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Question Number 35798 by ajfour last updated on 23/May/18

Commented by ajfour last updated on 23/May/18

△ABC and △CDE are congruent. Find the area of quadrilateral APQC in terms of a, b, and c ; the sides of △ABC .

$$\bigtriangleup{ABC}\:{and}\:\bigtriangleup{CDE}\:{are}\:{congruent}. \\ $$$${Find}\:{the}\:{area}\:{of}\:{quadrilateral} \\ $$$${APQC}\:{in}\:{terms}\:{of}\:{a},\:{b},\:{and}\:{c}\:;\:{the} \\ $$$${sides}\:{of}\:\bigtriangleup{ABC}\:. \\ $$

Commented by ajfour last updated on 23/May/18

Commented by ajfour last updated on 23/May/18

considering △ADP ((AP)/(sin ∠ADP)) = ((AD)/(sin ∠APD)) ⇒ ((c−BP)/(sin B)) = ((c−b)/(sin (θ+B))) θ = π−∠A so BP=c−(((c−b)sin B)/(sin (A−B))) ....................... considering △CEQ ((CQ)/(sin ∠CEQ)) = ((CE)/(sin ∠CQE)) ⇒ ((a−BQ)/(sin C)) = (b/(sin θ(=sin A))) BQ=a−((bsin C)/(sin A)) ......................... Area(△PBQ)=((BP×BQsin B)/2) Area(APQC)=Area(△ABC) −Area(△PBQ).

$${considering}\:\bigtriangleup{ADP} \\ $$$$\frac{{AP}}{\mathrm{sin}\:\angle{ADP}}\:=\:\frac{{AD}}{\mathrm{sin}\:\angle{APD}} \\ $$$$\Rightarrow\:\:\:\frac{{c}−{BP}}{\mathrm{sin}\:{B}}\:=\:\frac{{c}−{b}}{\mathrm{sin}\:\left(\theta+{B}\right)} \\ $$$$\theta\:=\:\pi−\angle{A} \\ $$$${so}\:\:\:\:\:{BP}={c}−\frac{\left({c}−{b}\right)\mathrm{sin}\:{B}}{\mathrm{sin}\:\left({A}−{B}\right)} \\ $$$$\:\:\:\:\:....................... \\ $$$${considering}\:\bigtriangleup{CEQ} \\ $$$$\frac{{CQ}}{\mathrm{sin}\:\angle{CEQ}}\:=\:\frac{{CE}}{\mathrm{sin}\:\angle{CQE}} \\ $$$$\Rightarrow\:\:\frac{{a}−{BQ}}{\mathrm{sin}\:{C}}\:=\:\frac{{b}}{\mathrm{sin}\:\theta\left(=\mathrm{sin}\:{A}\right)} \\ $$$${BQ}={a}−\frac{{b}\mathrm{sin}\:{C}}{\mathrm{sin}\:{A}} \\ $$$$\:\:\:\:\:\:\:......................... \\ $$$${Area}\left(\bigtriangleup{PBQ}\right)=\frac{{BP}×{BQ}\mathrm{sin}\:{B}}{\mathrm{2}} \\ $$$${Area}\left({APQC}\right)={Area}\left(\bigtriangleup{ABC}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−{Area}\left(\bigtriangleup{PBQ}\right). \\ $$

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