Question

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

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

Answer 1

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

In $△MCA$, using Pythagoras theorem,
$M{C}^2+A{C}^2=M{A}^2$
$(x+h{)}^2+d^2=M{A}^2$ ........ $\left(1\right)$
Given, $BM+MA=BC+CA$
$MA=h+d-x$
Put this value in $\left(1\right),$ we get
$(x+h{)}^2+d^2=(h+d-x{)}^2$
$x^2+h^2+2xh+d^2=h^2+d^2+x^2+2hd-2xd-2xh$
$4xh+2xd=2hd$
$2xh+xd=hd$
$\Rightarrow x=\frac{hd}{2h+d}$