Equilateral triangle ABC has side length 1. A line is drawn from the midpoint, M, of BC to the opposite vertex A. Which of the following is true? Check all that apply. AM = 1 AM = 1/2 AM = squareroot 3/2 BM = 1 BM = 1/2 BM = squareroot 3/2

Accepted Solution

Answer:[tex]AM=BM=\frac{\sqrt{3}}{2}[/tex]Step-by-step explanation:Consider the following figure,We know that equilateral triangle is a triangle in which in all sides are equal.So, AB = BC = AC = 1Also, in equilateral triangle, altitude and median are the same.As AM is the median, M is the midpoint of BC, [tex]CM=\frac{1}{2}BC=\frac{1}{2}[/tex]In the figure, AM is a median as well as an altitude.In [tex]\bigtriangleup AMC[/tex], [tex]\angle AMC=90^{\circ}[/tex]Using Pythagoras theorem: [tex](Hypotenuse)^2=(base)^2+(perpendicular)^2[/tex][tex]AC^2=AM^2+CM^2\Rightarrow AM^2=AC^2-CM^2[/tex][tex]AM^2=1^2-\left ( \frac{1}{2} \right )^2=1-\frac{1}{4}=\frac{3}{2}\\\Rightarrow AM=\frac{\sqrt{3}}{2}[/tex]Similarly, in [tex]\bigtriangleup AMB[/tex],[tex]BM=\frac{\sqrt{3}}{2}[/tex]