The area of the triangle ABC is 10sqrt{3} tex | vedclass

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(D) Given: $	ext{Area} = 10\sqrt{3} 	ext{ cm}^2$,$\angle B = 60^{\circ}$,and $a+b+c = 20 	ext{ cm}$.Using the area formula: $	ext{Area} = \frac{1}

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$7$

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(D) Given: $	ext{Area} = 10\sqrt{3} 	ext{ cm}^2$,$\angle B = 60^{\circ}$,and $a+b+c = 20 	ext{ cm}$.Using the area formula: $	ext{Area} = \frac{1}{2}ac \sin B$.$10\sqrt{3} = \frac{1}{2}ac \sin 60^{\circ} \Rightarrow 10\sqrt{3} = \frac{1}{2}ac \left(\frac{\sqrt{3}}{2}ight)$.$10\sqrt{3} = \frac{ac\sqrt{3}}{4} \Rightarrow ac = 40$.Using the Law of Cosines: $b^2 = a^2 + c^2 - 2ac \cos B$.$b^2 = (a+c)^2 - 2ac - 2ac \cos 60^{\circ}$.Since $a+c = 20-b$,we have $b^2 = (20-b)^2 - 2(40) - 2(40)(0.5)$.$b^2 = 400 + b^2 - 40b - 80 - 40$.$0 = 280 - 40b$.$40b = 280 \Rightarrow b = 7 	ext{ cm}$.Thus,$\ell(AC) = b = 7 	ext{ cm}$.

The area of the triangle $ABC$ is $10\sqrt{3} \text{ cm}^2$,angle $B$ is $60^{\circ}$ and its perimeter is $20 \text{ cm}$. Then $\ell(AC) = $ (in $\text{ cm}$)

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