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Question-172918




Question Number 172918 by mnjuly1970 last updated on 03/Jul/22
Commented by som(math1967) last updated on 03/Jul/22
yes i got ((16)/3)cm
$$\boldsymbol{{yes}}\:\boldsymbol{{i}}\:\boldsymbol{{got}}\:\frac{\mathrm{16}}{\mathrm{3}}\boldsymbol{{cm}} \\ $$
Commented by infinityaction last updated on 03/Jul/22
((16)/3) ???
$$\frac{\mathrm{16}}{\mathrm{3}}\:??? \\ $$
Commented by som(math1967) last updated on 03/Jul/22
Commented by som(math1967) last updated on 03/Jul/22
AE∥PQ∥BC  △ADQ∼△ACB   ∴((DQ)/(BC))=((AQ)/(AB))=(4/(12))=(1/3)  ⇒DQ=(1/3)BC=(8/3)   PD=(2/3)AE=(8/3)  ∴PQ=(8/3)+(8/3)=((16)/3)cm
$${AE}\parallel{PQ}\parallel\boldsymbol{{B}}{C} \\ $$$$\bigtriangleup{ADQ}\sim\bigtriangleup{ACB} \\ $$$$\:\therefore\frac{{DQ}}{{BC}}=\frac{{AQ}}{{AB}}=\frac{\mathrm{4}}{\mathrm{12}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow{DQ}=\frac{\mathrm{1}}{\mathrm{3}}{BC}=\frac{\mathrm{8}}{\mathrm{3}} \\ $$$$\:{PD}=\frac{\mathrm{2}}{\mathrm{3}}{AE}=\frac{\mathrm{8}}{\mathrm{3}} \\ $$$$\therefore{PQ}=\frac{\mathrm{8}}{\mathrm{3}}+\frac{\mathrm{8}}{\mathrm{3}}=\frac{\mathrm{16}}{\mathrm{3}}{cm} \\ $$
Commented by mnjuly1970 last updated on 03/Jul/22
excellent sir
$$\mathrm{excellent}\:\mathrm{sir} \\ $$
Commented by Tawa11 last updated on 03/Jul/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by infinityaction last updated on 04/Jul/22
Commented by infinityaction last updated on 04/Jul/22
AF = 12 and EF  =  8  △ADF ∼ APE    ((PE)/4) = (4/(12))   ⇒ PE = (4/3)  PQ = PE + EQ  PQ = (4/3)+4 = ((16)/3)
$${AF}\:=\:\mathrm{12}\:{and}\:{EF}\:\:=\:\:\mathrm{8} \\ $$$$\bigtriangleup{ADF}\:\sim\:{APE} \\ $$$$\:\:\frac{{PE}}{\mathrm{4}}\:=\:\frac{\mathrm{4}}{\mathrm{12}}\:\:\:\Rightarrow\:{PE}\:=\:\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${PQ}\:=\:{PE}\:+\:{EQ} \\ $$$${PQ}\:=\:\frac{\mathrm{4}}{\mathrm{3}}+\mathrm{4}\:=\:\frac{\mathrm{16}}{\mathrm{3}} \\ $$
Commented by Tawa11 last updated on 06/Jul/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

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