find the area of the following shape. you must show all work to receive credit.

Answer

Explanation:

Step1: Decompose the shape

The shape can be divided into a rectangle and a triangle. The rectangle has length from ( x = -1 ) to ( x = -4 ) (length ( 3 )) and height from ( y = 0 ) to ( y = 2 ) (height ( 2 ))? Wait, no, let's check coordinates. Wait, the rectangle is from ( x=-1 ) to ( x=-4 )? No, wait the bottom horizontal line is from ( x=-4 ) to ( x=-1 ) (length ( 3 )), vertical from ( x=-1 ) up to ( y=2 ) (height ( 2 )), so rectangle area: ( length \times height = 3 \times 2 = 6 )? Wait no, wait the vertical side: from ( y=0 ) to ( y=2 ), so height is ( 2 ), length is ( 3 ) (from ( x=-4 ) to ( x=-1 ), that's ( 3 ) units). Then the triangle: the top part. The base of the triangle: from ( x=-1 ) to ( x=2 )? Wait no, the triangle has vertices at ( (0,4) ), ( (2,2) ), and ( (-1,2) )? Wait no, let's list the vertices. The blue shape: let's find all vertices. From the graph: the points are ( (-4,0) ), ( (-1,0) ), ( (-1,2) ), ( (2,2) ), ( (0,4) ), back to ( (-4,0) )? Wait, no, connecting ( (-4,0) ) to ( (-1,0) ) (horizontal), ( (-1,0) ) to ( (-1,2) ) (vertical), ( (-1,2) ) to ( (2,2) ) (horizontal), ( (2,2) ) to ( (0,4) ) (line), ( (0,4) ) to ( (-4,0) ) (line). So we can split into a rectangle and a triangle. The rectangle is ( (-4,0) ), ( (-1,0) ), ( (-1,2) ), ( (x_1,2) )? Wait, no, the rectangle is from ( x=-4 ) to ( x=-1 ), ( y=0 ) to ( y=2 ): that's a rectangle with length ( 3 ) (from -4 to -1 is 3 units) and height ( 2 ) (from 0 to 2), area ( 3 \times 2 = 6 ). Then the triangle: the top part, with vertices ( (-1,2) ), ( (2,2) ), ( (0,4) ). The base of the triangle is the distance between ( (-1,2) ) and ( (2,2) ), which is ( 3 ) units (from -1 to 2 is 3 units), and the height is the vertical distance from ( (0,4) ) to ( y=2 ), which is ( 2 ) units. So triangle area: ( \frac{1}{2} \times base \times height = \frac{1}{2} \times 3 \times 2 = 3 ). Wait, but wait, is the triangle's base correct? Wait, the three points of the triangle: ( (-1,2) ), ( (2,2) ), ( (0,4) ). The base is between ( (-1,2) ) and ( (2,2) ), length ( 3 ), height is ( 4 - 2 = 2 ), so area ( \frac{1}{2} \times 3 \times 2 = 3 ). Then total area: rectangle area + triangle area = ( 6 + 3 = 9 )? Wait, no, wait another way: use the shoelace formula. Let's list the coordinates in order: ( (-4,0) ), ( (-1,0) ), ( (-1,2) ), ( (2,2) ), ( (0,4) ), back to ( (-4,0) ). Shoelace formula: ( \frac{1}{2} |\sum_{i=1}^{n} (x_i y_{i+1} - x_{i+1} y_i)| ).

Calculating:

( x_1=-4, y_1=0 )

( x_2=-1, y_2=0 )

( x_3=-1, y_3=2 )

( x_4=2, y_4=2 )

( x_5=0, y_5=4 )

( x_6=-4, y_6=0 )

Compute sum of ( x_i y_{i+1} ):

( (-4)(0) + (-1)(2) + (-1)(2) + (2)(4) + (0)(0) = 0 - 2 - 2 + 8 + 0 = 4 )

Compute sum of ( y_i x_{i+1} ):

( (0)(-1) + (0)(-1) + (2)(2) + (2)(0) + (4)(-4) = 0 + 0 + 4 + 0 - 16 = -12 )

Subtract: ( 4 - (-12) = 16 ), take absolute value and multiply by ( \frac{1}{2} ): ( \frac{1}{2} \times 16 = 8 )? Wait, that's different. Wait, maybe I listed the coordinates wrong. Let's recheck the vertices. Let's trace the shape:

Start at ( (-4, 0) ), go to ( (-1, 0) ) (right along x-axis), then up to ( (-1, 2) ) (vertical), then right to ( (2, 2) ) (horizontal), then up to ( (0, 4) ) (line), then left down to ( (-4, 0) ) (line). So the vertices are ( A(-4,0) ), ( B(-1,0) ), ( C(-1,2) ), ( D(2,2) ), ( E(0,4) ), back to ( A(-4,0) ).

