Question

In figure, if the length of $\overline{BC}$ equals one-half the length of $\overline{AB}$, the sum of $BC$ and $CD$ equals three-fourth the length of $\overline{AD}$, and the length of $\overline{AD}$ equals $16$, calculate the distance between point C and the midpoint of $\overline{AD}$.

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is A $2$

Given$BC=\frac{1}{2}AB$ and AD=16

$BC+CD=\frac{3}{4}AD$

$AB+BC+CD=AD$

$\Rightarrow AB+\frac{3}{4}AD=AD$

$\Rightarrow AB+\frac{3}{4}\times 16=16$

$\Rightarrow AB=16-12=4$

$BC=\frac{1}{2}\times 4=2$

$AC=4+2=6$

Then distance of point $C$ is $6$

Then mid point of $AD=$ $\frac{AD}{2}=\frac{16}{2}=8$

Then distance between point and mid point $AD=8-6=2$.