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Question Number 207691 by cherokeesay last updated on 23/May/24

Answered by mr W last updated on 24/May/24

Commented by mr W last updated on 23/May/24

$$DE=\sqrt{{10}^2-{(4+\frac{7}{2})}^2}=\frac{5\sqrt{7}}{2}$$$$\sin \theta =\frac{5\sqrt{7}}{2\times 10}=\frac{\sqrt{7}}{4}$$$$DC=\sqrt{{10}^2-{(4+\frac{7}{2})}^2+{\left(\frac{7}{2}\right)}^2}=2\sqrt{14}$$$$R=\frac{2\sqrt{14}}{2\times \frac{\sqrt{7}}{4}}=4\sqrt{2}$$

Commented by Tawa11 last updated on 21/Jun/24

$$\mathrm{many}\mathrm{things}\mathrm{to}\mathrm{learn}$$

Answered by A5T last updated on 23/May/24

$$(2\times 10\times 11-2\times 4\times 10)cos\theta ={10}^2+{11}^2-4^2-{10}^2$$$$\Rightarrow cos\theta =\frac{105}{140}=\frac{3}{4}\Rightarrow sin\theta =\frac{\sqrt{7}}{4};cos2\theta =\frac{1}{8};sin2\theta =\frac{3\sqrt{7}}{8}$$$$AC=\sqrt{{11}^2+4^2-2\times 11\times 4cos2\theta }=3\sqrt{14}$$$$\frac{4\times 11\times 3\sqrt{14}}{4R}=\frac{4\times 11\times \frac{3\sqrt{7}}{8}}{2}\Rightarrow \frac{\sqrt{2}}{R}=\frac{1}{4}\Rightarrow R=4\sqrt{2}$$

Commented by cherokeesay last updated on 23/May/24

$$nice!thankyou.$$

Answered by mr W last updated on 23/May/24

$${AD}^2=4^2+{10}^2-2\times 4\times 10\cos \theta$$$${DC}^2={11}^2+{10}^2-2\times 11\times 10\cos \theta$$$$AD=DC$$$$4^2+{10}^2-2\times 4\times 10\cos \theta ={11}^2+{10}^2-2\times 11\times 10\cos \theta$$$$2(11-4)\times 10\cos \theta ={11}^2-4^2$$$$\cos \theta =\frac{11+4}{2\times 10}=\frac{3}{4}\Rightarrow \sin \theta =\frac{\sqrt{7}}{4}$$$$DC=\sqrt{{11}^2+{10}^2-2\times 11\times 10\times \frac{3}{4}}=2\sqrt{14}$$$$R=\frac{DC}{2\sin \theta }=\frac{2\sqrt{14}}{2\times \frac{\sqrt{7}}{4}}=4\sqrt{2}✓$$

Commented by cherokeesay last updated on 23/May/24

$$nicethankyoumaster.$$

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