Question

Find the perimeter P of ABCD with vertices A(3,1), B(6,2), C(6,-2), and D(3,-3). Round your answer to the nearest tenth, if necessary. ​

Answer 1

Explanation:

the distance between 2 points is per Pythagoras

c² = a² + b²

with c being the Hypotenuse (the baseline opposite of the 90° angle, which is the direct distance between the points). a and b are the legs (the x and y coordinate differences).

so,

AB² = (3-6)² + (1-2)² = 9+1 = 10

AB = sqrt(10) = 3.16227766...

BC² = (6-6)² + (2 - -2)² = 0² + 4² = 16

BC = sqrt(16) = 4

CD² = (6-3)² + (-2 - -3)² = 3² + 1² = 10

CD = sqrt(10) = 3.16227766...

DA² = (3-3)² + (-3 - 1)² = 0² + 4² = 16

DA = sqrt(16) = 4

so, the perimeter is

2×4 + 2×sqrt(10) = 8 + 6.32455532... = 14.32455532... ≈

≈ 14.3