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find the area of the polygon: 24.6 square units 25.8 square units 26.3 …

Question

find the area of the polygon:

24.6 square units
25.8 square units
26.3 square units
27.5 square units

Explanation:

Identify the coordinates of the vertices

Using the Coordinate Geometry Area knowledge point
The vertices of the polygon \(HIKJL\) are:

  • \(H(0, 0)\)
  • \(I(-5, -5)\)
  • \(J(-8, -1)\)
  • \(K(-8, 4)\)
  • \(L(-5, 1)\)

Calculate the area using the Shoelace Formula

Using the Coordinate Geometry Area knowledge point
List the coordinates in counterclockwise order starting from \(H(0,0)\):

$$ (x_1, y_1) = (0, 0) $$
$$ (x_2, y_2) = (-5, -5) $$
$$ (x_3, y_3) = (-8, -1) $$
$$ (x_4, y_4) = (-8, 4) $$
$$ (x_5, y_5) = (-5, 1) $$

Apply the Shoelace Formula:

$$ \text{Area} = \frac{1}{2} |(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_5 + x_5y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_5 + y_5x_1)| $$

Calculate the diagonal products:

$$ x_1y_2 + x_2y_3 + x_3y_4 + x_4y_5 + x_5y_1 = (0)(-5) + (-5)(-1) + (-8)(4) + (-8)(1) + (-5)(0) $$
$$ = 0 + 5 - 32 - 8 + 0 = -35 $$

Calculate the reverse diagonal products:

$$ y_1x_2 + y_2x_3 + y_3x_4 + y_4x_5 + y_5x_1 = (0)(-5) + (-5)(-8) + (-1)(-8) + (4)(-5) + (1)(0) $$
$$ = 0 + 40 + 8 - 20 + 0 = 28 $$

Compute the absolute difference and divide by 2:

$$ \text{Area} = \frac{1}{2} |-35 - 28| = \frac{1}{2} |-63| = 31.5 $$

Alternative calculation by decomposing the polygon

Using the Coordinate Geometry Area knowledge point
Decompose the polygon into simpler shapes using a bounding box from \(x = -8\) to \(x = 0\) and \(y = -5\) to \(y = 4\).
The area of the bounding box is:

$$ \text{Area}_{\text{box}} = 8 \times 9 = 72 $$

Subtract the areas of the outer right triangles and shapes:

  1. Triangle below \(HI\): vertices \((0,0)\), \((-5,-5)\), \((0,-5)\).
$$ \text{Area}_1 = \frac{1}{2} \times 5 \times 5 = 12.5 $$
  1. Trapezoid below \(IJ\): vertices \((-5,-5)\), \((-8,-1)\), \((-8,-5)\), \((-5,-5)\).
$$ \text{Area}_2 = \frac{1}{2} \times (1 + 5) \times 3 = 9 $$
  1. Triangle above \(HKL\):

We can split the region above the polygon into:

  • Triangle above \(HL\): vertices \((0,0)\), \((-5,1)\), \((-5,0)\) has area:
$$ \text{Area}_{3a} = \frac{1}{2} \times 5 \times 1 = 2.5 $$
  • Trapezoid above \(LK\): vertices \((-5,1)\), \((-8,4)\), \((-8,1)\), \((-5,1)\) has area:
$$ \text{Area}_{3b} = \frac{1}{2} \times (3 + 0) \times 3 = 4.5 \text{ (using horizontal/vertical lines)} $$

Let's calculate the exact area of the region above the polygon bounded by \(y=4\) and \(x=0\):

  • Triangle with vertices \((0,0)\), \((0,4)\), \((-8,4)\) has area:
$$ \text{Area}_{\text{top\_right}} = \frac{1}{2} \times 8 \times 4 = 16 $$
  • Subtract the area of the region between the path \(H-L-K\) and the top-right boundary:

The polygon's upper boundary is \(H(0,0) \to L(-5,1) \to K(-8,4)\).
The area under this boundary (above the x-axis) is:

  • Under \(HL\): triangle with base 5, height 1 \(\implies 2.5\)
  • Under \(LK\): trapezoid with bases 1 and 4, width 3 \(\implies \frac{1+4}{2} \times 3 = 7.5\)

Total area under the upper boundary above \(y=0\) is \(2.5 + 7.5 = 10\).
Therefore, the area above the upper boundary inside the \([-8,0] \times [0,4]\) rectangle (which has area \(8 \times 4 = 32\)) is:
\[
\text{Area}_{\text{a…

Answer:

  • 24.6 square units
  • 25.8 square units
  • 26.3 square units
  • 27.5 square units (Correct answer)