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Question-80386




Question Number 80386 by gny last updated on 02/Feb/20
Commented by kaivan.ahmadi last updated on 02/Feb/20
h_(AB^△ D) =h_(AB^Δ C) =8⇒  S_(AB^△ D) =(1/2)×21×8=84  S_(AB^Δ C) =(1/2)×12×8=48  ⇒S_(BC^Δ D) =S_(AB^Δ D) −S_(AB^Δ C) =84−48=36
$${h}_{{A}\overset{\bigtriangleup} {{B}D}} ={h}_{{A}\overset{\Delta} {{B}C}} =\mathrm{8}\Rightarrow \\ $$$${S}_{{A}\overset{\bigtriangleup} {{B}D}} =\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{21}×\mathrm{8}=\mathrm{84} \\ $$$${S}_{{A}\overset{\Delta} {{B}C}} =\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{12}×\mathrm{8}=\mathrm{48} \\ $$$$\Rightarrow{S}_{{B}\overset{\Delta} {{C}D}} ={S}_{{A}\overset{\Delta} {{B}D}} −{S}_{{A}\overset{\Delta} {{B}C}} =\mathrm{84}−\mathrm{48}=\mathrm{36} \\ $$
Commented by gny last updated on 02/Feb/20
thank you so so much sir  you are really perfect
$${thank}\:{you}\:{so}\:{so}\:{much}\:{sir} \\ $$$${you}\:{are}\:{really}\:{perfect} \\ $$
Answered by sandy_delta last updated on 03/Feb/20
h_(BCD) =h_(ABC) =(√(10^2 −6^2 )) =8  S_(BCD) =(1/2)×8×9=36
$${h}_{{BCD}} ={h}_{{ABC}} =\sqrt{\mathrm{10}^{\mathrm{2}} −\mathrm{6}^{\mathrm{2}} }\:=\mathrm{8} \\ $$$${S}_{{BCD}} =\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{8}×\mathrm{9}=\mathrm{36} \\ $$

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