Find the perimeter of the triangle whose vertices are the following specified points in the plane. (4,−7),(−10,−4) and (3,−7)
1 answer:
Answer:
√205 +√178 +1 ≈ 28.6595
Step-by-step explanation:
The distance formula tells you the distance between points (x1, y1) and (x2, y2) is ...
d = √((x2 -x1)^2 +(y2-y1)^2)
Taken pairwise, the distances between the given points are ...
d1 = √((-10-4)^2 +(-4-(-7))^2) = √(196+9) = √205
d2 = √((3-(-10))^2 +(-7-(-4))^2) = √(169+9) = √178
d3 = √((4-3)^2 +(-7-(-7))^2) = √1 = 1
Then the sum of the distances gives the perimeter:
P = d1 +d2 +d3 = √205 +√178 +1 ≈ 28.6595
_____
In the attached figure, the distance "a" is the sum of the two long sides of the triangle.
You might be interested in
To find the area of a rectangle, we are going to use the formula: 
where
is the area
is the length
is the width
From our problem we can infer that
and 
. Lets replace those values in our formula to find the ara of our rectangle:
We can conclude that the area of our rectangle is.
where
From our problem we can infer that
We can conclude that the area of our rectangle is