Question-178001

Question Number 178001 by Tawa11 last updated on 11/Oct/22

Commented by Tawa11 last updated on 11/Oct/22

(b) The area of the shaded part .

Answered by aleks041103 last updated on 11/Oct/22

B D i s t a n g e n t t o t h e c i r c l e a t T

\Rightarrow A T ⊥ B D a n d A T = r = A Q = A P

S_{A B C D} = 2 S_{A B D} = 2 (\frac{1}{2} B D . A T) = B D . A T

B u t A B = A D a n d ∡ A B D = 60 ° \Rightarrow B D = A B

S_{A B C D} = A B . A D . s i n (A) = {A B}^{2} \frac{\sqrt{3}}{2} = A B . A T

\Rightarrow A T = \frac{\sqrt{3}}{2} A B

(a) P_{s h} = 4 A B - 2 r + \frac{π}{3} r = (4 - (2 - \frac{π}{3}) \frac{\sqrt{3}}{2}) A B

\Rightarrow P_{s h} = (4 + \frac{π}{2 \sqrt{3}} - \sqrt{3}) 15

(b) S_{s h} = S_{r h o m b} - \frac{π}{6} r^{2} =

= \frac{\sqrt{3}}{2} {A B}^{2} - \frac{π}{6} \frac{3}{4} {A B}^{2} =

= \frac{{A B}^{2}}{2} (\sqrt{3} - \frac{π}{4})

\Rightarrow S_{s h} = (\sqrt{3} - \frac{π}{4}) \frac{225}{2}

Commented by Tawa11 last updated on 12/Oct/22

God bless you sir . I appreciate

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