tell you what. i'll find the length of the medians and you can fill in the blanks where they belong.
br
b = (2a,0)
r = (a/2,b/2) remember these things are medians. they go to the 1/2 way point of the line opposite the vertex they face.
br^2 = (2a - a/2)^2 + (0 - b/2)^2
br^2 = (3/2 a) ^2 + b^2 / 4
br^2 = 9/4 a^2 + b^2 / 4 we need to find some relationship between a and b.
let's try ab = bc
ab = 2a
bc = (2a - a)^2 + (b - 0)^2
bc = sqrt(a^2 + b^2)
ab = bc
2a = sqrt(a^2 + b^2) square both sides.
4a^2 = a^2 + b^2 subtract a^2 from both sides.
sqrt(3a^2) = sqrt(b^2)
sqrt(3)a = b
let's put b into br^2
br^2 = 9/4 a^2 + 3b^2 / 4
br^2 = 12 a^2 / 4
br^2 = 3a^2
br = sqrt(3) * a
cp
c = (a,b) ; p = (a,0) this is another application of the distance formula, and it is a good one.
cp^2 = (a - a)^2 + (b - 0)^2
cp^2 = 0 + b^2
cp^2 = b^2 which you can see from the diagram.
cp = sqrt(b^2)
cp = b but b = sqrt(3) * a
cp = sqrt(3)*a
aq
a = (0,0)
q = (3/2) a, b/2
aq^2 = (3a/2 -0)^2 + (b/2 - 0)^2
aq^2 = 9a^2/4 + b^2/4
but b = sqrt(3) * a
aq^2 = 9a^2 / 4 + (sqrt(3)*a)^2 /4
aq^2 = 9 a^2/ 4 + 3 a^2 / 4
aq^2 = 12 a^2/4
aq^2 = 3 a^2
aq = (sqrt 3) * a which agrees with the other two.
