The area of a triangle is 10sqrt{3} text{ cm} | vedclass

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(B) Given: Area $\Delta = 10\sqrt{3}$,$C = 60^{\circ}$,and perimeter $a + b + c = 20$.Using the area formula $\Delta = \frac{1}{2}ab \sin C$:$10\sqrt{

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$7$

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(B) Given: Area $\Delta = 10\sqrt{3}$,$C = 60^{\circ}$,and perimeter $a + b + c = 20$.Using the area formula $\Delta = \frac{1}{2}ab \sin C$:$10\sqrt{3} = \frac{1}{2}ab \sin 60^{\circ} = \frac{1}{2}ab \left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{4}ab$.Thus,$ab = 40$ ...$(i)$.From the Law of Cosines,$\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \cos 60^{\circ} = \frac{1}{2}$.So,$a^2 + b^2 - c^2 = ab$.We know $(a+b)^2 = a^2 + b^2 + 2ab$,so $a^2 + b^2 = (a+b)^2 - 2ab$.Substituting this into the cosine equation:$(a+b)^2 - 2ab - c^2 = ab \Rightarrow (a+b)^2 - c^2 = 3ab$.Since $a+b = 20-c$,we have:$(20-c)^2 - c^2 = 3(40)$.$400 - 40c + c^2 - c^2 = 120$.$400 - 40c = 120$.$40c = 280$.$c = 7$.

The area of a triangle is $10\sqrt{3} \text{ cm}^2$,angle $C = 60^{\circ}$,and its perimeter is $20 \text{ cm}$. Find the length of side $c$.

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