Question Number 176942 by Ar Brandon last updated on 28/Sep/22
Answered by som(math1967) last updated on 28/Sep/22
$${HE}={GF}=\frac{\mathrm{1}}{\mathrm{2}}×{AC}=\frac{\mathrm{17}}{\mathrm{2}}{cm} \\ $$$${EF}={HG}=\frac{\mathrm{1}}{\mathrm{2}}×{DB}=\frac{\mathrm{13}}{\mathrm{2}}{cm} \\ $$$${perimeter}\:{EFGH}=\mathrm{2}\left(\frac{\mathrm{17}}{\mathrm{2}}\:+\frac{\mathrm{13}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{30}{cm} \\ $$
Commented by Ar Brandon last updated on 28/Sep/22
Thank you, Sir