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JEE Mains · Maths · STD 12 - 8. Application and integration

વર્તુળ \(x^2+y^2=169\) ના, રેખા \(5 x-y=13\) ની નીચે આવેલા ભાગનું ક્ષેત્રફળ (ચો.એકમમાં) \(\frac{\pi \alpha}{2 \beta}-\frac{65}{2}+\frac{\alpha}{\beta} \sin ^{-1}\left(\frac{12}{13}\right)\) છે., જ્યાં \(\alpha, \beta\) પરસ્પર અવિભાજ્ય છે. તો \(\alpha+\beta=\) ...........

  1. A \(137\)
  2. B \(711\)
  3. C \(271\)
  4. D \(171\)
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Answer & Solution

Correct Answer

(D) \(171\)

Step-by-step Solution

Detailed explanation

\( \text { Area }=\int_{-13}^{12} \sqrt{169-y^2} d y-\frac{1}{2} \times 25 \times 5 \) \( =\frac{\pi}{2} \times \frac{169}{2}-\frac{65}{2}+\frac{169}{2} \sin ^{-1} \frac{12}{13} \) \( \therefore \alpha+\beta=171\)
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