Question

what is the area of trapezoid defg with coordinates d (2, 3), e (4, 3), f (6, 1), and g (2, 1)? (4 points) 12 square units 8 square units 6 square units 3 square units

Explanation:

Step1: Identify the bases and height

First, we need to find the lengths of the two parallel sides (bases) and the height of the trapezoid.
- The coordinates of \( D(2, 3) \), \( E(4, 3) \), \( F(6, 1) \), and \( G(2, 1) \).
- The side \( DE \) is horizontal (since \( y \)-coordinates are the same, \( 3 \)). The length of \( DE \) is \( |4 - 2| = 2 \) (using the distance formula for horizontal lines \( \text{distance} = |x_2 - x_1| \)).
- The side \( FG \): Wait, actually, \( DG \) is vertical? Wait, no. Wait, \( D(2,3) \) and \( G(2,1) \) have the same \( x \)-coordinate, so \( DG \) is vertical. The length of \( DG \) is \( |3 - 1| = 2 \). Wait, no, maybe I made a mistake. Wait, the two parallel sides of the trapezoid: let's check the slopes.
- Slope of \( DE \): \( \frac{3 - 3}{4 - 2} = 0 \) (horizontal line).
- Slope of \( FG \): Wait, \( F(6,1) \) and \( G(2,1) \): \( \frac{1 - 1}{2 - 6} = 0 \)? Wait, no, \( G(2,1) \) and \( F(6,1) \): \( x \) from 2 to 6, \( y \) is 1. So \( FG \) is also horizontal? Wait, no, \( D(2,3) \), \( E(4,3) \), \( G(2,1) \), \( F(6,1) \). Wait, actually, \( DE \) is from (2,3) to (4,3), length 2. \( FG \) is from (6,1) to (2,1)? Wait, no, \( G(2,1) \) and \( F(6,1) \): length is \( |6 - 2| = 4 \)? Wait, no, \( G(2,1) \) and \( F(6,1) \): \( x \) difference is \( 6 - 2 = 4 \), so length 4. Wait, maybe the two parallel sides are \( DE \) (length 2) and \( FG \) (length 4)? Wait, no, maybe \( DE \) and \( FG \) are not parallel. Wait, maybe \( DE \) and \( FG \) are not the parallel sides. Wait, let's check \( DG \) and \( EF \). Wait, \( DG \) is from (2,3) to (2,1), vertical line (slope undefined). \( EF \) is from (4,3) to (6,1). Slope of \( EF \): \( \frac{1 - 3}{6 - 4} = \frac{-2}{2} = -1 \). Hmm, maybe I messed up the sides. Wait, maybe the trapezoid has \( DE \) and \( FG \) as the two parallel sides? Wait, no, \( DE \) is horizontal (slope 0), \( FG \) is from (6,1) to (2,1): wait, \( G(2,1) \) and \( F(6,1) \): that's a horizontal line (slope 0). Oh! Wait, \( G(2,1) \) and \( F(6,1) \): \( y \)-coordinate is 1, so that's a horizontal line. So \( DE \) is from (2,3) to (4,3) (length 2), \( FG \) is from (2,1) to (6,1) (length 4). Then the height is the vertical distance between these two horizontal lines, which is \( |3 - 1| = 2 \). Wait, no, the vertical distance between \( y = 3 \) and \( y = 1 \) is \( 3 - 1 = 2 \).

Wait, the formula for the area of a trapezoid is \( A = \frac{1}{2}(b_1 + b_2)h \), where \( b_1 \) and \( b_2 \) are the lengths of the two parallel sides, and \( h \) is the height (the perpendicular distance between them).

