To the nearest tenth, find the perimeter of ΔABC with vertices A(2,4), B(-2,1) and C(2,1). Show your work.
This is easy, since it's a right triangle. In fact, if you plot the points, you can easily see that it is a 3-4-5 triangle.
So, the perimeter is ...
To find the perimeter of a triangle, we need to measure the sum of the lengths of all three sides. In this case, we have the coordinates of the three vertices A(2,4), B(-2,1), and C(2,1).
Step 1: Find the lengths of the three sides of the triangle.
The length of a side can be calculated using the distance formula:
d = √[(x2 - x1)² + (y2 - y1)²]
Let's calculate the lengths of the sides AB, BC, and AC.
AB:
x1 = 2, y1 = 4 (coordinates of A)
x2 = -2, y2 = 1 (coordinates of B)
AB = √[(-2 - 2)² + (1 - 4)²]
= √[(-4)² + (-3)²]
= √[16 + 9]
= √25
= 5
BC:
x1 = -2, y1 = 1 (coordinates of B)
x2 = 2, y2 = 1 (coordinates of C)
BC = √[(2 - (-2))² + (1 - 1)²]
= √[(2 + 2)² + (0)²]
= √[(4)² + (0)²]
= √[16 + 0]
= √16
= 4
AC:
x1 = 2, y1 = 4 (coordinates of A)
x2 = 2, y2 = 1 (coordinates of C)
AC = √[(2 - 2)² + (1 - 4)²]
= √[(0)² + (-3)²]
= √[0 + 9]
= √9
= 3
Step 2: Add up the lengths of all three sides.
Perimeter = AB + BC + AC
= 5 + 4 + 3
= 12
Therefore, the perimeter of ΔABC is 12 units (to the nearest tenth).