Question Number 156534 by otchereabdullai@gmail.com last updated on 12/Oct/21 | ||
Commented by otchereabdullai@gmail.com last updated on 12/Oct/21 | ||
$$\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{with}\:\mathrm{these}\:\mathrm{question} \\ $$ | ||
Answered by puissant last updated on 12/Oct/21 | ||
$$\left.{b}\right) \\ $$$$\mathscr{A}\:=\:\mathscr{A}_{\mathrm{1}} +\mathscr{A}_{\mathrm{2}} +\mathscr{A}_{\mathrm{3}} \\ $$$$\mathscr{A}_{\mathrm{1}} =\frac{\left({G}_{{b}} +{P}_{{b}} \right)×{h}}{\mathrm{2}}=\frac{\left(\mathrm{8}+\mathrm{6}\right)×\mathrm{4}}{\mathrm{2}}=\mathrm{28}.. \\ $$$$\mathscr{A}_{\mathrm{2}} =\frac{\left(\mathrm{8}+\mathrm{2}\right)×\mathrm{2}}{\mathrm{2}}=\mathrm{10} \\ $$$$\mathscr{A}_{\mathrm{3}} ={C}×{C}=\mathrm{2}×\mathrm{2}=\mathrm{4}.. \\ $$$$\Rightarrow\:\:\mathscr{A}\:=\:\mathrm{28}+\mathrm{10}+\mathrm{4}\:=\:\mathrm{42}\:{cm}^{\mathrm{2}} ....\: \\ $$$$\Rightarrow\:\mathscr{A}\:=\:\mathrm{42}\:{cm}^{\mathrm{2}} .. \\ $$ | ||
Commented by otchereabdullai@gmail.com last updated on 12/Oct/21 | ||
$$\mathrm{God}\:\mathrm{richly}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}!\:\mathrm{am}\:\mathrm{much}\: \\ $$$$\mathrm{much}\:\mathrm{grateful}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}\:! \\ $$ | ||
Commented by otchereabdullai@gmail.com last updated on 12/Oct/21 | ||
$$\mathrm{Can}\:\mathrm{you}\:\mathrm{please}\:\mathrm{define}\:\mathrm{the}\:\mathrm{terms}\:\mathrm{for}\: \\ $$$$\mathrm{me}\:\mathrm{G}_{\mathrm{b}\:} ,\:\mathrm{P}_{\mathrm{b}} \\ $$ | ||
Commented by puissant last updated on 13/Oct/21 | ||
$${G}_{{b}} ={large}\:{base}\:\:{and}\:\:{P}_{{b}} ={small}\:{base}.. \\ $$ | ||
Commented by otchereabdullai@gmail.com last updated on 13/Oct/21 | ||
$$\mathrm{well}\:\mathrm{understood}\:\mathrm{sir}!\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you} \\ $$ | ||
Answered by puissant last updated on 12/Oct/21 | ||
$$\left.{c}\right) \\ $$$$\mathscr{A}\:=\:\mathrm{48}−\left(\mathrm{9}+\mathrm{8}+\mathrm{6}+\mathrm{4}\right)\:=\:\mathrm{48}−\mathrm{27}\:=\:\mathrm{21} \\ $$$$\Rightarrow\:\mathscr{A}\:=\:\mathrm{21}\:{cm}^{\mathrm{2}} .. \\ $$ | ||
Commented by otchereabdullai@gmail.com last updated on 12/Oct/21 | ||
$$\mathrm{well}\:\mathrm{understood}\:\mathrm{thanks}\:\mathrm{soo}\:\mathrm{much}! \\ $$ | ||