Question

Consider the triangle ABC shown in the following figure where $BC=12cm,DB=9cm,$ $CD=6cm$ and $\angle BCD=\angle BAC$.
What is the ratio of the perimeter of $△ADC$ to that of $△BDC$?

• A. $\frac{7}{9}$
• B. $\frac{8}{9}$
• C. $\frac{6}{9}$
• D. $\frac{5}{9}$

Answer 1

The correct option is A $\frac{7}{9}$

In $△BCD$ and $△ABC$,

$\angle DBC=\angle ABC$ ..... (Common angle)

$\angle BCD=\angle BAC$ ....... (Given)

$BC=BC$ ...... (Common side)

$\therefore △BCD\sim △ABC$ ........ (Using ASA Rule)

$\Rightarrow \frac{AC}{DC}=\frac{BC}{BD}=\frac{AB}{BC}$

$\Rightarrow \frac{AB}{BC}=\frac{BC}{BD}$

$\Rightarrow \frac{AB}{12}=\frac{12}{9}$

$\Rightarrow AB=16$

Similarly,

$\frac{AC}{DC}=\frac{BC}{BD}$

$\Rightarrow \frac{AC}{6}=\frac{12}{9}$

$\Rightarrow AC=8$
Hence, $AD=7$ and $AC=8$Therefore,

$\frac{Perimeterof△ADC}{Perimeterof△BDC}$

$=\frac{6+7+8}{9+6+12}$

$=\frac{21}{27}$

$=\frac{7}{9}$