Math Problem Statement

Find the area and perimeter of the figure shown below. The polygon has vertices A(0, 0), B(2, 4), C(6, 2), D(4, -2), E(-2, -2).

Solution

Let's solve the problem step by step. The given shape is a polygon with vertices labeled A,B,C,D,EA, B, C, D, E, and the coordinates are:

  • A(0,0)A(0, 0)
  • B(2,4)B(2, 4)
  • C(6,2)C(6, 2)
  • D(4,2)D(4, -2)
  • E(2,2)E(-2, -2)

Step 1: Finding the Area

We will use the Shoelace Theorem (or surveyor's formula) for finding the area of a polygon with known coordinates. The formula is:

Area=12x1y2+x2y3++xny1(y1x2+y2x3++ynx1)\text{Area} = \frac{1}{2} \left| x_1y_2 + x_2y_3 + \cdots + x_ny_1 - (y_1x_2 + y_2x_3 + \cdots + y_nx_1) \right|

Substituting the points A(0,0),B(2,4),C(6,2),D(4,2),E(2,2)A(0, 0), B(2, 4), C(6, 2), D(4, -2), E(-2, -2):

Area=12(0)(4)+(2)(2)+(6)(2)+(4)(2)+(2)(0)[(0)(2)+(4)(6)+(2)(4)+(2)(2)+(2)(0)]\text{Area} = \frac{1}{2} \left| (0)(4) + (2)(2) + (6)(-2) + (4)(-2) + (-2)(0) - [(0)(2) + (4)(6) + (2)(4) + (-2)(-2) + (-2)(0)] \right|

Calculating step by step:

  • First part: (0)(4)+(2)(2)+(6)(2)+(4)(2)+(2)(0)=0+4128+0=16(0)(4) + (2)(2) + (6)(-2) + (4)(-2) + (-2)(0) = 0 + 4 - 12 - 8 + 0 = -16
  • Second part: (0)(2)+(4)(6)+(2)(4)+(2)(2)+(2)(0)=0+24+8+4+0=36(0)(2) + (4)(6) + (2)(4) + (-2)(-2) + (-2)(0) = 0 + 24 + 8 + 4 + 0 = 36

Now subtract and take the absolute value:

Area=121636=12×52=26\text{Area} = \frac{1}{2} \left| -16 - 36 \right| = \frac{1}{2} \times 52 = 26

Thus, the area is 26 square units.

Step 2: Finding the Perimeter

The perimeter is the sum of the lengths of the sides. To find the length of each side, we will use the distance formula:

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

  • ABAB: Distance=(20)2+(40)2=4+16=20=25\text{Distance} = \sqrt{(2 - 0)^2 + (4 - 0)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}
  • BCBC: Distance=(62)2+(24)2=16+4=20=25\text{Distance} = \sqrt{(6 - 2)^2 + (2 - 4)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}
  • CDCD: Distance=(46)2+(22)2=4+16=20=25\text{Distance} = \sqrt{(4 - 6)^2 + (-2 - 2)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}
  • DEDE: Distance=(4(2))2+(2(2))2=36+0=6\text{Distance} = \sqrt{(4 - (-2))^2 + (-2 - (-2))^2} = \sqrt{36 + 0} = 6
  • EAEA: Distance=(0(2))2+(0(2))2=4+4=8=22\text{Distance} = \sqrt{(0 - (-2))^2 + (0 - (-2))^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}

Now, sum up all the lengths:

Perimeter=25+25+25+6+22\text{Perimeter} = 2\sqrt{5} + 2\sqrt{5} + 2\sqrt{5} + 6 + 2\sqrt{2}

This simplifies to:

Perimeter=65+6+22\text{Perimeter} = 6\sqrt{5} + 6 + 2\sqrt{2}

So, the perimeter is 65+22+66\sqrt{5} + 2\sqrt{2} + 6 units.

Final Answer:

  • Area = 26 square units
  • Perimeter = 65+22+66\sqrt{5} + 2\sqrt{2} + 6 units

Let me know if you need any clarifications or further details!

Here are 5 questions to expand on this topic:

  1. What is the Shoelace Theorem and how does it work for polygons?
  2. How would you find the perimeter of a polygon if it had more sides?
  3. Can the Shoelace Theorem be used for any type of polygon, and why?
  4. What is the difference between exact values like 5\sqrt{5} and decimal approximations?
  5. How would you approach calculating the area of a shape that is not on a grid?

Tip: When calculating the perimeter of polygons, always ensure to use the distance formula for each side, especially if the polygon is irregular or has diagonals.

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Math Problem Analysis

Mathematical Concepts

Geometry
Coordinate Geometry
Polygons

Formulas

Shoelace Theorem for area: Area = 1/2 * |x1y2 + x2y3 + ... + xny1 - (y1x2 + y2x3 + ... + ynx1)|
Distance formula for perimeter: Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)

Theorems

Shoelace Theorem

Suitable Grade Level

Grades 9-12