Question

what is the area of parallelogram abcd?
16 square units
20 square units
24 square units
25 square units

Explanation:

Step1: Identify base length

The base $\overline{AD}$ spans from $x=-4$ to $x=1$, so length $= |1 - (-4)| = 5$ units.

Step2: Identify height length

The height is the vertical distance from base $\overline{AD}$ to $\overline{BC}$, which is 4 units (from $y=0$ to $y=5$? No, correction: vertical height between the horizontal sides: the y-coordinate difference is $5-0=5$? No, correction: use base $\overline{AB}$ which is horizontal, length $|1 - (-4)|=5$? Wait no, correct method: area of parallelogram = base × height. