Given: The signal board is an equilateral triangle with perimeter \(180~\text{cm}\)

Find the length of each side

$$\begin{aligned} a &= \frac{\text{Perimeter}}{3} \\&= \frac{180}{3} \\&= 60~\text{cm} \end{aligned}$$

Find the semi-perimeter

$$\begin{aligned} s &= \frac{\text{Perimeter}}{2} \\&= \frac{180}{2} \\&= 90~\text{cm} \end{aligned}$$

Calculate the area using Heron's formula

$$ \begin{aligned} \text{Area} &= \small\sqrt{s(s-a)(s-b)(s-c)} \\ &= \scriptsize\sqrt{90 \times (90-60) \times (90-60) \times (90-60)} \\ &= \sqrt{90 \times 30 \times 30 \times 30} \\ &= \sqrt{90 \times 27,000} \\ &= \sqrt{2,430,000} \\ &= 900 \sqrt{3}~\text{cm}^2 \end{aligned} $$

Answer: The area of the signal board is $$900\sqrt{3}~\text{cm}^2$$

Given: The sides of the triangular wall are $$\begin{aligned} a &= 122\,\text{m}\\b &= 22\,\text{m}\\ c &= 120\,\text{m} \end{aligned}$$

Calculate the semi-perimeter
$$\begin{aligned} s &= \frac{a + b + c}{2} \\&= \frac{122 + 22 + 120}{2} \\&= \frac{264}{2} \\&= 132\,\text{m} \end{aligned}$$

Calculate $$ \begin{aligned} s - a &= 132 - 122 = 10\,\text{m} \\ s - b &= 132 - 22 = 110\,\text{m} \\ s - c &= 132 - 120 = 12\,\text{m} \end{aligned} $$

Calculate area using Heron's formula
$$ \begin{aligned} A &= \sqrt{s(s-a)(s-b)(s-c)} \\ &= \sqrt{132 \times 10 \times 110 \times 12} \\ &= \sqrt{132 \times 10 \times 110 \times 12} \\ &= \sqrt{1,742,400} \\ &= 1,320\,\text{m}^2 \end{aligned} $$

Calculate advertising cost for 3 months
Yearly rate per \( = ₹5000 /\text{m}^2\)

So, cost for 3 months: $$\begin{aligned} \text{Cost} &= \text{Area} \times \text{Rate} \times \frac{3}{12} \\\\&= 1,320 \times 5,000 \times \frac{3}{12}\\\\ &= 1,320 \times 5,000 \times \frac{1}{4} \\\\ &= 1,320 \times 1,250 \\ &= ₹1,650,000 \end{aligned} $$

Answer: The company paid ₹1,650,000 for 3 months’ rent.

Given: The sides of the wall are $$\begin{aligned} a &= 15\, \text{m}\\b& = 11\, \text{m}\\c &= 6\, \text{m}\end{aligned}$$

Calculate the semi-perimeter
$$\begin{aligned} s &= \frac{a + b + c}{2} \\&= \frac{15 + 11 + 6}{2} \\&= \frac{32}{2} \\&= 16\, \text{m} \end{aligned}$$

Calculate
$$ \begin{aligned} s - a &= 16 - 15 = 1\\ s - b &= 16 - 11 = 5\\ s - c &= 16 - 6 = 10 \end{aligned} $$

Find the area using Heron's formula
$$ \begin{aligned} A &= \sqrt{s(s-a)(s-b)(s-c)}\\ &= \sqrt{16 \times 1 \times 5 \times 10}\\ &= \sqrt{16 \times 50}\\ &= \sqrt{800}\\ &= \sqrt{16 \times 50} \\ &= 4 \sqrt{50}\\ &= 4 \times \sqrt{25 \times 2} \\ &= 4 \times 5 \sqrt{2}\\ &= 20 \sqrt{2}\, \text{m}^2 \end{aligned} $$

Answer: The area painted in colour is $$20 \sqrt{2}\, \text{m}^2$$

Given: Two sides of the triangle are $$\begin{aligned}a &= 18\,\text{cm and } \\b &= 10\,\text{cm}\end{aligned}$$ with perimeter $$P = 42\,\text{cm}$$

Find the third side
$$\begin{aligned} c &= P - (a + b) \\&= 42 - (18 + 10) \\&= 42 - 28 \\&= 14\,\text{cm} \end{aligned}$$

Step 2: Calculate the semi-perimeter
$$ s = \frac{P}{2} = \frac{42}{2} = 21\,\text{cm} $$

Calculate \((s-a),\,(s-b)\text{ and } (s-c)\)
$$ \begin{aligned} s - a &= 21 - 18 = 3\,\text{cm}\\ s - b &= 21 - 10 = 11\,\text{cm}\\ s - c &= 21 - 14 = 7\,\text{cm} \end{aligned} $$

Calculate the area using Heron's formula
$$ \begin{aligned} A &= \sqrt{s(s-a)(s-b)(s-c)} \\ &= \sqrt{21 \times 3 \times 11 \times 7} \\ &= \sqrt{21 \times 3 \times 7 \times 11} \\ &= \sqrt{21 \times 21 \times 11} \\ &= 21\sqrt{11}\,\text{cm}^2 \end{aligned} $$

Answer: The area of the triangle is $$21\sqrt{11}\,\text{cm}^2$$

Given: The sides of the triangle are in the ratio $$12 : 17 : 25$$ and the perimeter is $$540\,\text{cm}$$

Let the sides be $$12x, 17x, 25x$$
Sum of ratios: $$12 + 17 + 25 = 54$$

Find the value of \(x\)
$$\begin{aligned} 12x + 17x + 25x &= 54x\\54x &= 540\\ \\\implies x &= \frac{540}{54} \\&= 10 \end{aligned}$$

Find the lengths of the sides
$$ \begin{aligned} a &= 12x = 12 \times 10 = 120\,\text{cm}\\ b &= 17x = 17 \times 10 = 170\,\text{cm}\\ c &= 25x = 25 \times 10 = 250\,\text{cm} \end{aligned} $$

Calculate the semi-perimeter
$$\begin{aligned} s &= \frac{a + b + c}{2} \\&= \frac{540}{2} \\&= 270\,\text{cm} \end{aligned}$$

Find \((s - a),\, (s - b),\, (s - c)\)
$$ \begin{aligned} s-a &= 270 - 120 = 150\,\text{cm}\\ s-b &= 270 - 170 = 100\,\text{cm}\\ s-c &= 270 - 250 = 20\,\text{cm} \end{aligned} $$

Calculate the area using Heron's formula
$$ A = \sqrt{ 270 \times 150 \times 100 \times 20 } $$ On calculating: $$ 270 \times 150 = 40,500 \\ 40,500 \times 100 = 4,050,000 \\ 4,050,000 \times 20 = 81,000,000 \\ A = \sqrt{81,000,000 } = 9,000\,\text{cm}^2 $$

Answer: The area of the triangle is $$9,000\,\text{cm}^2$$