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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 $\pi {r^2}$.

Complete step-by-step answer:

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,

$\dfrac{\theta }{{{{360}^ \circ }}} \times \pi {r^2} \Rightarrow \dfrac{{{{60}^ \circ }}}{{{{360}^ \circ }}} \times \dfrac{{22}}{7} \times {4^2} = 8.38c{m^2}$

Note: There are lots of variations of this question but the core concept is the same. We’ll be having some values, then we need to use the correct formula to get the value of known.

Complete step-by-step answer:

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,

$\dfrac{\theta }{{{{360}^ \circ }}} \times \pi {r^2} \Rightarrow \dfrac{{{{60}^ \circ }}}{{{{360}^ \circ }}} \times \dfrac{{22}}{7} \times {4^2} = 8.38c{m^2}$

Note: There are lots of variations of this question but the core concept is the same. We’ll be having some values, then we need to use the correct formula to get the value of known.