Question Number 216515 by Jubr last updated on 09/Feb/25
Answered by Jubr last updated on 09/Feb/25
Commented by Jubr last updated on 09/Feb/25
$${Find}\:{the}\:{surface}\:{area}\:{of}\:{the}\:{figure}. \\ $$$${All}\:{units}\:{are}\:{in}\:\:{cm}. \\ $$
Commented by Rasheed.Sindhi last updated on 09/Feb/25
$${Bottom}:\:\mathrm{20}×\mathrm{5}=\mathrm{100} \\ $$$${Top}:\:\mathrm{20}×\mathrm{2}=\mathrm{40} \\ $$$${Front}\:: \\ $$$${length}\:\mathrm{20},\:{width}\:\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} }\:=\mathrm{3}\sqrt{\mathrm{5}}\: \\ $$$${area}=\mathrm{20}×\mathrm{3}\sqrt{\mathrm{5}}\:=\mathrm{60}\sqrt{\mathrm{5}}\: \\ $$$${Rear}:\mathrm{20}×\mathrm{6}=\mathrm{120} \\ $$$${Sides}: \\ $$$$\left(\mathrm{6}×\mathrm{2}+\frac{\mathrm{3}×\mathrm{6}}{\mathrm{2}}\:\right)\:\:\:\:×\mathrm{2}=\mathrm{42} \\ $$$${Total}\:{surface}\:{area} \\ $$$$={Botto}+{Top}+{Front}+{Rear}+{Sides} \\ $$$$=\mathrm{100}+\mathrm{40}+\mathrm{60}\sqrt{\mathrm{5}}\:+\mathrm{120}+\mathrm{42} \\ $$$$=\mathrm{302}+\mathrm{60}\sqrt{\mathrm{5}}\: \\ $$
Commented by Jubr last updated on 10/Feb/25
$${Thanks}\:{sir} \\ $$
Answered by mr W last updated on 09/Feb/25
$${base}\:{area}\:=\frac{\left(\mathrm{2}+\mathrm{5}\right)×\mathrm{6}}{\mathrm{2}}=\mathrm{21}\:{cm}^{\mathrm{2}} \\ $$$${base}\:{perimeter}\:=\mathrm{2}+\mathrm{6}+\mathrm{5}+\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{13}+\mathrm{3}\sqrt{\mathrm{5}}\:{cm} \\ $$$${height}\:=\mathrm{20}\:{cm} \\ $$$${surface}\:{area}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}×\mathrm{21}+\left(\mathrm{13}+\mathrm{3}\sqrt{\mathrm{5}}\right)×\mathrm{20} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{302}+\mathrm{60}\sqrt{\mathrm{5}}\approx\mathrm{436}.\mathrm{16}\:{cm}^{\mathrm{2}} \\ $$
Commented by Jubr last updated on 10/Feb/25
$${Thanks}\:{sir}. \\ $$