For an equilateral triangle of side $a$, $s=\dfrac{3a}{2}$. By Heron's formula, area $=\sqrt{s(s-a)(s-a)(s-a)}=\sqrt{\dfrac{3a}{2}\left(\dfrac a2\right)^3}=\dfrac{\sqrt3}{4}a^2$. If the perimeter is $180$ cm, then $a=60$ cm, so area $=\dfrac{\sqrt3}{4}\times60^2=900\sqrt3\text{ cm}^2$.
The area is $\dfrac{\sqrt3}{4}a^2$. If the perimeter is $180$ cm, the area is $900\sqrt3\text{ cm}^2$.
Since $22^2+120^2=122^2$, the wall is a right triangle. Its area is $\dfrac12\times22\times120=1320\text{ m}^2$. The rate for $3$ months is $\dfrac14$ of $₹5000$, i.e. $₹1250$ per $\text{m}^2$. Therefore, rent $=1320\times1250=₹16,50,000$.
The company paid $₹16,50,000$.
Here $s=\dfrac{15+11+6}{2}=16$. Area $=\sqrt{16(16-15)(16-11)(16-6)}=\sqrt{16\times1\times5\times10}=20\sqrt2\text{ m}^2$.
The painted area is $20\sqrt2\text{ m}^2$, approximately $28.3\text{ m}^2$.
The third side is $42-(18+10)=14$ cm, and $s=21$ cm. Area $=\sqrt{21(21-18)(21-10)(21-14)}=\sqrt{21\times3\times11\times7}=21\sqrt{11}\text{ cm}^2$.
$21\sqrt{11}\text{ cm}^2$, approximately $69.6\text{ cm}^2$.
Let the sides be $12x,17x,25x$. Then $54x=540$, so $x=10$ and the sides are $120$ cm, $170$ cm and $250$ cm. Now $s=270$ cm. Area $=\sqrt{270(150)(100)(20)}=9000\text{ cm}^2$.
$9000\text{ cm}^2$.
The third side is $30-12-12=6$ cm, and $s=15$ cm. Area $=\sqrt{15(15-12)(15-12)(15-6)}=\sqrt{15\times3\times3\times9}=9\sqrt{15}\text{ cm}^2$.
$9\sqrt{15}\text{ cm}^2$, approximately $34.9\text{ cm}^2$.