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Question Number 138658 by ajfour last updated on 16/Apr/21

Commented by ajfour last updated on 16/Apr/21

$$Findmaximumareaof△AOB.$$$$a=1.$$

Commented by mr W last updated on 16/Apr/21

$$ifbisgiven:$$$$\sin \theta =\frac{\sqrt{a^2-{\left(\frac{b}{2}\right)}^2}}{a}=\sqrt{1-\frac{b^2}{4a^2}}$$$$A=\frac{a\times 2b\times \sin \theta }{2}=ab\sqrt{1-\frac{b^2}{4a^2}}$$

Answered by ajfour last updated on 16/Apr/21

$$△=\frac{\sin \theta }{2}\times 4\cos \theta$$$${△}_{max}=1$$$$\frac{Areaofcircle}{{△}_{max}}=\pi$$

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