A-triangle-is-inscribed-in-a-circle-the-vertices-of-the-triangle-divided-the-circumference-of-the-circle-into-three-area-of-length-6-8-10-units-then-the-area-of-triangle-is-equal-to-a-64-3-

FacebookTweetPin

Question Number 140533 by liberty last updated on 09/May/21

$$\mathrm{A}\mathrm{triangle}\mathrm{is}\mathrm{inscribed}\mathrm{in}\mathrm{a}\mathrm{circle}.$$$$\mathrm{the}\mathrm{vertices}\mathrm{of}\mathrm{the}\mathrm{triangle}\mathrm{divided}$$$$\mathrm{the}\mathrm{circumference}\mathrm{of}\mathrm{the}\mathrm{circle}$$$$\mathrm{into}\mathrm{three}\mathrm{area}\mathrm{of}\mathrm{length}6,8,10$$$$\mathrm{units}\mathrm{then}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{triangle}$$$$\mathrm{is}\mathrm{equal}\mathrm{to}\dots$$$$\left(\mathrm{a}\right)\frac{64\sqrt{3}(\sqrt{3}+1)}{{\pi }^2}\left(\mathrm{c}\right)\frac{36\sqrt{3}(\sqrt{3}-1)}{{\pi }^2}$$$$\left(\mathrm{b}\right)\frac{72\sqrt{3}(\sqrt{3}+1)}{{\pi }^2}\left(\mathrm{d}\right)\frac{36\sqrt{3}(\sqrt{3}+1)}{{\pi }^2}$$

Answered by mr W last updated on 09/May/21

$$perimeterofcircle=6+8+10=24$$$$radiusofcircle=\frac{24}{2\pi }=\frac{12}{\pi }$$$$areaoftriangle:$$$$\frac{1}{2}{r}^2(\sin \frac{6\times 2\pi }{24}+\sin \frac{8\times 2\pi }{24}+\sin \frac{10\times 2\pi }{24})$$$$=\frac{1}{2}\times \frac{144}{{\pi }^2}\times (\sin \frac{\pi }{2}+\sin \frac{2\pi }{3}+\sin \frac{5\pi }{6})$$$$=\frac{1}{2}\times \frac{144}{{\pi }^2}\times (1+\frac{\sqrt{3}}{2}+\frac{1}{2})$$$$=\frac{36\sqrt{3}(\sqrt{3}+1)}{{\pi }^2}$$

Commented by mr W last updated on 09/May/21

$$noansweriscorrect!maybetypoin$$$$thequestion,since\left(c\right)and\left(d\right)arethe$$$$same.$$

Commented by liberty last updated on 09/May/21

$$\mathrm{oo}\mathrm{yes}.\mathrm{right}$$

Terms of Service
Privacy Policy
Contact: info@tinkutara.com

FacebookTweetPin