Question Number 190370 by cortano12 last updated on 02/Apr/23
Answered by som(math1967) last updated on 02/Apr/23
Commented by som(math1967) last updated on 02/Apr/23
$${ar}\:{of}\bigtriangleup{FDC}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{10}×\mathrm{5}=\mathrm{25}{cm}^{\mathrm{2}} \\ $$$${ar}\:{of}\:{ABFD}=\mathrm{100}−\mathrm{25}=\mathrm{75}{cm}^{\mathrm{2}} \\ $$$${ar}\:\bigtriangleup{AEF}=\bigtriangleup{BEF}=\frac{\mathrm{5}×\mathrm{5}}{\mathrm{2}}=\mathrm{12}.\mathrm{5}{cm}^{\mathrm{2}} \\ $$$${ar}\bigtriangleup{AFD}=\mathrm{75}−\mathrm{25}=\mathrm{50}{cm}^{\mathrm{2}} \\ $$$${ar}\:{of}\:{shaded}\:=\frac{\mathrm{50}}{\mathrm{2}}+\mathrm{12}.\mathrm{5}=\mathrm{37}.\mathrm{5}{cm}^{\mathrm{2}} \\ $$