Now apply shoelace formula:

List the coordinates in order:

( A(-4,0) ), ( B(-1,0) ), ( C(-1,2) ), ( D(2,2) ), ( E(0,4) ), ( A(-4,0) )

Compute ( \sum x_i y_{i+1} ):

( (-4)(0) + (-1)(2) + (-1)(2) + (2)(4) + (0)(0) = 0 - 2 - 2 + 8 + 0 = 4 )

Compute ( \sum y_i x_{i+1} ):

( (0)(-1) + (0)(-1) + (2)(2) + (2)(0) + (4)(-4) = 0 + 0 + 4 + 0 - 16 = -12 )

Now, ( |\sum x_i y_{i+1} - \sum y_i x_{i+1}| = |4 - (-12)| = 16 ), then area is ( \frac{1}{2} \times 16 = 8 ). Wait, but maybe splitting into rectangle and triangle is better. Let's split into rectangle ( A(-4,0) ), ( B(-1,0) ), ( C(-1,2) ), ( F(-4,2) ) (a rectangle with length 3, height 2, area 6) and then the polygon ( F(-4,2) ), ( C(-1,2) ), ( D(2,2) ), ( E(0,4) ), ( A(-4,0) )? No, that's not right. Wait, maybe the top part is a triangle and a trapezoid? Wait, no, let's use another method. The shape can be seen as a rectangle plus a triangle. Wait, the rectangle is from ( x=-4 ) to ( x=-1 ), ( y=0 ) to ( y=2 ): area ( 3 \times 2 = 6 ). Then the upper part: from ( x=-1 ) to ( x=2 ), ( y=2 ) to ( y=4 ). Wait, the upper part is a triangle? Wait, the points ( (-1,2) ), ( (2,2) ), ( (0,4) ). So that's a triangle with base ( 3 ) (from -1 to 2 is 3 units) and height ( 2 ) (from 2 to 4 is 2 units). Area of triangle: ( \frac{1}{2} \times 3 \times 2 = 3 ). Then total area: ( 6 + 3 = 9 ). But shoelace gave 8. There's a mistake. Wait, let's check the coordinates again. Wait, the point ( (0,4) ) to ( (-4,0) ): let's see the slope. From ( (0,4) ) to ( (-4,0) ), slope is ( (0-4)/(-4-0) = (-4)/(-4) = 1 ), so equation ( y = x + 4 ). At ( x=-1 ), ( y = -1 + 4 = 3 ), but the point at ( x=-1 ) is ( (-1,2) ). Wait, so my vertex list was wrong. Oh! I see the mistake. The vertical line is from ( (-1,0) ) to ( (-1,2) ), then horizontal to ( (2,2) ), then the line from ( (2,2) ) to ( (0,4) ), then the line from ( (0,4) ) to ( (-4,0) ), and the line from ( (-4,0) ) to ( (-1,0) ). So the correct vertices are: ( (-4,0) ), ( (-1,0) ), ( (-1,2) ), ( (2,2) ), ( (0,4) ), and back to ( (-4,0) ). Wait, but at ( x=-1 ), the y-coordinate is 2, and the line from ( (0,4) ) to ( (-4,0) ) passes through ( x=-1 ), ( y=3 ), but the shape has a vertical line at ( x=-1 ) up to ( y=2 ), so the shape is a rectangle (from ( -4,0 ) to ( -1,0 ) to ( -1,2 ) to ( x_1,2 ) to ( -4,0 ))? No, the horizontal line from ( -1,2 ) to ( 2,2 ) is correct, then from ( 2,2 ) to ( 0,4 ), then from ( 0,4 ) to ( -4,0 ), and from ( -4,0 ) to ( -1,0 ), and from ( -1,0 ) to ( -1,2 ). So the area can be calculated as the area of the rectangle (from ( -4,0 ) to ( -1,0 ) to ( -1,2 ) to ( -4,2 ) to ( -4,0 )) plus the area of the trapezoid or triangle above the rectangle. Wait, the rectangle is ( 3 ) (length) by ( 2 ) (height), area ( 6 ). Then the area above the rectangle: the region from ( y=2 ) to ( y=4 ), from ( x=-4 ) to ( x=2 ), but bounded by the line ( y = x + 4 ) (from ( (0,4) ) to ( (-4,0) )) and the line from ( (2,2) ) to ( (0,4) ) (slope ( (4-2)/(0-2) = 2/(-2) = -1 ), equation ( y = -x + 4 )). Wait, at ( x=2 ), ( y = -2 + 4 = 2 ), correct. At ( x=-1 ), ( y = 1 + 4 = 3 ) (from ( y = x + 4 )) and ( y = 1 + 4 = 3 ) (from ( y = -x + 4 ) at ( x=-1 ) is ( y = 1 + 4 = 5 )? No, wait ( y = -x + 4 ) at ( x=-1 ) is ( y = 1 + 4 = 5 ), which is not correct. I'm getting confused. Let's use the grid method. Each square is 1x1. Let's count the squares or use the formula for composite shapes.

