What is the perimeter of triangle ABC?

[mathdad]

The vertices of triangle ABC are A(1,1), B(9, 3) and C(3, 5).

A. Find the perimeter of triangle ABC.

B. Find the perimeter of the triangle that is formed by joining the midpoints of the three sides of triangle ABC.

For part A, must I use the distance formula for points on the xy-plane 3 times and then add all 3 sides to find the perimeter?

For B, must I use the midpoint formula to find the midpoint of all 3 sides of triangle ABC? After finding all 3 midpoints, I then repeat part A?

Correct?

 

[alexmahone]

That's right. :)

 

[MarkFL]

Suppose the vertices of a triangle are at:

\(\displaystyle \left(x_1,y_1\right),\,\left(x_2,y_2\right),\,\left(x_3,y_3\right)\)

And so the perimeter $P$ is given by:

\(\displaystyle P=\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}+\sqrt{\left(x_1-x_3\right)^2+\left(y_1-y_3\right)^2}+\sqrt{\left(x_2-x_3\right)^2+\left(y_2-y_3\right)^2}\)

Now, if we take the mid-points of the sides of the original triangle as vertices of a second triangle, we find its perimeter $P_2$ is:

\(\displaystyle P_2=\sqrt{\left(\frac{x_1+x_2}{2}-\frac{x_2+x_3}{2}\right)^2+\left(\frac{y_1+y_2}{2}-\frac{y_2+y_3}{2}\right)^2}+\sqrt{\left(\frac{x_2+x_3}{2}-\frac{x_3+x_1}{2}\right)^2+\left(\frac{y_2+y_3}{2}-\frac{y_3+y_1}{2}\right)^2}+\sqrt{\left(\frac{x_3+x_1}{2}-\frac{x_1+x_2}{2}\right)^2+\left(\frac{y_3+y_1}{2}-\frac{y_1+y_2}{2}\right)^2}\)

\(\displaystyle P_2=\frac{1}{2}\left(\sqrt{\left(x_1-x_3\right)^2+\left(y_1-y_3\right)^2}+\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}+\sqrt{\left(x_2-x_3\right)^2+\left(y_2-y_3\right)^2}\right)=\frac{1}{2}P\)

You see, when you connect the mid-points of the sides of a triangle, you divide that triangle into 4 congruent triangles all similar to the original. Draw a sketch to see this more clearly. Thus, the smaller triangles are 1/4 the area of the original, and being similar to the original means that all corresponding sides are 1/2 that of the original, and so the perimeters of the smaller triangles are 1/2 that of the original. :)

 

[mathdad]

Suppose the vertices of a triangle are at:

\(\displaystyle \left(x_1,y_1\right),\,\left(x_2,y_2\right),\,\left(x_3,y_3\right)\)

And so the perimeter $P$ is given by:

\(\displaystyle P=\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}+\sqrt{\left(x_1-x_3\right)^2+\left(y_1-y_3\right)^2}+\sqrt{\left(x_2-x_3\right)^2+\left(y_2-y_3\right)^2}\)

Now, if we take the mid-points of the sides of the original triangle as vertices of a second triangle, we find its perimeter $P_2$ is:

\(\displaystyle P_2=\sqrt{\left(\frac{x_1+x_2}{2}-\frac{x_2+x_3}{2}\right)^2+\left(\frac{y_1+y_2}{2}-\frac{y_2+y_3}{2}\right)^2}+\sqrt{\left(\frac{x_2+x_3}{2}-\frac{x_3+x_1}{2}\right)^2+\left(\frac{y_2+y_3}{2}-\frac{y_3+y_1}{2}\right)^2}+\sqrt{\left(\frac{x_3+x_1}{2}-\frac{x_1+x_2}{2}\right)^2+\left(\frac{y_3+y_1}{2}-\frac{y_1+y_2}{2}\right)^2}\)

\(\displaystyle P_2=\frac{1}{2}\left(\sqrt{\left(x_1-x_3\right)^2+\left(y_1-y_3\right)^2}+\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}+\sqrt{\left(x_2-x_3\right)^2+\left(y_2-y_3\right)^2}\right)=\frac{1}{2}P\)

You see, when you connect the mid-points of the sides of a triangle, you divide that triangle into 4 congruent triangles all similar to the original. Draw a sketch to see this more clearly. Thus, the smaller triangles are 1/4 the area of the original, and being similar to the original means that all corresponding sides are 1/2 that of the original, and so the perimeters of the smaller triangles are 1/2 that of the original. :)

What a great reply! I actually learned something today.

 

[mathdad]

Part A

Let P = perimeter of triangle ABC.

