Question Number 171165 by MathsFan last updated on 09/Jun/22
$${A}\:{triangle}\:{QRS}\:{is}\:{to}\:{be}\:{constructed}\:{from} \\ $$$${a}\:{line}\:{segment}\:{of}\:{lenght}\:\mathrm{15}{cm}.\: \\ $$$${Construct}\:{the}\:{triangle}\:{using}\:{the} \\ $$$${division}\:{of}\:{the}\:{line}\:{segment}\:{into}\:{the} \\ $$$${ratio}\:\mathrm{5}:\mathrm{4}:\mathrm{3}\:{such}\:{that}\:{QS}\:{and}\:{RS}\:{are} \\ $$$${laegest}\:{and}\:{smallest}\:{ratio}\:{respectively}. \\ $$$${circumscribe}\:{the}\:{triangle}\:{by}\:{locating} \\ $$$${the}\:{circumcenter}. \\ $$
Answered by aleks041103 last updated on 09/Jun/22
$${From}\:{the}\:{problem}\:{statement} \\ $$$${we}\:{can}\:{say}\:{that}: \\ $$$${QS}=\mathrm{5}{x},\:{QR}=\mathrm{4}{x},\:{RS}=\mathrm{3}{x} \\ $$$${But}: \\ $$$${QS}^{\mathrm{2}} =\mathrm{25}{x}^{\mathrm{2}} =\mathrm{16}{x}^{\mathrm{2}} +\mathrm{9}{x}^{\mathrm{2}} =\left(\mathrm{4}{x}\right)^{\mathrm{2}} +\left(\mathrm{3}{x}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{QS}^{\mathrm{2}} ={QR}^{\mathrm{2}} +{RS}^{\mathrm{2}} \\ $$$$\Rightarrow{the}\:\bigtriangleup{QRS}\:{is}\:{a}\:{right}\:{triangle}\:{with} \\ $$$${hypotenuse}\:{QS}. \\ $$$${It}\:{is}\:{known},\:{that}\:{the}\:{center}\:{of}\:{the} \\ $$$${circumscribed}\:{circle}\:{is}\:{in}\:{the}\:{middle} \\ $$$${of}\:{the}\:{right}\:{triangle}'{s}\:{hypotenuse}. \\ $$$$\Rightarrow{The}\:{answer}\:{is}\:{the}\:{middle}\:{of}\:{QS}. \\ $$$${As}\:{for}\:{the}\:{radius}:\:{R}=\frac{{QS}}{\mathrm{2}}=\frac{\mathrm{5}}{\mathrm{2}}{x} \\ $$$${QS}+{RS}+{QR}=\mathrm{5}{x}+\mathrm{3}{x}+\mathrm{4}{x}=\mathrm{12}{x}=\mathrm{15} \\ $$$$\Rightarrow{x}=\frac{\mathrm{15}}{\mathrm{12}}=\frac{\mathrm{5}}{\mathrm{4}}{cm} \\ $$$$\Rightarrow{R}=\frac{\mathrm{25}}{\mathrm{8}}{cm}=\mathrm{3},\mathrm{125}{cm} \\ $$
Commented by MathsFan last updated on 09/Jun/22
$${thanks} \\ $$