Question

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

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

Answer 1

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

Here, $\angle ACB=c+{180}^{\circ }-(2c-b)={180}^{\circ }-(b+c)$

Corresponding angles in $△BCD$ and & $△ABC$ are equal

So, We can say that $△BCD$ and $△ABC$ are similar.

According to property of similarity,

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

Hence,

$AB=16$

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

$AC=8$

Hence, $AD=7$ and $AC=8$

Now,

$\frac{Perimeterof△ADC}{Perimeterof△BDC}=\frac{6+7+8}{9+6+12}$

$=\frac{21}{27}=\frac{7}{9}$