Question

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

Answer 1

Answer:

$AM=BM=\frac{\sqrt{3}}{2}$

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 = 1

Also, in equilateral triangle, altitude and median are the same.

As AM is the median, M is the midpoint of BC, $CM=\frac{1}{2}BC=\frac{1}{2}$

In the figure, AM is a median as well as an altitude.

In $\bigtriangleup AMC$, $\angle AMC=90^{\circ}$

Using Pythagoras theorem: $(Hypotenuse)^2=(base)^2+(perpendicular)^2$

$AC^2=AM^2+CM^2\Rightarrow AM^2=AC^2-CM^2$

$AM^2=1^2-\left ( \frac{1}{2} \right )^2=1-\frac{1}{4}=\frac{3}{2}\\\Rightarrow AM=\frac{\sqrt{3}}{2}$

Similarly, in $\bigtriangleup AMB$,

$BM=\frac{\sqrt{3}}{2}$