Question

Determine the perimeter of the triangle whose sides have the midpoints (1,0), (2, -1) and (3,1)

Answer 1

where did the 2 came from? i mean the constant.

Answer 2

(1,0), (2, -1) and (3,1)
first two
sqrt((-1-0)^2+(2-1)^2) = sqrt(2)
second 2
sqrt((1-1)^2+(3-2)^2) = 1
last and first one
sqrt((1-0)^2+(3-1)^2) = sqrt(5)
so sqrt(5) +sqrt(2) + 1

Answer 3

owww. okok. thanks :))

Answer 4

can you also help me find the vertices of those midpoints.

Answer 5

@zzr0ck3r can you also help me find the vertices of those midpoints.

Answer 6

A line segment that joins the mid-points of the sides of a triangle is parallel to the third side. Therefore you can use two of the given mid-points to find the slope of the segment that joins the mid points using the formula
\[m=\frac{y _{1}-y _{2}}{x _{1}-x _{2}}\]
then using this slope and the third mid-point you can find the equation of the line that contains a side of the triangle using the point-slope form of the equation of a line:
\[y-y _{1}=m(x-x _{1})\]

Answer 7

yea, a thing to note for the initial question of finding perimeter :
What zz solved was the perimeter of triangle formed by joining midpoints. You need to take twice of that for the actual perimeter of the triangle

Answer 8

so if the perimeter of the midpoints is n should i multiply it by 2??

Answer 9

Yes ! the side joining midpoints is half the parallel side

Answer 10

Then you need to do the above process once more using a different set of mid-points.
Now you have have two equations forming a system of equations with a solution set that is the ordered pair that defines one vertex of the triangle.

Answer 11

ok.thanks for the help :))

Answer 12

:) continue wid finding vertices... it seems complex

Answer 13

Yes, multiply by 2.

It's an interesting effect when you take a triangle and draw an triangle inside it at the outside triangle's midpoints:

You effectively make a similar triangle as the original, except that it's rotated 180 degrees and is half the size.

Answer 14

And oh, I forgot, when you draw that inner triangle, all 4 resultant triangles are congruent.