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# AB is the chord of a circle of centre O and radius 4 cm. AB is of length 4 cm. Find the area of the sector of the circle formed by chord AB. Verified
Hint: Use the information, area of the sector $= \dfrac{\theta }{{{{360}^ \circ }}} \times \pi {r^2}$ and area of the circle with radius r is $\pi {r^2}$.
Given the radius of the circle $r = 4cm$ and length of chord $= 4cm$ $\therefore$ the triangle formed is an equilateral triangle. $\therefore \theta = {60^ \circ }$ because every angle of the equilateral triangle is ${60^ \circ }$. Now, we know that the area of the sector $= \dfrac{\theta }{{{{360}^ \circ }}} \times \pi {r^2}$. Where, $\theta$ is angle subtended at a centre and $\pi {r^2}$ is the area of the circle. On putting the values,
$\dfrac{\theta }{{{{360}^ \circ }}} \times \pi {r^2} \Rightarrow \dfrac{{{{60}^ \circ }}}{{{{360}^ \circ }}} \times \dfrac{{22}}{7} \times {4^2} = 8.38c{m^2}$