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Question Number 113114 by bemath last updated on 11/Sep/20

Commented by bemath last updated on 11/Sep/20

$$\mathrm{find}\mathrm{AD}$$

Answered by mr W last updated on 11/Sep/20

$$AD=2r$$$$6^2-{(r-\frac{14}{2})}^2=r^2-{\left(\frac{14}{2}\right)}^2$$$$r^2-7r-18=0$$$$(r+2)(r-9)=0$$$$\Rightarrow r=9$$$$\Rightarrow AD=18$$

Commented by bemath last updated on 11/Sep/20

$$\mathrm{sir},\mathrm{it}\mathrm{by}\mathrm{cosine}\mathrm{rule}?$$

Commented by mr W last updated on 11/Sep/20

Commented by mr W last updated on 11/Sep/20

$${BE}^2={AB}^2-{AE}^2={OB}^2-{OE}^2$$$$\Rightarrow 6^2-{(r-7)}^2=r^2-7^2$$

Commented by bemath last updated on 11/Sep/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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