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

In equilateral triangle ABC ,

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

MB=
`(a)/(2)`

In Right Triangle AMB

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

AM² = AB²- MB²

AM²= a²- \frac{a^2}{4}[/tex]

AM²=
`(\sqrt3)/(4) * a^2`

AM=
`(\sqrt3)/(2)* a`

Also, MP = b

Again using pythagorean theorem In Right Δ APB

BP²= AP² - AB²

=
`((√(3)a)/(2) + b)^2 -a^2\\\\ b^2 + √(3) a b -(a^2)/(4)`

BP=
`\sqrt{ b^2 + √(3) a b -(a^2)/(4)}`

Perimeter of Triangle ABP = AB + AP + BP

= a +
`(\sqrt3)/(2)* a` +b +
`\sqrt{ b^2 + √(3) a b -(a^2)/(4)}`