Question

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 $\ell(\text{AC}) =$

• A. 10 cm
• B. 8 cm
• C. 5 cm
• D. 7 cm

Hint:

Whenever you see a system of equations where you have values for both $ac$ and $a+c$, look for ways to eliminate them using perfect square expansions. This avoids having to solve for the individual side lengths $a$ and $c$ separately!

Answer 1

The Correct Option is D

Step 1: Translate the area into a side product.
Area $= \dfrac{1}{2}ac\sin B$. With area $10\sqrt{3}$ and $B = 60^\circ$ (so $\sin B = \dfrac{\sqrt{3}}{2}$): $10\sqrt{3} = \dfrac{1}{2}ac\cdot\dfrac{\sqrt{3}}{2}$.
Step 2: Solve for $ac$.
$10\sqrt{3} = \dfrac{\sqrt{3}}{4}ac \Rightarrow ac = 40$.
Step 3: Use the perimeter.
$a + b + c = 20$, so $a + c = 20 - b$.
Step 4: Write the Cosine Rule for $B$.
$\cos 60^\circ = \dfrac{a^2 + c^2 - b^2}{2ac}$, and $\cos 60^\circ = \dfrac{1}{2}$. Use $a^2 + c^2 = (a+c)^2 - 2ac$.
Step 5: Substitute the known values.
$\dfrac{1}{2} = \dfrac{(20-b)^2 - 2(40) - b^2}{2(40)}$, so $40 = (400 - 40b + b^2) - 80 - b^2$.
Step 6: Solve for $b = \ell(AC)$.
The $b^2$ terms cancel: $40 = 320 - 40b \Rightarrow 40b = 280 \Rightarrow b = 7$ cm.
\[ \boxed{\ell(AC) = 7\ \text{cm}} \]