Question

In triangle ABC, with A=90°, the bisector s of the angles B and C meet at P. The distance from P to the hypotenuse is 4√2. The distance AP is

Answer 1

Given:

∴ Incircle is formed a radius = $4 \sqrt{2}$

STEP BY STEP EXPLANATION:

PE = r = $4 \sqrt{2}$

PE = r = $4 \sqrt{2}$ also PE = AE

∴ Δ APE , ${(AP)}^{2}$ = ${(AE)}^{2}$ + ${(PE)}^{2}$

= ${( 4\sqrt{2}) }^{2}$ + ${( 4\sqrt{2}) }^{2}$

= 64

ANSWER:

∴ AP = 8

Hence verified !