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A chord PQ of length 16 cm is drawn in a circle of radius 10 cm. Calculate: a) The distance of the chord from the centre. b) The angle subtended by the chord at the centre.

Mathematics

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2 parts
a

16 cm16 \text{ cm}

b

106.26∘106.26^\circ

Step-by-step solution

8 steps

Here's how to solve the problem:

Part a· 4 steps

Calculate the length of the chord PQ.

  1. Identify the right-angled triangle.

    The radius of the circle (rr), the distance from the center to the midpoint of the chord (dd), and half the length of the chord (L2\frac{L}{2}) form a right-angled triangle.
    Given: r=10 cmr = 10 \text{ cm} and d=6 cmd = 6 \text{ cm}.

  2. Apply the Pythagorean theorem.

    r2=d2+(L2)2r^2 = d^2 + \left(\frac{L}{2}\right)^2 (10 cm)2=(6 cm)2+(L2)2(10 \text{ cm})^2 = (6 \text{ cm})^2 + \left(\frac{L}{2}\right)^2 100 cm2=36 cm2+(L2)2100 \text{ cm}^2 = 36 \text{ cm}^2 + \left(\frac{L}{2}\right)^2

  3. Solve for half the length of the chord.

    (L2)2=100 cm2−36 cm2\left(\frac{L}{2}\right)^2 = 100 \text{ cm}^2 - 36 \text{ cm}^2 (L2)2=64 cm2\left(\frac{L}{2}\right)^2 = 64 \text{ cm}^2 L2=64 cm2\frac{L}{2} = \sqrt{64 \text{ cm}^2} L2=8 cm\frac{L}{2} = 8 \text{ cm}

  4. Calculate the full length of the chord PQ.

    L=2×8 cmL = 2 \times 8 \text{ cm} L=16 cmL = 16 \text{ cm}

    The length of the chord PQ is 16 cm\boxed{16 \text{ cm}}.

Part b

106.26∘106.26^\circ

Calculate the angle chord PQ subtends at the centre of the circle.

  1. Consider the right-angled triangle formed by the center O, the midpoint M of the chord, and one endpoint P of the chord.

    Let θ\theta be the angle subtended by the chord PQ at the center O (i.e., ∠POQ\angle POQ). The line segment from O to M bisects this angle, so ∠POM=θ2\angle POM = \frac{\theta}{2}.
    In △OMP\triangle OMP:
    Hypotenuse OP=r=10 cmOP = r = 10 \text{ cm}
    Adjacent side OM=d=6 cmOM = d = 6 \text{ cm}

  2. Use trigonometry to find half the central angle.

    We can use the cosine function:

    cos⁡(θ2)=adjacenthypotenuse=OMOP\cos\left(\frac{\theta}{2}\right) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{OM}{OP} cos⁡(θ2)=6 cm10 cm\cos\left(\frac{\theta}{2}\right) = \frac{6 \text{ cm}}{10 \text{ cm}} cos⁡(θ2)=0.6\cos\left(\frac{\theta}{2}\right) = 0.6

  3. Calculate the value of θ2\frac{\theta}{2}.

    θ2=arccos⁡(0.6)\frac{\theta}{2} = \arccos(0.6) θ2≈53.13∘\frac{\theta}{2} \approx 53.13^\circ

  4. Calculate the full central angle θ\theta.

    θ=2×53.13∘\theta = 2 \times 53.13^\circ θ≈106.26∘\theta \approx 106.26^\circ

    The angle chord PQ subtends at the center is 106.26∘\boxed{106.26^\circ} (to two decimal places).

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