Question Number 223783 by mr W last updated on 04/Aug/25
Commented by mr W last updated on 04/Aug/25
$${find}\:{shaded}\:{area}\:{in}\:{rectangle}. \\ $$
Answered by fantastic last updated on 04/Aug/25
Commented by fantastic last updated on 04/Aug/25
$$ \\ $$$$\mathrm{In}\:\mathrm{both}\:\mathrm{triangle}{s}\:\mathrm{the}\:\mathrm{height} \\ $$$$\mathrm{ar}{e}\:\mathrm{same}.\:\mathrm{so}\:\mathrm{the}\:\mathrm{ratios}\:\mathrm{of}\: \\ $$$$\mathrm{their}\:\mathrm{bases}\:\mathrm{will}\:\mathrm{be}\:\mathrm{the}\:\mathrm{same}\: \\ $$$$\mathrm{as}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{their}\:\mathrm{areas}. \\ $$$${So}\:{EO}:{OC}\:=\mathrm{5}:\mathrm{10}=\mathrm{1}:\mathrm{2} \\ $$$$ \\ $$$$\mathrm{sam}{e}\:\mathrm{goes}\:\mathrm{for}\:\mathrm{this}\:\mathrm{triangle} \\ $$$$\mathrm{the}{i}\mathrm{r}\:\mathrm{height}\:\mathrm{are}\:\mathrm{same}\:{so} \\ $$$$\mathrm{th}{e}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{their}\:\mathrm{area}\:\mathrm{will}\:\mathrm{be} \\ $$$$\mathrm{sa}{m}\mathrm{e}\:\mathrm{as}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{their}\: \\ $$$$\mathrm{bases}. \\ $$$${So}\:\bigtriangleup{BEO}:\bigtriangleup{BOC}=\mathrm{1}:\mathrm{2}\Rightarrow{t}\:{and}\:\mathrm{2}{t} \\ $$$${let}\:\bigtriangleup{ABE}={x} \\ $$$${So}\:{x}+{t}+\mathrm{5}=\mathrm{2}{t}+\mathrm{10} \\ $$$${x}=\mathrm{5}+{t} \\ $$$${Now}\:\bigtriangleup{BEC}=\bigtriangleup{ABE}+\bigtriangleup{CDE} \\ $$$$\mathrm{2}{t}+{t}=\mathrm{5}+{t}+\mathrm{15} \\ $$$${or}\:{t}=\mathrm{10} \\ $$$${shaded}\:{area}={t}+\mathrm{5}+{t}=\mathrm{25}{cm}^{\mathrm{2}} \\ $$
Commented by mr W last updated on 04/Aug/25
��
Answered by mr W last updated on 04/Aug/25
Commented by mr W last updated on 04/Aug/25
$${C}+{B}={C}+\mathrm{10}\:\Rightarrow{B}=\mathrm{10} \\ $$$$\frac{{C}}{\mathrm{10}}=\frac{{B}}{\mathrm{5}}\:\Rightarrow{C}=\mathrm{20} \\ $$$${A}+{B}+\mathrm{5}={C}+\mathrm{10}\:\Rightarrow{A}=\mathrm{15} \\ $$$${shaded}\:{area}\:={A}+{B}=\mathrm{25}\:\checkmark \\ $$