm}^{2} \\ &=21 \sqrt{11} \mathrm{cm}^{2} \end{aligned}\) 

Ans : Let the common ratio between the sides of the given triangle be x. 
Therefore, the side Of the triangle will be 12x, 17x, and 25x. 
Perimeter of this triangle = 540 cm 
12x + 17x + 25x = 540 cm 
54x = 540 cm 
x = 10 cm 
Sides of the triangle will be 120 cm, 170 cm, and 250 cm. 
\(s=\frac{\text { Perimeter of triangle }}{2}=\frac{540 \mathrm{cm}}{2}=270 \mathrm{cm}\)
By Heron's formula, 
\(\begin{aligned} \text { Area of triangle } &=\sqrt{s(s-a)(s-b)(s-c)} \\ &=[\sqrt{270(270-120)(270-170)(270-250)}] \mathrm{cm}^{2} \\ &=[\sqrt{270 \times 150 \times 100 \times 20}] \mathrm{cm}^{2} \end{aligned}\)
Therefore, the area Of this triangle is 9000  \(cm^{2}\) .