A geometry problem by Ajay Sambhriya

Geometry Level 3

We are given that $BC = 12$ , $DC = 6$ , $DB = 9$ , and $\angle BCD = \angle BAC$ .

What is the ratio of the perimeter of the triangle $ADC$ to the that of the triangle $BDC$ ?

3:2

7:9

4:5

4:3

1 solution

Marta Reece
Jun 13, 2017

Triangle $ACB$ is similar to triangle $CDB$ so $\dfrac{x}{12}=\dfrac 69$ , which results in $x=8$

Also $\dfrac{AB}{12}=\dfrac{12}{9}$ , which gives $AB=16$

$AD=AB-9=7$

The perimeter of triangle $ADC$ is $7+6+8=21$

The perimeter of triangle $BDC$ is $9+6+12=27$

The ratio of the perimeters is $\dfrac {21}{27}=\boxed{7:9}$