Alternative approach: Divide the shape into two parts: a rectangle and a triangle.

Rectangle: from ( x = -4 ) to ( x = -1 ) (length 3), ( y = 0 ) to ( y = 2 ) (height 2). Area: ( 3 \times 2 = 6 ).
Triangle: The top part. The base of the triangle is from ( x = -1 ) to ( x = 2 ) (length 3), and the height is from ( y = 2 ) to ( y = 4 ) (height 2). Wait, no, the triangle has vertices at ( (-1,2) ), ( (2,2) ), and ( (0,4) ). The base is 3 (from -1 to 2), height is 2 (from 2 to 4), so area ( \frac{1}{2} \times 3 \times 2 = 3 ). But wait, the line from ( (0,4) ) to ( (-4,0) ) passes through ( x=-1 ), ( y=3 ), but the shape has a vertical line at ( x=-1 ) up to ( y=2 ), so there's a small rectangle or a mistake? Wait, no, the correct way is to use the shoelace formula with correct coordinates. Let's list the correct vertices:

( (-4, 0) )
( (-1, 0) )
( (-1, 2) )
( (2, 2) )
( (0, 4) )
( (-4, 0) )

Now apply shoelace formula:

First, list the coordinates in order:

( (-4, 0) )
( (-1, 0) )
( (-1, 2) )
( (2, 2) )
( (0, 4) )
( (-4, 0) )

Compute ( \sum x_i y_{i+1} ):

( (-4)(0) + (-1)(2) + (-1)(2) + (2)(4) + (0)(0) = 0 - 2 - 2 + 8 + 0 = 4 )

Compute ( \sum y_i x_{i+1} ):

( (0)(-1) + (0)(-1) + (2)(2) + (2)(0) + (4)(-4) = 0 + 0 + 4 + 0 - 16 = -12 )

Now, the shoelace formula is ( \frac{1}{2} |\sum x_i y_{i+1} - \sum y_i x_{i+1}| )

So ( |4 - (-12)| = 16 ), ( \frac{1}{2} \times 16 = 8 ). Wait, but why the discrepancy? Let's check with another method. Let's use the area of the rectangle plus the area of the triangle above the rectangle. The rectangle is from ( -4,0 ) to ( -1,0 ) to ( -1,2 ) to ( -4,2 ) to ( -4,0 ): area ( 3 \times 2 = 6 ). Then the area above the rectangle: from ( x=-4 ) to ( x=2 ), ( y=2 ) to ( y=4 ), bounded by the two lines: ( y = x + 4 ) (from ( -4,0 ) to ( 0,4 )) and ( y = -x + 4 ) (from ( 2,2 ) to ( 0,4 )). This is a triangle with base from ( x=-4 ) to ( x=2 ), length ( 6 ), and height ( 2 ) (from ( y=2 ) to ( y=4 )), but wait, no, because at ( x=-1 ), the vertical line is at ( y=2 ), so actually, the area above the rectangle is a trapezoid? Wait, no, the shape above the rectangle (from ( y=2 ) to ( y=4 )) is a triangle with vertices at ( (-4,2) ), ( (2,2) ), ( (0,4) ). Wait, ( (-4,2) ) is on the line ( y = x + 4 ) (since ( 2 = -4 + 4 = 0 )? No, ( y = x + 4 ) at ( x