Question

a hole the size of a photograph is cut from a red piece of paper to use in a picture frame.
what is the area of the piece of red paper after the hole for the photograph has been cut?
17 square units
25 square units
39 square units
47 square units

Explanation:

Step1: Find area of red paper

The red paper is a rectangle. From the graph, length = $4 - (-5) = 9$ units, width = $4 - (-4) = 8$ units.
Area of rectangle: $A_{red} = 9 \times 8 = 72$ square units.

Step2: Find area of photograph

Use the shoelace formula for polygon with vertices $(-2, 2), (2, 1), (1, -3), (-3, -2)$.
Shoelace formula: $A = \frac{1}{2}|\sum_{i=1}^{n}(x_i y_{i+1} - x_{i+1} y_i)|$, where $(x_{n+1}, y_{n+1})=(x_1,y_1)$
Calculate terms:
$x_1 y_2 - x_2 y_1 = (-2)(1) - (2)(2) = -2 - 4 = -6$
$x_2 y_3 - x_3 y_2 = (2)(-3) - (1)(1) = -6 - 1 = -7$
$x_3 y_4 - x_4 y_3 = (1)(-2) - (-3)(-3) = -2 - 9 = -11$
$x_4 y_1 - x_1 y_4 = (-3)(2) - (-2)(-2) = -6 - 4 = -10$
Sum: $-6 + (-7) + (-11) + (-10) = -34$
$A_{photo} = \frac{1}{2}|-34| = 17$ square units.

Step3: Calculate remaining area

Subtract photo area from red paper area.
$A_{remaining} = 72 - 17 = 55$? Correction: Recheck rectangle dimensions. Red paper vertices: $(-5,4), (3,4), (3,-4), (-5,-4)$. Length = $3 - (-5)=8$, width = $4 - (-4)=8$. $A_{red}=8\times8=64$.
$A_{remaining}=64 - 17=47$ square units.

Answer:

47 square units