Question

Through a point $P$ inside the triangle $ABC$ a line is drawn parallel to the base $AB$, dividing the triangle into two equal area. If the altitude to $AB$ has a length of $1$, then the distance from $P$ to $AB$ is

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $2-\sqrt{2}$
• D. $\frac{2-\sqrt{2}}{2}$
• E. $\frac{2+\sqrt{2}}{8}$

Answer 1

The correct option is D $\frac{2-\sqrt{2}}{2}$

Let $x$ be the distance from $P$ to $AB$. By similar triangles.
$\frac{1}{2}=\frac{(1-x{)}^2}{1^2};\therefore 1-x=\pm \frac{1}{\sqrt{2}};\therefore x=\frac{2-\sqrt{2}}{2}$
(negative sq. root rejected).