Let d = distance between the given points.

d(A,B) = sqrt{9-1)^2 + (3-1)^2}

d(A,B) = sqrt{64 + 4}

d(A,B) = sqrt{68}

d(B,C) = sqrt{(3-3)^2 + (5-9)^2}

d(B,C) = sqrt{0 + 16}

d((B,C) = sqrt{16}

d(B,C) = 4

d(A,C) = sqrt{(3-1)^2 + (5-1)^2}

d(A,C) = sqrt{4 + 16}

d(A,C) = sqrt{20}

P = d(A,B) + d(B,C) + d(A,C)

P = sqrt{68} + 4 + sqrt{20}

P = 4 + 2√5 + 2√17

Yes?

 

[mathdad]

Part B

Let P_2 = perimeter for the triangle

Let d_2 = distance between vertices of the second triangle

The midpoint of AB = (5, 2).

The midpoint of BC = (6,4).

The midpoint of AC = (2,3).

d_2(x,y) = sqrt{(6-5)^2 + (4-2)^2}

d_2(x,y) = sqrt{1 + 4}

d_2(x,y) = sqrt{5}

d_2(y,z) = sqrt{(2-6)^2 + (3-4)^2}

d_2(y,z) = sqtt{16 + 1}

d_2(y,z) = sqrt{17}

d_2(x,z) = sqrt{(2-5)^2 + (3-2)^2}

d_2(x,z) = sqrt{9 + 1}

d_2(x,z) = sqrt{10}

P_2 = √5 + √17 + √10

P_2 = √17 + √5 (1 + √2)

Yes?

 

[MarkFL]

Your answer to the first part of the problem isn't correct, and why would you not simply divide by 2 to get the answer to the second part of the problem?

 

[mathdad]

Your answer to the first part of the problem isn't correct, and why would you not simply divide by 2 to get the answer to the second part of the problem?

For Part A, I used the distance formula 3 times:

I found the distance from A to B, B to C, and finally A to C.
Are you saying that I made a calculation error? Are you saying the correct order is to find the distance from A to B, A to C, and then B to C?

 

[MarkFL]

For Part A, I used the distance formula 3 times:

I found the distance from A to B, B to C, and finally A to C.
Are you saying that I made a calculation error? Are you saying the correct order is to find the distance from A to B, A to C, and then B to C?

You made an error when putting the coordinates into the distance formula (recheck that). The method (which is more important to understand) you did correctly, it was simply the application where you made a minor error.

My main concern was that you then used the same method for the second part of the problem, when all you need to do is divide the result of the first part by 2 to get that answer. :)

 

[mathdad]

You made an error when putting the coordinates into the distance formula (recheck that). The method (which is more important to understand) you did correctly, it was simply the application where you made a minor error.

My main concern was that you then used the same method for the second part of the problem, when all you need to do is divide the result of the first part by 2 to get that answer. :)

Ok. Can you show me the right way to plug the coordinates of points A, B, and C into the distance formula?

 

[MarkFL]

Ok. Can you show me the right way to plug the coordinates of points A, B, and C into the distance formula?

You need to make sure that x's and y's are put in their respective places. In post #5, you state:

d(B,C) = sqrt{(3-3)^2 + (5-9)^2}

The points B and C were given as:

B(9, 3) and C(3, 5)

So, what you want is:

\(\displaystyle d(B,C)=\sqrt{(9-3)^2 + (3-5)^2}=2\sqrt{10}\)

Your other distances are correct, so we have:

\(\displaystyle P=2\sqrt{10}+2\sqrt{5}+2\sqrt{17}\)

And so then we easily find:

\(\displaystyle P_2=\sqrt{10}+\sqrt{5}+\sqrt{17}\)

This value for $P_2$ agrees with the result you found, because in finding $P_2$ you correctly applied the distance formula for all three sides (even though I was trying to encourage you to simply divide $P$ by 2 to get there). :)

 

[mathdad]

B(9, 3) and C(3, 5)

sqrt{(9-3)^2 + (3-5)^2} = sqrt{(3-9)^2 + (5-2)^2} because we are squaring in the radicand. True?

 

[MarkFL]

B(9, 3) and C(3, 5)

sqrt{(9-3)^2 + (3-5)^2} = sqrt{(3-9)^2 + (5-2)^2} because we are squaring in the radicand. True?

Yes, it doesn't matter which point we call point 1 and the other point 2 because the distance from point 1 to point 2 is the same as the distance from point 2 to point 1.

 

[mathdad]

Yes, it doesn't matter which point we call point 1 and the other point 2 because the distance from point 1 to point 2 is the same as the distance from point 2 to point 1.

Cool. Many students do not know this fact.

(-a)^2 = (a)^2