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A chord of length $5$ cm subtends an angle of $60^\circ$ at the centre of a circle. The length, in cm, of a chord that subtends an angle of $120^\circ$ at the centre of the same circle is

1. $8$
2. $6\sqrt2$
3. $5\sqrt3$
4. $2\pi$

Given that, a chord length $5 \; \text{cm}$ subtends an angle of $60^{\circ}$ at the centre of a circle.

Let the radius of a circle be $`r\text{’} \; \text {cm}.$

So, $\boxed{OA = OB = OC = r}$ In $\triangle OAB,$ two sides are equal $(OA = OB = r).$

So, $\angle A = \angle B = 60^{\circ}$

Thus, $\triangle ABC$ is an equilateral triangle.

$OABC$ is a rhombus.

Area of rhombus $= \frac{d_{1} \times d_{2}}{2} \; \text{sq. units},$ where $d_{1}, d_{2}$ are the diagonal of the rhombus. Diagonals are perpendicular, to each other. $\boxed{d^{2}_{1} + d^{2}_{2} = 4a^{2}}$

So, $(OB)^{2} + (AC)^{2} = 4r^{2}$

$\Rightarrow (5)^{2} + (AC)^{2} = 4(5)^{2}$

$\Rightarrow 25 + (AC)^{2} = 100$

$\Rightarrow (AC)^{2} = 75$

$\Rightarrow AC = \sqrt{75} = \sqrt{25 \times 3}$

$\Rightarrow AC = 5 \sqrt{3} \; \text{cm}$

$\therefore$ The length of a chord that subtends an angle of $120^{\circ}$ at the center of the circle is $5 \sqrt{3} \; \text{cm}.$

Correct Answer $:\text{C}$

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