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Question Number 146000 by mnjuly1970 last updated on 10/Jul/21

Commented by mr W last updated on 10/Jul/21

$$ithinktheareaof\mathrm{\Delta }ABC{can}^{\prime }tbe$$$$uniquelydeterminedwiththegiven$$$$condition.pleaserecheck.$$

Commented by mr W last updated on 10/Jul/21

$$AC=a$$$$BC=b$$$$AB=c=\sqrt{a^2+b^2}$$$$r=\frac{ab}{a+b+c}$$$$AD=a-r$$$$BD=b-r$$$$AD\times BD={DE}^2=h^2(herewithh=1)$$$$(a-r)(b-r)=h^2$$$$ab\left(\frac{a+c}{a+b+c}\right)\left(\frac{b+c}{a+b+c}\right)=h^2$$$$\frac{ab(a+c)(b+c)}{{(a+b+c)}^2}=h^2$$$$ab=(1+\frac{b}{a+c})(1+\frac{a}{b+c})h^2$$$$A_{\mathrm{\Delta }ABC}=(1+\frac{b}{a+c})(1+\frac{a}{b+c})h^2\ne unique$$

Commented by mr W last updated on 10/Jul/21

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