To find the perimeter and area of quadrilateral $ABCD$ with vertices $A(3, 5)$, $B(6, 5)$, $C(4, -1)$, and $D(1, -1)$, follow these steps:

Perimeter:

1. Calculate the distance between each pair of consecutive vertices using the distance formula:

   • $AB = \sqrt{(6-3)^2 + (5-5)^2} = \sqrt{3^2 + 0} = \sqrt{9} = 3$
   • $BC = \sqrt{(4-6)^2 + (-1-5)^2} = \sqrt{(-2)^2 + (-6)^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}$
   • $CD = \sqrt{(1-4)^2 + (-1+1)^2} = \sqrt{(-3)^2 + 0} = \sqrt{9} = 3$
   • $DA = \sqrt{(3-1)^2 + (5+1)^2} = \sqrt{2^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}$

2. Add the distances to find the perimeter:

   $\text{Perimeter} = AB + BC + CD + DA = 3 + 2\sqrt{10} + 3 + 2\sqrt{10} = 6 + 4\sqrt{10}$

   Approximating $\sqrt{10} \approx 3.16$:

   $\text{Perimeter} \approx 6 + 4(3.16) = 6 + 12.64 = 18.64$

Area:

Since the quadrilateral is not a standard shape like a rectangle or parallelogram, use the Shoelace Theorem (or surveyor's formula):

$\text{Area} = \frac{1}{2} \left| x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1) \right|$

Plug in the coordinates:

$= \frac{1}{2} \left| 3\cdot5 + 6\cdot(-1) + 4\cdot(-1) + 1\cdot5 - (5\cdot6 + 5\cdot4 + (-1)\cdot1 + (-1)\cdot3) \right|$

$= \frac{1}{2} \left| 15 - 6 - 4 + 5 - (30 + 20 - 1 - 3) \right|$

$= \frac{1}{2} \left| 10 - 46 \right| = \frac{1}{2} \times 36 = 18$

Final Answers:

• Perimeter: 18.64 units
• Area: 18 square units