Question

please give the solution of this question in detail please help please

Answer 1

In triangle PQR,

Let the central angle be x.


$\cos(x) = \frac{pq}{pr} \\ \cos(x) = \frac{7 \sqrt{3} }{14 \sqrt{3} } \\ \cos(x) = \frac{1}{2} \\ x = 60°$


Shaded Region:

ar(PQR) - area of sector PQTS

$\frac{1}{2} \times ( \sqrt{196 \times 3 - 49 \times 3} ) \times 7 \sqrt{3} - \frac{60°}{360°} \times \pi \times 49 \times 3 \\ = \frac{147 \sqrt{3} - 49\pi}{2} \\ = 50.3 {cm}^{2}$