Math Problem Statement

Use the diagram to find the perimeter and the area of quadrilateral $ABCD$ . Round your answers to the nearest hundredth.

Coordinate plane shows a polygon. The vertical axis extends from negative 6 to 5. Horizontal axis extends from negative 4 to 6. Coordinates of triangle A B F. Ordered pair A negative 5 comma 4. Ordered pair B 0 comma 3. Ordered pair F negative 2 comma 1. Coordinates of rectangle B C E F. Ordered pair B 0 comma 3. Ordered pair C 4 comma negative 1. Ordered pair E 2 comma negative 3. Ordered pair F negative 2 comma 1. Coordinates of triangle C D E. Ordered pair C 4 comma negative 1. Ordered pair D 4 comma negative 5. Ordered pair E 2 comma negative 3.

The perimeter is about units.

The area is square units.

Solution

To solve for the perimeter and the area of quadrilateral ABCDABCD, we can use the coordinates provided and apply mathematical formulas.

1. Coordinates of the vertices:

The coordinates of the quadrilateral ABCDABCD are:

  • A(5,4)A(-5, 4)
  • B(0,3)B(0, 3)
  • C(4,1)C(4, -1)
  • D(4,5)D(4, -5)

2. Perimeter of the quadrilateral

The perimeter is the sum of the lengths of the sides of the quadrilateral. We will use the distance formula to calculate the lengths of the sides.

The distance formula between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is:

d=(x2x1)2+(y2y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Side ABAB:

AB=(0(5))2+(34)2=(5)2+(1)2=25+1=265.10AB = \sqrt{(0 - (-5))^2 + (3 - 4)^2} = \sqrt{(5)^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26} \approx 5.10

Side BCBC:

BC=(40)2+(13)2=(4)2+(4)2=16+16=325.66BC = \sqrt{(4 - 0)^2 + (-1 - 3)^2} = \sqrt{(4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} \approx 5.66

Side CDCD:

CD=(44)2+(5(1))2=(0)2+(4)2=16=4CD = \sqrt{(4 - 4)^2 + (-5 - (-1))^2} = \sqrt{(0)^2 + (-4)^2} = \sqrt{16} = 4

Side DADA:

DA=(4(5))2+(54)2=(9)2+(9)2=81+81=16212.73DA = \sqrt{(4 - (-5))^2 + (-5 - 4)^2} = \sqrt{(9)^2 + (-9)^2} = \sqrt{81 + 81} = \sqrt{162} \approx 12.73

Now, we can sum the lengths of the sides to find the perimeter:

Perimeter=AB+BC+CD+DA=5.10+5.66+4+12.73=27.49 units\text{Perimeter} = AB + BC + CD + DA = 5.10 + 5.66 + 4 + 12.73 = 27.49 \text{ units}

3. Area of the quadrilateral

The area of a polygon with vertices (x1,y1),(x2,y2),,(xn,yn)(x_1, y_1), (x_2, y_2), \dots, (x_n, y_n) can be calculated using the shoelace formula:

Area=12x1y2+x2y3++xny1(y1x2+y2x3++ynx1)\text{Area} = \frac{1}{2} \left| x_1 y_2 + x_2 y_3 + \cdots + x_n y_1 - (y_1 x_2 + y_2 x_3 + \cdots + y_n x_1) \right|

Using the coordinates of A(5,4),B(0,3),C(4,1),D(4,5)A(-5, 4), B(0, 3), C(4, -1), D(4, -5):

Area=12(5)(3)+(0)(1)+(4)(5)+(4)(4)(4(0)+3(4)+(1)(4)+(5)(5))\text{Area} = \frac{1}{2} \left| (-5)(3) + (0)(-1) + (4)(-5) + (4)(4) - \left(4(0) + 3(4) + (-1)(4) + (-5)(-5)\right) \right| =1215+020+16(0+124+25)= \frac{1}{2} \left| -15 + 0 - 20 + 16 - \left(0 + 12 - 4 + 25\right) \right| =121933=12×52=26 square units= \frac{1}{2} \left| -19 - 33 \right| = \frac{1}{2} \times 52 = 26 \text{ square units}

Final Answers:

  • The perimeter of quadrilateral ABCDABCD is approximately 27.49 units.
  • The area of quadrilateral ABCDABCD is 26 square units.

Would you like more details or have any questions? Here are some related questions:

  1. How do we generalize the distance formula in 3D coordinates?
  2. What are other methods to calculate the area of irregular polygons?
  3. How do we derive the shoelace formula?
  4. Can we use vector cross products to find the area of polygons?
  5. What happens to the perimeter if we change one vertex position?

Tip: Always double-check your calculations by revisiting each step, especially in complex geometrical problems.

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Math Problem Analysis

Mathematical Concepts

Geometry
Coordinate Geometry
Perimeter
Area

Formulas

Distance formula: d = √((x2 - x1)^2 + (y2 - y1)^2)
Shoelace formula for area: Area = 1/2 * |x1y2 + x2y3 + ... + xn y1 - (y1x2 + y2x3 + ... + yn x1)|

Theorems

Coordinate Geometry Distance Theorem
Shoelace Theorem for Area

Suitable Grade Level

Grades 9-12