Solution

Side of traffic signal board $=a$

Perimeter of traffic signal board $=3 \times a$

$2 s=3 a \Rightarrow s=\frac{3}{2} a$

By Heron’s formula,

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

Area of given triangle $=\sqrt{\frac{3}{2} a(\frac{3}{2} a-a)(\frac{3}{2} a-a)(\frac{3}{2} a-a)}$

$$ \begin{align} & =\sqrt{(\frac{3}{2} a)(\frac{a}{2})(\frac{a}{2})(\frac{a}{2})} \\ & =\frac{\sqrt{3}}{4} a^{2} \tag{1} \end{align} $$

Perimeter of traffic signal board $=180 cm$

Side of traffic signal board

$ (a)=(\frac{180}{3}) cm=60 cm $

Using equation (1), area of traffic signal board

$ =\frac{\sqrt{3}}{4}(60 cm)^{2} $

$=(\frac{3600}{4} \sqrt{3}) cm^{2}=900 \sqrt{3} cm^{2}$

Solution

The sides of the triangle (i.e., a, b, c) are of $122 m, 22 m$, and $120 m$ respectively.

Perimeter of triangle $=(122+22+120) m$

$2 s=264 m s=$

$132 m$

By Heron’s formula,

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

Area of given triangle $=[\sqrt{132(132-122)(132-22)(132-120)}] m^{2}$

$ =[\sqrt{132(10)(110)(12)}] m^{2}=1320 m^{2} $

Rent of $1 m^{2}$ area per year $=$ Rs 5000

Rent of $1 m^{2}$ area per month $=Rs \frac{\frac{5000}{12}}{12}$

$Rs(\frac{5000}{12} \times 3 \times 1320)$

Rent of $1320 m^{2}$ area for 3 months $=$

$ Rs(\frac{5000}{12} \times 3 \times 1320) $

$=Rs(5000 \times 330)=Rs 1650000$

Therefore, the company had to pay Rs 1650000 .

Solution

Here $a=15 \mathrm{~m}, \mathrm{~b}=11 \mathrm{~m}, \mathrm{c}=6 \mathrm{~m}$ $$ \begin{aligned} & s=\frac{a+b+c}{2} \ & s=\frac{15+11+6}{2}=16 \mathrm{~m} \end{aligned} $$

Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ $$ \begin{aligned} & =\sqrt{16(16-15)(16-11)(16-6)} \\ & =\sqrt{16 \times 1 \times 5 \times 10} \\ & =20 \sqrt{2} \mathrm{~m}^2 \end{aligned} $$

Hence, the area painted in colour $=20 \sqrt{2} m^2$

Solution

Let the third side of the triangle be $x$.

Perimeter of the given triangle $=42 cm$

$18 cm+10 cm+x=42 x=$

$14 cm$

$s=\frac{\text{ Perimeter }}{2}=\frac{42 cm}{2}=21 cm$

By Heron’s formula.

$ \begin{aligned} & \text{ 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)}) cm^{2} \\ & =(\sqrt{21(3)(11)(7)}) cm^{2} \\ & =21 \sqrt{11} cm^{2} \end{aligned} \end{aligned} $

Solution

Let the common ratio between the sides of the given triangle be $x$.

Therefore, the side of the triangle will be $12 x, 17 x$, and $25 x$.

Perimeter of this triangle $=540 cm$

$12 x+17 x+25 x=540 c m$

$54 x=540 cm x=$

$10 cm$

Sides of the triangle will be $120 cm, 170 cm$, and $250 cm$.

$s=\frac{\text{ Perimeter of triangle }}{2}=\frac{540 cm}{2}=270 cm$

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)}] cm^{2} \\ & =[\sqrt{270 \times 150 \times 100 \times 20}] cm^{2} \\ & =9000 cm^{2} \end{aligned} $

Therefore, the area of this triangle is $9000 cm^{2}$.

Solution

Let the third side of this triangle be $x$.

Perimeter of triangle $=30 cm$

$12 cm+12 cm+x=30 cm$

$ \begin{aligned} & x=6 cm \\ & s=\frac{\text{ Perimeter of triangle }}{2}=\frac{30 cm}{2}=15 cm \end{aligned} $

By Heron’s formula,

$ \begin{aligned} \text{ Area of triangle } & =\sqrt{s(s-a)(s-b)(s-c)} \\ & =[\sqrt{15(15-12)(15-12)(15-6)}] cm^{2} \\ & =[\sqrt{15(3)(3)(9)}] cm^{2} \\ & =9 \sqrt{15} cm^{2} \end{aligned} $