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Preeti Dabral 1 month, 1 week ago

Similar as(side=14cm):

As per the given figure in question,

∠ABC = ∠BAC = ∠ACB = 60 ….[In an equilateral triangle all angles are of 60° each]

Now θ = 60°, r = {tex}{14 \over 2}{/tex} = 7 cm

Area of shaded region = ar(ΔABC) – 3{tex}\times{/tex} (area of minor sector)

{tex}{ = \frac{{\sqrt 3 }}{4}{{\left( {Side} \right)}^2}\; - \;3 \times \frac{\emptyset }{{360}}\;\pi {{\text{r}}^2}}{/tex}

{tex}{ = \;\;\frac{{1.73}}{4}\; \times 14 \times 14\; - \;3 \times \frac{{60}}{{360}}\; \times \;\frac{{22}}{7} \times 7 \times 7}{/tex}

{tex}{ = \;1.73\; \times 7 \times 7\; - \;3\; \times \frac{1}{6} \times 22 \times 7}{/tex}

{tex}{ = \;84.77\; - \;\frac{1}{2} \times 22 \times 7}{/tex}

{tex}{ = \;84.77\; - \;\;11 \times 7}{/tex}

= 84.77 - 77

= 7.77 cm

^{2}Hence, the area of the shaded region = 7.77 cm

^{2}0Thank You