**1** **A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is $180 \ cm,$ what will be the area of the signal board?**

Side of traffic signal board =$ a$$\\$ Perimeter of traffic signal board = $3 * a$$\\$ $2s=3a \Rightarrow s=\dfrac{3}{2}a$$\\$ By Heron’s formula, Area of triangle =$\\$ $\sqrt{s(s-a)(s-b)(s-c)}$$\\$ Area of given triangle=$\\$ $\sqrt{\dfrac{3}{2}a(\dfrac{3}{2}a-a)(\dfrac{3}{2}a-a)(\dfrac{3}{2}a-a)}$$\\$ $=\sqrt{\dfrac{3}{2}a(\dfrac{a}{2})(\dfrac{a}{2})(\dfrac{a}{2})}$$\\$ $=\dfrac{\sqrt{3}}{2}a^2....(1)$$\\$ Perimeter of traffic signal board $= 180 cm$$\\$ Side of traffic signal board $(a) =(\dfrac{180}{3}) cm =60 cm$$\\$ Using Equation (1), area of traffic signal board $=\dfrac{\sqrt{3}}{2}(60 cm)^2$$\\$ $=(\dfrac{3600}{4}\sqrt{3}) cm^2$$\\$ $=900\sqrt{3}cm^2$

**2** **The triangular side walls of a flyover have been used for advertisements. The sides of the walls are $122 m, 22 m,$ and $120 m$ (see the given figure). The advertisements yield an earning of Rs. $5000 $ per $m^2$ per year. A company hired one of its walls for $3$ months. How much rent did it pay?**

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$$\\$ By Heron’s formula,$\\$ Area of triangle =$\sqrt{s(s-a)(s-b)(s-c)}$$\\$ Area of given triangle =$[\sqrt{132(132-122)(132-22)(132-120)}]m^2$$\\$ $=[\sqrt{132(10)(110)(12)}]$$\\$ Rent of $1 m^2$ area per year = Rs. $5000$$\\$ Rent of $1 m^2$ area per month = Rs.$(\dfrac{5000}{15}*3*1320)$$\\$ $Rs.(5000*330)=Rs 1650000$$\\$ Therefore, the company had to pay Rs. $1650000.$

**3** **The floor of a rectangular hall has a perimeter $250 m$. If the cost of panting the four walls at the rate of Rs. $10 $ per $m^2$ is Rs.$15000,$ find the height of the hall. [Hint: Area of the four walls = Lateral surface area.]**

Let length, breadth, and height of the rectangular hall be $l$ m, $b$ m, and $h$ m respectively.$\\$ Area of four walls = $2lh + 2bh = 2(l + b) h$$\\$ Perimeter of the floor of hall =$ 2(l + b) = 250 m$$\\$ $\therefore $ Area of four walls = $2(l + b) h = 250h m^2$$\\$ Cost of painting per $m^2$ area = Rs. $10$$\\$ Cost of painting $250h m^2$ area = Rs.$ (250h × 10) = Rs. 2500h$$\\$ However, it is given that the cost of paining the walls is Rs. $15000.$$\\$ $\therefore 15000 = 2500h$$\\$ $h = 6$$\\$ Therefore, the height of the hall is $6 m.$

**4** **Find the area of a triangle two sides of which are $18 cm$ and $10 cm$ and the perimeter is $42 cm.$**

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=\dfrac{\text{Perimeter}}{2}=\dfrac{42 cm}{2}\\ =21 cm$$\\$ By Heron’s formula, Area of triangle =$\sqrt{s(s-a)(s-b)(s-c)}$$\\$ Area of given triangle = $\\$ $[\sqrt{21(21-18)(21-10)(21-14)}]cm^2$$\\$ $[\sqrt{21(3)(11)(7)}]$$\\$ $=21\sqrt{11}cm^2$$\\$

**5** **Sides of a triangle are in the ratio of $12:17:25 $ and its perimeter is $540 cm$. Find its area.**

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=\dfrac{\text{Perimeter of triangle}}{2}\\ \dfrac{540 cm}{2}\\ =270 cm$$\\$ By Heron’s formula,$\\$ Area of triangle =$\\$ $\sqrt{s ( s - a )( s - b )( s - c )}$$\\$ Area of given triangle = $\sqrt{270 ( 270 -120 )(270 - 170 )( 270 - 250 )} cm ^2$$\\$ $\sqrt{ 270 ( 150 )(100 )( 20 )} cm^ 2$$\\$ $9000cm^2$$\\$ Therefore, the area of this triangle is $9000 cm ^2$ .