Question Number 35798 by ajfour last updated on 23/May/18
Commented by ajfour last updated on 23/May/18
$$\bigtriangleup{ABC}\:{and}\:\bigtriangleup{CDE}\:{are}\:{congruent}. \\ $$$${Find}\:{the}\:{area}\:{of}\:{quadrilateral} \\ $$$${APQC}\:{in}\:{terms}\:{of}\:{a},\:{b},\:{and}\:{c}\:;\:{the} \\ $$$${sides}\:{of}\:\bigtriangleup{ABC}\:. \\ $$
Commented by ajfour last updated on 23/May/18
Commented by ajfour last updated on 23/May/18
$${considering}\:\bigtriangleup{ADP} \\ $$$$\frac{{AP}}{\mathrm{sin}\:\angle{ADP}}\:=\:\frac{{AD}}{\mathrm{sin}\:\angle{APD}} \\ $$$$\Rightarrow\:\:\:\frac{{c}−{BP}}{\mathrm{sin}\:{B}}\:=\:\frac{{c}−{b}}{\mathrm{sin}\:\left(\theta+{B}\right)} \\ $$$$\theta\:=\:\pi−\angle{A} \\ $$$${so}\:\:\:\:\:{BP}={c}−\frac{\left({c}−{b}\right)\mathrm{sin}\:{B}}{\mathrm{sin}\:\left({A}−{B}\right)} \\ $$$$\:\:\:\:\:………………….. \\ $$$${considering}\:\bigtriangleup{CEQ} \\ $$$$\frac{{CQ}}{\mathrm{sin}\:\angle{CEQ}}\:=\:\frac{{CE}}{\mathrm{sin}\:\angle{CQE}} \\ $$$$\Rightarrow\:\:\frac{{a}−{BQ}}{\mathrm{sin}\:{C}}\:=\:\frac{{b}}{\mathrm{sin}\:\theta\left(=\mathrm{sin}\:{A}\right)} \\ $$$${BQ}={a}−\frac{{b}\mathrm{sin}\:{C}}{\mathrm{sin}\:{A}} \\ $$$$\:\:\:\:\:\:\:……………………. \\ $$$${Area}\left(\bigtriangleup{PBQ}\right)=\frac{{BP}×{BQ}\mathrm{sin}\:{B}}{\mathrm{2}} \\ $$$${Area}\left({APQC}\right)={Area}\left(\bigtriangleup{ABC}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−{Area}\left(\bigtriangleup{PBQ}\right). \\ $$