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Mathematics, 22.06.2019 21:00 batista68

What is the length of the shortest side of a triangle that has vertices at (-3, -2), (1, 6), and (5, 3)?

Answers

ansver
Answer from: mariana2006

)-2,2)

Explanation:

ansver
Answer from: donterriuscollier

5

Step-by-step explanation:

Figuring the short side, it becomes a nice 3, 4, 5 triangle

so the short side is 5 units.

Hope this helped

:)

ansver
Answer from: rehel5106

5

Step-by-step explanation:

This is a 3, 4, 5 triangle

ansver
Answer from: guardhic7165

5

Step-by-step explanation:

We can use the distance formula with 3 different vertices to figure out the shortest of the three sides.

The distance formula is \sqrt{(y_2-y_1)^2+(x_2-x_1)^2}

Where (x_1,y_1) is the first points and (x_2,y_2) is the second set of points, respectively.

Now let's figure out the length of 3 sides.

1. The length between (-6,-5) & (-5,6):

\sqrt{(y_2-y_1)^2+(x_2-x_1)^2} \\=\sqrt{(6--5)^2+(-5--6)^2}\\ =\sqrt{(6+5)^2+(-5+6)^2} \\=\sqrt{11^2+1^2} \\=\sqrt{122}

2. The length between (-6,-5) & (-2,2):

\sqrt{(y_2-y_1)^2+(x_2-x_1)^2} \\=\sqrt{(2--5)^2+(-2--6)^2} \\=\sqrt{(2+5)^2+(-2+6)^2}\\ =\sqrt{7^2+4^2}\\ =\sqrt{65}

3. The length between (-5,6) & (-2,2):

\sqrt{(y_2-y_1)^2+(x_2-x_1)^2} \\=\sqrt{(2-6)^2+(-2--5})^2}\\ =\sqrt{(-4)^2+(3)^2} \\=\sqrt{25} \\=5

Thus, length of the shortest side is 5.

ansver
Answer from: ellemarshall13

a

Step-by-step explanation:

ansver
Answer from: tynyiaawrightt

The length of shortest side is 5\ units

Step-by-step explanation:

Let

A(-3,-2),B(1,6),C(5,3)

we know that

the formula to calculate the distance between two points is equal to

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

step 1

Find the distance AB

A(-3,-2),B(1,6)

substitute in the formula

AB=\sqrt{(6+2)^{2}+(1+3)^{2}}

AB=\sqrt{(8)^{2}+(4)^{2}}

AB=\sqrt{80}=8.94\ units

step 2

Find the distance BC

B(1,6),C(5,3)

substitute in the formula

BC=\sqrt{(3-6)^{2}+(5-1)^{2}}

BC=\sqrt{(-3)^{2}+(4)^{2}}

BC=\sqrt{25}=5\ units

step 3

Find the distance AC

A(-3,-2),C(5,3)

substitute in the formula

AC=\sqrt{(3+2)^{2}+(5+3)^{2}}

AC=\sqrt{(5)^{2}+(8)^{2}}

AC=\sqrt{89}=9.43\ units

Compare the length sides

The length of shortest side is 5\ units

ansver
Answer from: tresajacko4931
Given 2 point P(a, b) and Q(c, d), the distance |PQ|

is given by |PQ|= \sqrt{ (a-c)^{2} + (b-d)^{2}}

i) the distance between points (-2, 5) and (-2, -7) is:

d_1= \sqrt{ (-2-(-2))^{2} + (5-(-7))^{2}}=\sqrt{ (0)^{2} + (12)^{2}}=12
units

ii) the distance between points (-6, -4) and (-2, -7) is:

d_2= \sqrt{ (-6-(-2))^{2} + (-4-(-7))^{2}}=\sqrt{ (-4)^{2} + (3)^{2}}=5
units

iii) the distance between points (-2, 5) and (-6, -4) is:

d_3=\sqrt{ (-2-(-6))^{2} + (5-(-4))^{2}}=\sqrt{ (4)^{2} + (9)^{2}}=\sqrt{ 16 + 81}= \sqrt{97} 

units.



the shortest distance is the distance between points (-6, -4) and (-2, -7)
ansver
Answer from: stef76
I think the answer is 3
ansver
Answer from: hanjonez
Ithink you’re answering is d
ansver
Answer from: orcawave9408
The distance of BC is going to be the shortest side. (see attached picture)
B(1,-2) C(0,-6)

Now plug the points into the distance formula
Distance= √(x2-x1)² + (y2-y1)²
Distance= √(0-1)² + (-6+2)²
Distance= √(-1)² + (4)²
Distance= √1 + 16
Distance= √17
Distance is approx 4.12

What is the length of the shortest side of a triangle that has vertices at(0,-6,(1,-2 and (-7,0?

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