MHB What is the perimeter of triangle ABC?

AI Thread Summary
To find the perimeter of triangle ABC with vertices A(1, 1), B(9, 3), and C(3, 5), the distance formula is used to calculate the lengths of all three sides, which are then summed. For the second part, the perimeter of the triangle formed by the midpoints of the sides is determined by first calculating the midpoints using the midpoint formula and then finding the distances between these midpoints. The ratio of the perimeters from part one to part two is set up as R = (perimeter of part 1)/(perimeter of part 2). It is noted that the desired ratio is 2:1, indicating that the perimeter of the triangle formed by the midpoints is half that of triangle ABC. This exercise combines principles of geometry and algebra, stemming from a precalculus context.
mathdad
Messages
1,280
Reaction score
0
The vertices of triangle ABC are A(1, 1), B(9, 3), and
C(3, 5).

1. Find the perimeter of triangle ABC.

I must use the distance formula for points on the xy-plane to find all three sides. I then add all three sides. Correct?

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

I am not too sure about part 2.

3. Compute the ratio of the perimeter in part 1 to the perimeter in part 2.

I will let R = ratio.

The set up for part 3 is

R = (perimeter of part 1)/(perimeter of part 2)

Correct? I cannot do part 3 without computing part 2, which I don't know how to do.
 
Mathematics news on Phys.org
RTCNTC said:
The vertices of triangle ABC are A(1, 1), B(9, 3), and
C(3, 5).

1. Find the perimeter of triangle ABC.

I must use the distance formula for points on the xy-plane to find all three sides. I then add all three sides. Correct?

Correct.

RTCNTC said:
2. Find the perimeter of the triangle that is formed by joining the midpoints of the three sides of triangle ABC.

A sheet of graph paper will be handy for this exercise. Mathematically, the coordinates of a midpoint may be found with

$$x_M=\frac{x_1+x_2}{2},\quad y_M=\frac{y_1+y_2}{2}$$

RTCNTC said:
3. Compute the ratio of the perimeter in part 1 to the perimeter in part 2.

I will let R = ratio.

The set up for part 3 is

R = (perimeter of part 1)/(perimeter of part 2)

Correct?

Correct. You may also write

$$P_1:P_2$$

As a hint, the desired ratio is 2:1.
 
1. Use the distance formula to find the distance between all 3 sides. Add all three sides. Adding all three sides yields perimeter 1.

2. Use the midpoint formula to find the midpoint of the distance between the three given points.

3. Find the distance between the 3 midpoints found in part 2 above. Add all 3 sides. This yields perimeter 2.

4. The ratio = (perimeter 1)/(perimeter 2)

This exercise is related more to geometry mixed with algebra. Correct? I did not post in the geometry forum because the question is from David Cohen's precalculus textbook.
 
Last edited:
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...

Similar threads

Replies
13
Views
4K
Replies
2
Views
2K
Replies
1
Views
957
Replies
2
Views
1K
Replies
1
Views
1K
Replies
5
Views
2K
Back
Top