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Question-176942




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

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