Question

In equilateral ∆ABC length of the side is a. Perpendicular to side AB at point B intersects extension of median in point P. What is the perimeter of ∆ABP, if MP = b?

Answer 1

The answer is a+6b, if you want to know the proof please comment on this post

Answer 2

Solution:

In equilateral triangle ABC ,

You must keep in mind that Median in an equilateral triangle works as a perpendicular bisector.

MB= $\frac{a}{2}$

In Right Triangle AMB

AM² + MB²=AB² →→→[By Pythagorean Theorem]

AM² = AB²- MB²

AM²= a²- \frac{a^2}{4}[/tex]

AM²=$\frac{\sqrt3}{4} \times a^2$

AM=$\frac{\sqrt3}{2}\times a$

Also, MP = b

Again using pythagorean theorem In Right Δ APB

BP²= AP² - AB²

     = $(\frac{\sqrt{3}a}{2} + b)^2 -a^2\\\\ b^2 + \sqrt{3} a b -\frac{a^2}{4}$

BP= $\sqrt{ b^2 + \sqrt{3} a b -\frac{a^2}{4}}$

Perimeter of Triangle ABP = AB + AP + BP

                                           = a  +$\frac{\sqrt3}{2}\times a$ +b + $\sqrt{ b^2 + \sqrt{3} a b -\frac{a^2}{4}}$