JAMB - Mathematics (1993 - No. 25)
Calculate the length in cm. of the arc of a circle of diameter 8cm which subtends an angle of 22\(\frac{1}{2}\)o at the centre of the circle
2\(\pi\)
\(\pi\)
\(\frac{2}{3}\)
\(\frac{\pi}{2}\)
Explanation
Diameter = 8cm
∴ Radius = 4cm
Length of arc = \(\frac{\theta}{360}\) x 2 \(\pi\)r but Q = 22\(\frac{1}{2}\)
∴ Length \(\frac{22\frac{1}{2}}{360}\) x 2 x \(\pi\) x 4
= \(\frac{22\frac{1}{2} \times 8\pi}{360}\)
= \(\frac{180}{360}\)
= \(\frac{\pi}{2}\)
∴ Radius = 4cm
Length of arc = \(\frac{\theta}{360}\) x 2 \(\pi\)r but Q = 22\(\frac{1}{2}\)
∴ Length \(\frac{22\frac{1}{2}}{360}\) x 2 x \(\pi\) x 4
= \(\frac{22\frac{1}{2} \times 8\pi}{360}\)
= \(\frac{180}{360}\)
= \(\frac{\pi}{2}\)
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