Practice: Find The Perimeter Of A Triangle

The distance between two points $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ in the Cartesian plane is given by $\sqrt{\left(x_2 - x_1\right)^{2} + \left(y_2-y_1\right)^{2}}$ .

Hence, the perimeter of the triangle is found by finding the distances between each pair of points and summing them:

Between $\left(-3, 0\right)$ and $\left(1, -3\right)$ :

$D_1 = \sqrt{\left(1 - (-3)\right)^{2} + \left((-3) - 0\right)^{2}}$

$\Rightarrow D_1 = 5$

Between $\left(-3, 0\right)$ and $\left(4, 1\right)$ :

$D_2 = \sqrt{\left(4 - (-3)\right)^{2} + \left(1 - 0\right)^{2}}$

$\Rightarrow D_2 = 5\sqrt{2}$

Between $\left(1, -3\right)$ and $\left(4, 1\right)$ :

$D_3 = \sqrt{\left(4 - 1\right)^{2} + \left(1 - (-3)\right)^{2}}$

$\Rightarrow D_3 = 5$

Hence, $Perimeter = D_1 + D_2 + D_3 = 5 + 5\sqrt{2} + 5 = 17.07... \approx \boxed{17}$

Let point $A$ at coordinates $( -3,0 )$

Let point $B$ at coordinates $( 1,-3 )$

Let point $C$ at coordinates $( 4,1 )$

To measure the distance between two point, use:

$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

Line AB = $\sqrt{(-3-1)^2+(0-3)^2} = \boxed{5}$

Line BC = $\sqrt{(1-4)^2+(-3-1)^2} = \boxed{5}$

Line AC = $\sqrt{(-3-4)^2+(0-1)^2} =\boxed{5\sqrt{2} }$

By using calculator, we get $5 \sqrt{2} = 7.07$

$Perimeter = 5+5+7.07 =17.07 \Rightarrow \boxed{17}$

So the answer is 17

We shall use here the distance formula, $d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

The vertices are $(-3,0) , (1,-3) , (4,1)$ . Using the distance formula, we get,

1st side $=\sqrt{(-3-1)^{2}+(0+3)^{2}} = \sqrt{(-4)^{2}+3^{2}} = \sqrt{16+9} = \sqrt{25} = 5$ units.

2nd side $=\sqrt{(4-1)^{2}+(1+3)^{2}} = \sqrt{3^{2}+4^{2}} = \sqrt{9+16} = \sqrt{25} = 5$ units.

3rd side $=\sqrt{(4+3)^{2}+(1-0)^{2}} = \sqrt{7^{2}+1^{2}} = \sqrt{49+1} = \sqrt{50} = 5\sqrt{2}$ units.

Now, Perimeter $=5+5+5\sqrt{2}=10+5\times 1.414 = 10+7.07 = \boxed{17.07}$

So, perimeter of the triangle to the nearest unit is $\boxed{17}$ units.

The point (-3, 0) lies in the 2nd quadrant.
The point (1, -3) lies in the 4th quadrant.
The point (4, 1) lies in the 3rd quadrant. Always the 'x' coordinate will be given at the first in a set of coordinates.
Consider any two sets (since, two points form a line i.e. the side of the triangle)
I've taken the first set as (-3, 0) and (1, -3), the second as (1, -3) and (4, 1), and the third as ((-3, 0) and (4, 1)
Let the 'x' and 'y' coordinates of the first set that you are considering be 'x1' and 'y1' and the coordinates of the second set be 'x2' and 'y2'
Use the formula to calculate the distance between the points:
√(x2 - x1)² + (y2 - y1)²
Substitute the values in the formula:
The distance between the first two points = √(1 + 3)² + (-3 - 0)²
= √4² + (-3)²
= √16 + 9
= √25 = 5
The distance between the next two points = √(4 - 1)² + (1 + 3)²
= √3² + 4²
= √9 + 16
= √25 = 5
The distance between the next set of points = √(4 + 3)² + (1 - 0)²
= √7² + 1²
= √49 + 1
= √50 = 7.07 ≈ 7
The distances are the lengths. Therefore, the perimeter of the triangle = 5 + 5 + 7 = 17

Assumt that it triangle ABC, first to side AB is sqrt((3-(-1))^2 + (0-(-3))^2 )= 5. Second, side BC = sqrt((1-(-3)^2 + (4-1)^2) = 5. Finally, side AC = sqrt((4-(-3)^2 + (1-0)^2) = 5 * sqrt(2) = 5 * 1,4 = 7. ---> Thus, the perimeter of this plane triangle is 5 + 5 + 7 = 17.(Answered)

I thought it said area when I read it through the first time, no wonder I kept getting it wrong. I thought I got it wrong when I did (1/2)(side 1)(side 2) so I tried (1/2)(base)(height) which took awhile cause I had to get the formula for two of the lines, then the formula for the line perpendicular to the base that went through the tip (the bottom point). Gah.

The area rounds to 13 units squared if anyone is curious :P

Using the Distance Formula get the sides of the triangle then add for the Perimeter :)

easiest way- use the Pythagoras theorem

Perimeter = sum of all the sides = 5 + 5 + sqrt(50) = 17 approximately

Distance b/w any two points can be found using the following formula from coordinate geometry :

\sqrt{(x1-x2)^{2}+(y2-y1)^{2}+(z2-z1)^{2}}

Using the above formula we find \sqrt{50} ,5 and 5, so perimeter which is the measure of the sum of the sides of tehe triangle so we get 17(around).

The method known is the find the sides of the triangle by using the formula to measure the distance between two points.

Hence, with that method, the first side is $\sqrt{1 + 49}$ = $\sqrt{50}$ = 12.07 Also, the other sides are 5 each. Hence the perimeter of the triangle is 5 +5 + 12.07 = 17.07

Distance between (-3,0) and (1,3) is 5; [sqrt(16+9)]; similarly other two sides are 5 and 5*1.732=7; So Perimeter is 5+5+7=17

Let, A = (-3,0) B = (1,-3) C = (4,1) AB = ((-3-1)^2 + (-3-0)^2)^1/2 = 25^1/2 = 5 BC = ((4-1)^2+(1--3)^2)^1/2 = 25^1/2 = 5 AC = ((-3-4)^2 +(1-0)^2)^1/2 = 50^1/2 Perimeter of triangle ABC = AB + BC+ AC = 5 + 5 + 7.07 = 17.07 = 17(Rounded Off)

By distance formula, get $5 + 5 + \sqrt{50}$ . We know $7 < \sqrt{50} < 8$ . Thus, the answer is $5 + 5 + 7 = \boxed {17}$ .

Let A = (-3,0) , B = (1,-3) , C = (4,1) By distance formula we get AB = 5, BC = 5 and AC = root 50 Perimeter = AB+BC+AC = 5+5+root50 = 17.02 units.

I wrote it down on a piece of paper, marked off the dots and drew the triangle(x,y). Then I used the pythagoran theorem to conclude that the upper line((-3,0)-(4,1)) were about 7 units(-3to4=7,0to1=1 sqrt(1^2+7^2)). Then I did the same on the other two and they were =5(the beautiful equation sqrt(3^2+4^2)=5). 7+5+5=17. DONE!

Use distance formula between two co-ordinates.

let A(-3,O) B(1,-3) C(4,1) THEN AB=5 UNITS

BC=5 UNITS

AC=7.07(APPROX BY UNDER ROOT 50)

AB+BC+AC=5+5+7.07

                   =17.07


                  =17 UNITS(NEAREST UNIT)