Answer:
Perimeter = (2 + √3)·a
Step-by-step explanation:
Given: ΔABC is equilateral and AB = a
The diagram is given below :
AM is a median , PB ⊥ AB , PM = b
Now, by using properties of equilateral triangle, median is perpendicular bisector and each angle is of 60°.
We get, ∠AMB = 90°. So, by linear pair ∠AMB + ∠PMB = 180° ⇒ ∠PMB = 90°. Also, ∠ABC = 60° and ∠ABP = 90° (given) So, ∠PBM = 30°
Since, AM is perpendicular bisector of BC. So,
[tex]MB = \frac{a}{2}[/tex]
Now in ΔAMB , By using Pythagoras theorem
[tex]AB^{2}=AM^{2}+MB^{2}\\AM^{2}=AB^{2}-MB^{2}\\AM^{2}=a^{2}-(\frac{a}{2})^{2}\\AM=\frac{\sqrt{3}\cdot a}{2}[/tex]
Now, in ΔBMP :
[tex]sin\thinspace 30^{o}=\frac{\text{Perpendicular}}{\text{Hypotenuse}}\\\\sin\thinspace 30^{o}=\frac{\text{MB}}{\text{PB}}\\\\PB=\frac{\text{MB}}{\text{sin 30}}\\\\PB=\frac{\frac{a}{2}}{\frac{1}{2}}\implies PB = a\\\\tan\thinspace 30^{o}=\frac{\text{Perpendicular}}{\text{Base}}\\\\tan\thinspace 30^{o}=\frac{\text{MB}}{\text{PM}}\\\\PM=\frac{\text{MB}}{\text{tan 30}}\\\\PM=\frac{\frac{a}{2}}{\frac{1}{\sqrt3}}\implies PM=b= \frac{\sqrt{3}\cdot a}{2}[/tex]
Perimeter of ABM = AB + PB + PM + AM
[tex]\text{Perimeter = }a+a+b+ \frac{\sqrt{3}\cdot a}{2}\\\\=2\cdot a + \frac{\sqrt{3}\cdot a}{2} +\frac{\sqrt{3}\cdot a}{2}\\\\=2\cdot a +\sqrt{3}\cdot a\\\\=(2+\sqrt3})\cdot a[/tex]
Hence, Perimeter of ΔABP = (2 + √3)·a units
