Question Number 223340 by fantastic last updated on 21/Jul/25
Commented by fantastic last updated on 21/Jul/25
$${This}\:{will}\:{take}\:{a}\:{lot}\:{of}\:{time} \\ $$
Answered by A5T last updated on 21/Jul/25
Commented by A5T last updated on 21/Jul/25
![Let HE=HG=a, AE=AC=R, CD=BD=r △HEA≈△AEB⇒(a/R)=(R/(4+3+2a))⇒7a+2a^2 =R^2 ...(i) △CBF≈△EBA ⇒((2r)/(7+2a))=(4/(2r+R))⇒2r^2 +Rr=14+4a...(ii) ⇒2ar^2 +aRr=14a+4a^2 =2R^2 [from (i)]...(iii) △CBF≈△ABH⇒((2r)/(2r+R))=(4/(7+a))⇒3r+ar=2R...(iv) ⇒3Rr+aRr=2R^2 ...(v) (iii)−(v)⇒2ar^2 =3Rr⇒2ar=3R⇒a=((3R)/(2r)) a in (iv)⇒r=(R/6)⇒ a=((3R)/(R/3))=9 a=9 in (i)⇒ 63+162=225=R^2 ⇒ R=15 [AEH]=((a×(√(R^2 −a^2 )))/2)=((9×12)/2)=54](https://www.tinkutara.com/question/Q223345.png)
$$\mathrm{Let}\:\mathrm{HE}=\mathrm{HG}=\mathrm{a},\:\mathrm{AE}=\mathrm{AC}=\mathrm{R},\:\mathrm{CD}=\mathrm{BD}=\mathrm{r} \\ $$$$\bigtriangleup\mathrm{HEA}\approx\bigtriangleup\mathrm{AEB}\Rightarrow\frac{\mathrm{a}}{\mathrm{R}}=\frac{\mathrm{R}}{\mathrm{4}+\mathrm{3}+\mathrm{2a}}\Rightarrow\mathrm{7a}+\mathrm{2a}^{\mathrm{2}} =\mathrm{R}^{\mathrm{2}} …\left(\mathrm{i}\right) \\ $$$$\bigtriangleup\mathrm{CBF}\approx\bigtriangleup\mathrm{EBA} \\ $$$$\Rightarrow\frac{\mathrm{2r}}{\mathrm{7}+\mathrm{2a}}=\frac{\mathrm{4}}{\mathrm{2r}+\mathrm{R}}\Rightarrow\mathrm{2r}^{\mathrm{2}} +\mathrm{Rr}=\mathrm{14}+\mathrm{4a}…\left(\mathrm{ii}\right) \\ $$$$\Rightarrow\mathrm{2ar}^{\mathrm{2}} +\mathrm{aRr}=\mathrm{14a}+\mathrm{4a}^{\mathrm{2}} =\mathrm{2R}^{\mathrm{2}} \left[\mathrm{from}\:\left(\mathrm{i}\right)\right]…\left(\mathrm{iii}\right) \\ $$$$\bigtriangleup\mathrm{CBF}\approx\bigtriangleup\mathrm{ABH}\Rightarrow\frac{\mathrm{2r}}{\mathrm{2r}+\mathrm{R}}=\frac{\mathrm{4}}{\mathrm{7}+\mathrm{a}}\Rightarrow\mathrm{3r}+\mathrm{ar}=\mathrm{2R}…\left(\mathrm{iv}\right) \\ $$$$\Rightarrow\mathrm{3Rr}+\mathrm{aRr}=\mathrm{2R}^{\mathrm{2}} …\left(\mathrm{v}\right) \\ $$$$\left(\mathrm{iii}\right)−\left(\mathrm{v}\right)\Rightarrow\mathrm{2ar}^{\mathrm{2}} =\mathrm{3Rr}\Rightarrow\mathrm{2ar}=\mathrm{3R}\Rightarrow\mathrm{a}=\frac{\mathrm{3R}}{\mathrm{2r}} \\ $$$$\mathrm{a}\:\mathrm{in}\:\left(\mathrm{iv}\right)\Rightarrow\mathrm{r}=\frac{\mathrm{R}}{\mathrm{6}}\Rightarrow\:\mathrm{a}=\frac{\mathrm{3R}}{\frac{\mathrm{R}}{\mathrm{3}}}=\mathrm{9} \\ $$$$\mathrm{a}=\mathrm{9}\:\mathrm{in}\:\left(\mathrm{i}\right)\Rightarrow\:\mathrm{63}+\mathrm{162}=\mathrm{225}=\mathrm{R}^{\mathrm{2}} \:\Rightarrow\:\mathrm{R}=\mathrm{15} \\ $$$$\left[\mathrm{AEH}\right]=\frac{\mathrm{a}×\sqrt{\mathrm{R}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }}{\mathrm{2}}=\frac{\mathrm{9}×\mathrm{12}}{\mathrm{2}}=\mathrm{54} \\ $$
Answered by HeferH24 last updated on 22/Jul/25
Commented by HeferH24 last updated on 22/Jul/25
$$\:\frac{\mathrm{3}}{\mathrm{4}}\:=\:\frac{\mathrm{4}{k}}{\left(\mathrm{3}{k}+\mathrm{7}\right)}\:\Rightarrow\:{k}\:=\:\mathrm{3}\: \\ $$$$\:\therefore\:{Red}\:=\:\mathrm{6}{k}^{\mathrm{2}} \:=\:\mathrm{54}\:{u}^{\mathrm{2}} \\ $$
Answered by mr W last updated on 22/Jul/25
Commented by mr W last updated on 22/Jul/25
$$\angle{DBC}=\alpha+\beta=\angle{GDF}=\frac{\angle{GEF}}{\mathrm{2}}=\mathrm{45}° \\ $$$$\Rightarrow{CB}={CD}=\mathrm{3} \\ $$$$\Rightarrow{AB}=\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }=\mathrm{5} \\ $$$${GE}={EB}={R} \\ $$$$\frac{{R}}{{R}+\mathrm{5}}=\frac{\mathrm{3}}{\mathrm{4}}\:\Rightarrow{R}=\mathrm{15} \\ $$$$\Rightarrow{GH}=\mathrm{9},\:{EH}=\mathrm{12} \\ $$$${A}_{{red}} =\frac{\mathrm{9}×\mathrm{12}}{\mathrm{2}}=\mathrm{54} \\ $$