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त्रिभुज का अधिकतम क्षेत्रफल कितना है?
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दिया गया,
\(AB + AC = 3\)
मान लीजिए \(AB = x\) और \(AC = 3 - x \) ।
तब,
\(BC = \sqrt{AC^2 - AB^2} = \sqrt{(3 - x)^2 - x^2} = \sqrt{9 - 6x} \)
त्रिभुज का क्षेत्रफल है,
\(A = \tfrac12\,x\,BC = \tfrac12\,x\,\sqrt{9 - 6x}\)
अधिकतम करने के लिए, \(x\) के संबंध में अवकलन करते हैं और शून्य पर सेट करते हैं:
\(\displaystyle \frac{d}{dx}\bigl(x\sqrt{9-6x}\bigr) = \sqrt{9-6x} \;-\;\frac{6x}{2\sqrt{9-6x}} = 0 \;\Longrightarrow\; x = 1 \)
\(x = 1\) पर, हमें \(BC = \sqrt{9 - 6} = \sqrt{3}\) प्राप्त होता है, अतः
\(A_{\max} = \tfrac12 \times 1 \times \sqrt{3} = \frac{\sqrt{3}}{2}\)
∴ अधिकतम क्षेत्रफल \(\frac{\sqrt{3}}{2}\) वर्ग इकाई है।
अतः, सही उत्तर विकल्प 1 है।
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