So here, \( b_1 = DE = 2 \) (from \( x=2 \) to \( x=4 \) at \( y=3 \)), \( b_2 = FG = 4 \) (from \( x=2 \) to \( x=6 \) at \( y=1 \)), and the height \( h \) is the vertical distance between \( y=3 \) and \( y=1 \), which is \( 3 - 1 = 2 \). Wait, but let's confirm the sides. Alternatively, maybe the two parallel sides are \( DE \) (length 2) and \( DG \)? No, \( DG \) is vertical. Wait, no, the correct parallel sides: let's plot the points. \( D(2,3) \), \( E(4,3) \), \( F(6,1) \), \( G(2,1) \). So connecting \( D \) to \( E \) to \( F \) to \( G \) to \( D \). So \( DE \) is from (2,3) to (4,3), \( EF \) from (4,3) to (6,1), \( FG \) from (6,1) to (2,1), \( GD \) from (2,1) to (2,3). So \( DE \) is horizontal (length 2), \( FG \) is horizontal (length 4), and the height is the vertical distance between \( y=3 \) and \( y=1 \), which is 2. Wait, but then the area would be \( \frac{1}{2}(2 + 4) \times 2 = \frac{1}{2}(6) \times 2 = 6 \). Wait, but let's check again. Wait, maybe the height is the horizontal distance? No, no, the height is the perpendicular distance between the two parallel sides. Since the two parallel sides are horizontal, the perpendicular distance is vertical, so the difference in \( y \)-coordinates.

Wait, another way: the trapezoid can be considered as having bases \( DE = 2 \) and \( FG = 4 \), and height \( h = 2 \) (the vertical distance between \( y=3 \) and \( y=1 \)). Then area \( A = \frac{1}{2}(b_1 + b_2)h = \frac{1}{2}(2 + 4) \times 2 = 6 \). Wait, but let's check with coordinates using the shoelace formula. Let's apply the shoelace formula to confirm.

Shoelace formula: for coordinates \( (x_1,y_1), (x_2,y_2), (x_3,y_3), (x_4,y_4) \), area is \( \frac{1}{2}|x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)| \).

Plugging in \( D(2,3) \), \( E(4,3) \), \( F(6,1) \), \( G(2,1) \):

\( x_1 = 2, y_1 = 3 \)

\( x_2 = 4, y_2 = 3 \)

\( x_3 = 6, y_3 = 1 \)

\( x_4 = 2, y_4 = 1 \)

Compute \( x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 \):

\( 2 \times 3 + 4 \times 1 + 6 \times 1 + 2 \times 3 = 6 + 4 + 6 + 6 = 22 \)

Compute \( y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1 \):

\( 3 \times 4 + 3 \times 6 + 1 \times 2 + 1 \times 2 = 12 + 18 + 2 + 2 = 34 \)

Wait, that can't be right. Wait, no, I think I messed up the order of the points. The shoelace formula requires the points to be ordered either clockwise or counter-clockwise. Let's order them correctly: \( D(2,3) \), \( E(4,3) \), \( F(6,1) \), \( G(2,1) \), back to \( D(2,3) \). Let's re-calculate:

\( x_1y_2 = 2 \times 3 = 6 \)

\( x_2y_3 = 4 \times 1 = 4 \)

\( x_3y_4 = 6 \times 1 = 6 \)

\( x_4y_1 = 2 \times 3 = 6 \)

Sum: \( 6 + 4 + 6 + 6 = 22 \)

\( y_1x_2 = 3 \times 4 = 12 \)

\( y_2x_3 = 3 \times 6 = 18 \)

\( y_3x_4 = 1 \times 2 = 2 \)

\( y_4x_1 = 1 \times 2 = 2 \)

Sum: \( 12 + 18 + 2 + 2 = 34 \)

Then area is \( \frac{1}{2}|22 - 34| = \frac{1}{2}| -12 | = 6 \). So that confirms the area is 6 square units.

Step2: Apply the trapezoid area formula

The formula for the area of a trapezoid is \( A = \frac{1}{2}(b_1 + b_2)h \), where \( b_1 \) and \( b_2 \) are the lengths of the two parallel sides, and \( h \) is the height (perpendicular distance between them).

From Step 1:
- \( b_1 = \) length of \( DE = 2 \)
- \( b_2 = \) length of \( FG = 4 \)
- \( h = \) vertical distance between \( DE \) and \( FG = 2 \) (since \( DE \) is at \( y = 3 \) and \( FG \) is at \( y = 1 \), the distance is \( 3 - 1 = 2 \))

Plugging into the formula:
\( A = \frac{1}{2}(2 + 4) \times 2 \)
\( A = \frac{1}{2}(6) \times 2 \)
\( A = 3 \times 2 \)
\( A = 6 \)

Answer:

6 square units (Option: 6 square units)