In a triangle with one angle of 120^circ,the | vedclass

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(B) Let the sides of the triangle be $a, b, c$ in $A.P.$ such that $a < b < c$. Given $c = 7 \ cm$.Since they are in $A.P.$,$2b = a + c$,so $a = 2b -

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$\frac{15\sqrt{3}}{4} \ cm^2$

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(B) Let the sides of the triangle be $a, b, c$ in $A.P.$ such that $a Since they are in $A.P.$,$2b = a + c$,so $a = 2b - 7$.The angle opposite to the greatest side $c$ is $120^\circ$. Using the Law of Cosines:$\cos(120^\circ) = \frac{a^2 + b^2 - c^2}{2ab}$$-\frac{1}{2} = \frac{(2b-7)^2 + b^2 - 7^2}{2(2b-7)b}$$-b(2b-7) = 4b^2 - 28b + 49 + b^2 - 49$$-2b^2 + 7b = 5b^2 - 28b$$7b^2 - 35b = 0$Since $b 
eq 0$,$b = 5 \ cm$.Then $a = 2(5) - 7 = 3 \ cm$.The area of the triangle is $\frac{1}{2}ab \sin(120^\circ) = \frac{1}{2} 	imes 3 	imes 5 	imes \frac{\sqrt{3}}{2} = \frac{15\sqrt{3}}{4} \ cm^2$.

In a triangle with one angle of $120^\circ$,the lengths of the sides form an $A.P.$ If the length of the greatest side is $7 \ cm$,the area of the triangle is:

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