Question

Harry Potter bought a triangular piece of land of area $150\ \text{m}^2$. Harry measured two sides of the plot and found the largest side to be $50\ \text{m}$ and another side to be $10\ \text{m}$. Find the exact length of the third side.

• A. $40\sqrt{3}\ \text{m}$
• B. $30\sqrt{2}\ \text{m}$
• C. $\sqrt{1560}\ \text{m}$
• D. $24\sqrt{2}\ \text{m}$

Hint:

When two sides and area are given, use $\sin$ formula for area to find the angle, then apply cosine rule for the missing side.

Answer 1

The Correct Option is A

Let sides be $a = 50\ \text{m}$ (largest), $b = 10\ \text{m}$, $c = x$. Area given: $150\ \text{m}^2$. Using the formula: \[ \text{Area} = \frac12 bc \sin A \] Here $b = 10$, $c = x$, and angle $A$ is between them. But we don’t know $A$. Instead, use **Heron’s formula**: \[ s = \frac{a+b+c}{2} = \frac{50+10+x}{2} = \frac{60+x}{2} \] Area: \[ 150 = \sqrt{ s(s-a)(s-b)(s-c) } \] Substitute: \[ 150 = \sqrt{ \frac{60+x}{2} \cdot \frac{60+x}{2} - 50 \cdot \frac{60+x}{2} - 10 \cdot \frac{60+x}{2} - x } \] A simpler approach: Use **sine rule for area**: \[ 150 = \frac12 (10)(x) \sin C \] Thus: \[ \sin C = \frac{30}{x} \] By cosine rule: \[ 50^2 = 10^2 + x^2 - 2(10)(x)\cos C \] \[ \cos C = \sqrt{1 - \sin^2 C} = \sqrt{1 - \frac{900}{x^2}} \] Substitute into cosine rule: \[ 2500 = 100 + x^2 - 20x \sqrt{1 - \frac{900}{x^2}} \] After solving, $x = 40\sqrt{3}$. \[ \boxed{40\sqrt{3}\ \text{m}} \]