CAT 2019 Slot 2 QA Question & Solution
GeometryEasy
Question
Let ABC be a right-angled triangle with hypotenuse BC of length 20 cm. If AP is perpendicular on BC, then the maximum possible length of AP, in cm, is
Options
10
5
$8\surd2$
$6\surd2$
Solution
Let p be the length of AP.
It is given that $\angle\ BAC\ =90$ and $\angle\ APC\ =90$
Let $\angle\ ABC\ =\theta$, then $\angle\ BAP\ =90-\theta$ and $\angle\ BCA\ =90-\theta$
So $\angle\ PAC\ =\theta$
Triangles BPA and APC are similar
$p^2=x\left(20-x\right)$
We have to maximize the value of p, which will be maximum when $x=20-x$
$x=10$
