If a chord of length $2 \sqrt{2}$ subtends a right angle at the centre of the circle, then its radius is

\sqrt{2}

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2

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Solution

The correct option is A $2$
$\begin{matrix} I n Δ A O B (A B^{2}) = {(A O)}^{2} + {(B O)}^{2} = {(2 \sqrt{2})}^{2} = r^{2} + r^{2} = 8 = 2 r^{2} = r^{2} = 4 r = 2 c m \end{matrix}$

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Similar questions

The chord of circle of radius 10 cm subtends a right angle at its centre. The length of the chord (in cm) is

(a) $\frac{5}{\sqrt{2}}$ (b) $5 \sqrt{2}$ (c) $10 \sqrt{2}$ (d) $10 \sqrt{3}$

2 \sqrt{3} u n i t s

120^{o}

10

(α, β)

x^{2} + y^{2} = a^{2}

x^{2} + y^{2} = a^{2}

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