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Question Number 19796 by ajfour last updated on 15/Aug/17

Commented by ajfour last updated on 15/Aug/17

$$\mathrm{Q}.19794\left(\mathrm{solution}\right)$$

Answered by ajfour last updated on 15/Aug/17

$$\mathrm{let}\mathrm{presently}\mathrm{AB}=2$$$$\mathrm{let}\mathrm{area}\mathrm{of}△\mathrm{ABC}=4\mathrm{\Delta }$$$$\mathrm{area}\mathrm{of}△\mathrm{AXZ}=\mathrm{\Delta }$$$$\mathrm{Area}\mathrm{of}△\mathrm{XYZ}=\left(\tan {}^2\theta \right)\mathrm{\Delta }$$$$(\mathrm{because}△\mathrm{XYZ}\sim △\mathrm{AXZ})$$$$\mathrm{Area}\mathrm{BXYC}=\mathrm{Area}\mathrm{CWXZ}+$$$$\mathrm{Area}△\mathrm{BXW}-\mathrm{Area}(△\mathrm{AXZ})$$$$=2\mathrm{\Delta }+\mathrm{\Delta }-\left(\tan {}^2\theta \right)\mathrm{\Delta }$$$$\frac{\mathrm{Area}\mathrm{BXYC}}{\mathrm{Area}△\mathrm{ABC}}=\frac{3\mathrm{\Delta }-\left(\tan {}^2\theta \right)\mathrm{\Delta }}{4\mathrm{\Delta }}=\frac{13}{18}$$$$\Rightarrow 3-\tan {}^2\theta =\frac{26}{9}$$$$\mathrm{So}\tan {}^2\theta =\frac{1}{9}$$$$\mathrm{Now}\tan \theta =\frac{1}{3}=\frac{\mathrm{BC}}{\mathrm{AC}}$$$$\Rightarrow \mathrm{AC}=3\left(\mathrm{BC}\right)=3\times 12=36.$$

Commented by Tinkutara last updated on 15/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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