Question

Find the area of the shaded region where a circular arc of radius 6cm has been drawn with vertex O of an equilateral traingle OAB of side 12 cm as centre.


I wonder..why tao and kris has done such kind of things.
that's really weird..
Tao is not of Kris Wu's level.



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No spamming​

Answer 1

$\huge \tt \fbox \green {5442/35 cm²}$

Step-by-step explanation:

Side of equilateral triangle = 12 cm

Radius = 6 cm

Construction : Join the point C & D to form a complete circle.

Since OAB is an equilateral triangle

Therefore Ang.AOB = ANG.OAB = Ang.OBA = 60⁰

So, in the circle Ang.COD = 60⁰

$Area \: of \: the \: sector \: OCD = \frac{ \alpha }{360} \times \pi {r}^{2} \\ \\ = > \frac{60}{360} \times \frac{22}{7} \times 6 \times 6 \\ \\ = > \frac{132}{7} \: {cm \: }^{2}$

$Area \: of \: the \: circle \: = \pi \: {r}^{2} \\ \\ = > \frac{22}{7} \times 6 \times 6 \\ \\ = > \frac{792}{7} \: {cm \: }^{2}$

$Area \: of \: the \: given \: circular \: arc \\ \\ = > \frac{792}{7} - \frac{132}{7} \\ \\ = > \frac{660}{7} \: {cm \: }^{2}$

$Area \:o f \: the \: equil \: triangle = \frac{ \sqrt{3} }{4} \times {a}^{2} \\ \\ = > 61.2 \: {cm \: }^{2}$

$Area \: of \: the \: shaded \: region \\ \\ = > 61.2 + \frac{660}{7} \\ \\ = > \frac{5442}{35} \: {cm}^{2}$

$\tt \fbox \red {Hope this helps you}$