What is the perimeter of a triangle with corners at #(3 ,3 )#, #(1 ,5 )#, and #(2 ,1 )#?

Answer 1

Perimeter is #9.1876#

As perimeter is sum of all the sides, let us find all the sides of triangle formed by #(3,3)#, #(1,5)# and #(2,1)#. This will be surely distance between pair of points, (which is given by #sqrt((x_2-x_1)^2+(y_2-y_1)^2)#. Hence the three sides are:

#a=sqrt((1-3)^2+(5-3)^2)=sqrt(4+4)=sqrt8=2.8284#

#b=sqrt((2-1)^2+(1-5)^2)=sqrt(1+16)=sqrt17=4.1231# and

#c=sqrt((2-3)^2+(1-3)^2)=sqrt(1+4)=sqrt5=2.2361#

Hence perimeter is #2.8284+4.1231+2.2361=9.1876#

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Answer 2

Claire Akin

To find the perimeter of a triangle with vertices at (3, 3), (1, 5), and (2, 1), you calculate the distance between each pair of vertices and then sum up those distances.

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ in a coordinate plane can be found using the distance formula:

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

Distance between (3, 3) and (1, 5): $d_1 = \sqrt{(1 - 3)^2 + (5 - 3)^2} = \sqrt{(-2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8}$
Distance between (1, 5) and (2, 1): $d_2 = \sqrt{(2 - 1)^2 + (1 - 5)^2} = \sqrt{(1)^2 + (-4)^2} = \sqrt{1 + 16} = \sqrt{17}$
Distance between (2, 1) and (3, 3): $d_3 = \sqrt{(3 - 2)^2 + (3 - 1)^2} = \sqrt{(1)^2 + (2)^2} = \sqrt{1 + 4} = \sqrt{5}$

Now, sum up the distances to find the perimeter: $\text{Perimeter} = d_1 + d_2 + d_3 = \sqrt{8} + \sqrt{17} + \sqrt{5}$

This is the perimeter of the triangle with the given vertices.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.