Question

Find the perimeter of a quadrilateral with vertices at C (−2, 1), D (2, 4), E (5, 0), and F (1, −3). Round your answer to the nearest hundredth when necessary.

Answer 1

Answer: 20

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Answer 2

Answer: 20 units

Step-by-step explanation:

The distance between two points P(a,b) and Q(c,d) is given by :-

$PQ=\sqrt{(d-b)^2+(c-a)^2}$

Given : The vertices of a quadrilateral = C (−2, 1), D (2, 4), E (5, 0), and F (1, −3).

Distance between points C (−2, 1) and D (2, 4) :-

$CD=\sqrt{(4-1)^2+(2-(-2))^2}\\\\\Rightarrow\ CD=\sqrt{(3)^2+(4)^2}\\\\\Rightarrow\ CD=\sqrt{9+16}=\sqrt{25}\\\\\Rightarrow\ CD=5\text{ units}$

Distance between points D (2, 4) and E (5, 0) :-

$DE=\sqrt{(4-0)^2+(2-5)^2}\\\\\Rightarrow\ DE=\sqrt{(4)^2+(-3)^2}\\\\\Rightarrow\ DE=\sqrt{16+9}=\sqrt{25}\\\\\Rightarrow\ DE=5\text{ units}$

Distance between points E (5, 0) and F (1, −3).:-

$EF=\sqrt{(-3-0)^2+(1-5)^2}\\\\\Rightarrow\ EF=\sqrt{(-3)^2+(-4)^2}\\\\\Rightarrow\ EF=\sqrt{9+16}=\sqrt{25}\\\\\Rightarrow\ EF=5\text{ units}$

Distance between points C (−2, 1) and F (1, −3).:-

$FC=\sqrt{(-3-1))^2+(1-(-2))^2}\\\\\Rightarrow\ FC=\sqrt{(-4)^2+(3)^2}\\\\\Rightarrow\ FC=\sqrt{16+9}=\sqrt{25}\\\\\Rightarrow\ FC=5\text{ units}$

Now, the perimeter of the quadrilateral = CD+DE+EF+FC

$=5+5+5+5=20\text{ units}$