Question Number 223383 by mr W last updated on 23/Jul/25
Commented by Frix last updated on 23/Jul/25
$$\mathrm{I}\:\mathrm{guess}\:\mathrm{it}'\mathrm{s}\:\mathrm{66} \\ $$
Commented by Frix last updated on 23/Jul/25
$$\mathrm{Yes}\:\mathrm{it}'\mathrm{s}\:\mathrm{possible}. \\ $$
Commented by fantastic last updated on 23/Jul/25
$${wait}.\:{I}\:{think}\:{I}\:{have}\:{solved}\:{it} \\ $$$$ \\ $$
Answered by mr W last updated on 23/Jul/25
Commented by mr W last updated on 23/Jul/25
$${d}={diameter}\:{of}\:{circle}={height}\:{of}\:{rectangle} \\ $$$${area}\:{of}\:{rectangle}\:=\left({a}+{d}\right){d} \\ $$$$\frac{{a}+{d}}{\mathrm{5}+\mathrm{6}}=\frac{\mathrm{6}}{{d}} \\ $$$$\left({a}+{d}\right){d}=\mathrm{6}×\left(\mathrm{5}+\mathrm{6}\right)=\mathrm{66}\:\checkmark \\ $$
Answered by Ghisom last updated on 23/Jul/25
$$\mathrm{it}\:\mathrm{must}\:\mathrm{be}\:\mathrm{constant}\:\mathrm{or}\:\mathrm{we}\:\mathrm{could}\:\mathrm{not}\:\mathrm{solve} \\ $$$$\mathrm{it}\:\Rightarrow \\ $$$$\mathrm{let}\:\mathrm{the}\:\mathrm{line}\:\mathrm{go}\:\mathrm{through}\:\mathrm{the}\:\mathrm{center}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{circle}\:\Rightarrow \\ $$$$\mathrm{the}\:\mathrm{rectangle}\:\mathrm{measures}\:\mathrm{11}×\mathrm{6}=\mathrm{66} \\ $$
Answered by fantastic last updated on 23/Jul/25
Commented by fantastic last updated on 23/Jul/25
$${Given}\:{GH}=\mathrm{5}\:{and}\:{HE}=\mathrm{6} \\ $$$${Let}\:{radius}\:{og}\:{thr}\:{circle}\:{be}\:{r} \\ $$$${So}\:{breadth}\:{of}\:{the}\:{rectangle}=\mathrm{2}{r} \\ $$$${Let}\:{the}\:{length}\:{of}\:{the}\:{rectangle}\:{be}\:{x} \\ $$$${So}\:{area}\:=\mathrm{2}{rx} \\ $$$${Now}\:{in}\:\bigtriangleup{GJE}\:{and}\bigtriangleup{HIE} \\ $$$$\measuredangle{GJE}=\measuredangle{IHE} \\ $$$$\measuredangle{GEJ}\:{is}\:{common} \\ $$$${So}\bigtriangleup{GJE}\sim\bigtriangleup{HIE} \\ $$$${so}\:\frac{\mathrm{5}+\mathrm{6}}{\mathrm{2}{r}}=\frac{{x}}{\mathrm{6}} \\ $$$${So}\:\mathrm{2}{rx}=\mathrm{11}×\mathrm{6}=\mathrm{66} \\ $$$${Area}\:{of}\:{the}\:{rectangle}=\mathrm{66}\:{u}^{\mathrm{2}} \\ $$