Find the length of a chord which subtends an angle of 90° at the centre of a circle whose radius is 8 cm.

1. Analyze the question

We are given a circle with a radius ($r$) of 8 cm. A chord is drawn such that it subtends an angle ($\theta$) of 90° at the center of the circle. We need to find the length of this chord.

2. Explain the concept

When a chord subtends an angle at the center of a circle, it forms an isosceles triangle with the two radii connecting the center to the endpoints of the chord.

• The two sides of the triangle are the radii ($r = 8$ cm).
• The angle between these two sides is 90°.
• The chord acts as the base of this triangle.

Since the angle is 90°, we have a right-angled isosceles triangle. We can find the length of the chord ($L$) using the Pythagorean theorem ($a^2 + b^2 = c^2$) or the trigonometric ratio for sine.

3. Show the solution

Method 1: Pythagorean Theorem In a right-angled triangle where the two legs are the radii ($r = 8$ cm): $L^2 = r^2 + r^2$ $L^2 = 8^2 + 8^2$ $L^2 = 64 + 64$ $L^2 = 128$ $L = \sqrt{128}$ $L = \sqrt{64 \times 2}$ $L = 8\sqrt{2}$ cm

Method 2: Trigonometry If we draw a line from the center to bisect the 90° angle, we create two smaller right-angled triangles with an angle of 45°. $\sin(45^\circ) = \frac{\text{opposite}}{\text{hypotenuse}}$ The chord is split into two equal parts, let's call each half $x$. $\sin(45^\circ) = \frac{x}{8}$ $\frac{1}{\sqrt{2}} = \frac{x}{8}$ $x = \frac{8}{\sqrt{2}}$ Rationalizing the denominator: $x = \frac{8\sqrt{2}}{2} = 4\sqrt{2}$ Since the full chord is $2x$: $L = 2 \times 4\sqrt{2} = 8\sqrt{2}$ cm

4. Identify the answer

Comparing our result to the given options, the correct choice is D.

5. Exam Tip

For any chord subtending an angle $\theta$ at the center, the length of the chord is given by the formula: $L = 2r \sin(\frac{\theta}{2})$ Using this for 90°: $L = 2(8) \sin(45^\circ) = 16 \times \frac{\sqrt{2}}{2} = 8\sqrt{2}$. This formula works for any angle!