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Question-64138




Question Number 64138 by Tawa1 last updated on 14/Jul/19
Answered by som(math1967) last updated on 14/Jul/19
△ADC∼△BAC  ∴((AC)/(DC))=((BC)/(AC))=((AB)/(AD))  ∴((AC)/(DC))=((AB)/(AD))=((14)/7)  ∴AC=2DC  Again ((AC)/(DC))=((BC)/(AC)) ⇒AC^2 =BC×DC  AC^2 =(12+DC)DC★  ⇒AC^2 =12DC+DC^2   ⇒AC^2 −DC^2 =12DC  ⇒(AC+DC)(AC−DC)=12DC  ⇒3DC(AC−DC)=12DC★★  AC−DC=4cm  ★BC=BD+DC=12+DC  ★★AC+DC=2DC+DC=3DC
$$\bigtriangleup{ADC}\sim\bigtriangleup{BAC} \\ $$$$\therefore\frac{{AC}}{{DC}}=\frac{{BC}}{{AC}}=\frac{{AB}}{{AD}} \\ $$$$\therefore\frac{{AC}}{{DC}}=\frac{{AB}}{{AD}}=\frac{\mathrm{14}}{\mathrm{7}}\:\:\therefore{AC}=\mathrm{2}{DC} \\ $$$${Again}\:\frac{{AC}}{{DC}}=\frac{{BC}}{{AC}}\:\Rightarrow{AC}^{\mathrm{2}} ={BC}×{DC} \\ $$$${AC}^{\mathrm{2}} =\left(\mathrm{12}+{DC}\right){DC}\bigstar \\ $$$$\Rightarrow{AC}^{\mathrm{2}} =\mathrm{12}{DC}+{DC}^{\mathrm{2}} \\ $$$$\Rightarrow{AC}^{\mathrm{2}} −{DC}^{\mathrm{2}} =\mathrm{12}{DC} \\ $$$$\Rightarrow\left({AC}+{DC}\right)\left({AC}−{DC}\right)=\mathrm{12}{DC} \\ $$$$\Rightarrow\mathrm{3}{DC}\left({AC}−{DC}\right)=\mathrm{12}{DC}\bigstar\bigstar \\ $$$${AC}−{DC}=\mathrm{4}{cm} \\ $$$$\bigstar{BC}={BD}+{DC}=\mathrm{12}+{DC} \\ $$$$\bigstar\bigstar{AC}+{DC}=\mathrm{2}{DC}+{DC}=\mathrm{3}{DC} \\ $$$$ \\ $$
Commented by Tawa1 last updated on 16/Jul/19
God bless you sir, i appreciate your effort.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{effort}. \\ $$

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