Question

In the right triangle shown the sum of the distance $\overline{BM}$ and $\overline{MA}$ is equal to the sum of the distance $\overline{BC}$ and $\overline{CA}$. If $\overline{MB}=x,\overline{CB}=h$ and $\overline{CA}=d$, then $x$ equals:

• A. $\frac{hd}{2h+d}$
• B. $d-h$
• C. $\frac{1}{2}d$
• D. $h+d-\sqrt{2d}$
• E. $\sqrt{h^2+d^2-h}$

Answer 1

The correct option is A $\frac{hd}{2h+d}$

Side $AM=\sqrt{(x+h{)}^2+d^2}$

Given condition is :

$x+\sqrt{(x+h{)}^2+d^2}=h+d$

$\Rightarrow \sqrt{(x+h{)}^2+d^2}=h+d-x$

Squaring on both sides ;

$\Rightarrow x^2+h^2+2xh+d^2=h^2+d^2+x^2+2hd-2dx-2hx$

$\Rightarrow 2xh+dx=hd$

$\Rightarrow x=\frac{hd}{2h+d}$