Question

In the following figure, $XY$ is parallel to $BC$, $AX=9cm$. $XB=4.5cm$ and $BC=18cm$. Find

$\frac{YC}{AC}$

• A. $\frac{2}{5}$
• B. $\frac{7}{6}$
• C. $\frac{1}{3}$
• D. $\frac{11}{4}$

Answer 1

The correct option is C $\frac{1}{3}$

In $△$s AXY and ABC,

$\angle XAY=\angle BAC$ (Common angle)

$\angle AXY=\angle ABC$ (Corresponding angles for parallel lines, XY II BC)

$\angle AYX=\angle ACB$ (Corresponding angles for parallel lines, XY II BC)

Thus, $△AXY\sim △ABC$

Hence, $\frac{AX}{AB}=\frac{AY}{AC}$ (Using similar triangle property)

$\frac{AX}{AX+XB}=\frac{AY}{AY+YC}$

$\frac{9}{9+4.5}=\frac{AY}{AY+YC}$

$9AY+9YC=13.5AY$

$9YC=4.5AY$

$\frac{AY}{YC}=2$

$\frac{AY}{YC}+1=2+1$

$\frac{AY+YC}{YC}=3$

$\therefore \frac{YC}{AC}=\frac{1}{3}$