ધારોકે \(C\) એ વર્તુળ \(x^{2}+y^{2}-x+2 y=\frac{11}{4}\) નું કેન્દ્ર છે અને \(P\) એ આ વર્તુળ પરનું એક બિંદુ છે. બિંદુ \(C\) માંથી પસાર થતી એક રેખા, એ રેખા \(CP\) સાથે \(\frac{\pi}{4}\) નો ખૂણો બનાવે છે અને આ વર્તુળને બિંદુઓ \(Q\) અને \(R\) માં છેદે છે. તો ત્રિકોણ \(PQR\) નું ક્ષેત્રફળ (એકમ\(^{2}\) માં) .................. છે.

A \(2\)
B \(2 \sqrt{2}\)
C \(8 \sin \left(\frac{\pi}{8}\right)\)
D \(8 \cos \left(\frac{\pi}{8}\right)\)

Verified Solution

Answer & Solution

Correct Answer

(B) \(2 \sqrt{2}\)

Step-by-step Solution

Detailed explanation

\(x ^{2}+ y ^{2}- x +2 y =\frac{11}{4}\) \(\left( x -\frac{1}{2}\right)^{2}+( y +1)^{2}=(2)^{2}\) Or \(\triangle PQR\) \(PR = QK \sin 2 \geq \frac{1}{3}\) \(=4 \cdot 6 \sin \frac{\pi}{8}\) \(PQ = QR \cos 22 \frac{1}{2}\) \(=4 \cos \frac{\pi}{8}\) As…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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