Question Number 223213 by mr W last updated on 18/Jul/25
Commented by mr W last updated on 18/Jul/25
$${find}\:{the}\:{area}\:{of}\:{rectangle}\:{ABCD}. \\ $$
Answered by gregori last updated on 18/Jul/25
$$\:{Let}\:{BC}\:=\:{a}\:,\:{CD}\:=\:{b}\: \\ $$$$\:\:{AE}\:=\frac{\mathrm{6}}{{b}}\:,\:{FD}\:=\:\frac{\mathrm{12}}{{b}} \\ $$$$\:\:\frac{{EF}}{{a}}\:=\:\sqrt{\frac{\mathrm{4}}{{ab}−\mathrm{9}}}\:\Rightarrow\:\frac{{a}−\frac{\mathrm{18}}{{b}}}{{a}}\:=\:\frac{\mathrm{2}}{\:\sqrt{{ab}−\mathrm{5}}} \\ $$$$\:\:\frac{{ab}−\mathrm{18}}{{ab}}\:=\:\frac{\mathrm{2}}{\:\sqrt{{ab}−\mathrm{5}}}\:,\:{ab}\:=\:{L} \\ $$$$\:\:\frac{\left({L}−\mathrm{18}\right)^{\mathrm{2}} }{{L}^{\mathrm{2}} }\:=\:\frac{\mathrm{4}}{{L}−\mathrm{5}} \\ $$$$\:{then}\:{we}\:{get}\:{L}\:=\:\mathrm{30}\: \\ $$$$\: \\ $$
Answered by mr W last updated on 18/Jul/25
Commented by mr W last updated on 18/Jul/25
![((6+3)/4)=(((FC)/(FG)))^2 ((x+6+3)/(x+4))=((FC)/(FG))=(√((6+3)/4))=(3/2) ⇒x=6 area of [ABCD]=2(x+6+2)=30 ✓](https://www.tinkutara.com/question/Q223223.png)
$$\frac{\mathrm{6}+\mathrm{3}}{\mathrm{4}}=\left(\frac{{FC}}{{FG}}\right)^{\mathrm{2}} \\ $$$$\frac{{x}+\mathrm{6}+\mathrm{3}}{{x}+\mathrm{4}}=\frac{{FC}}{{FG}}=\sqrt{\frac{\mathrm{6}+\mathrm{3}}{\mathrm{4}}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow{x}=\mathrm{6} \\ $$$${area}\:{of}\:\left[{ABCD}\right]=\mathrm{2}\left({x}+\mathrm{6}+\mathrm{2}\right)=\mathrm{30}\:\checkmark \\ $$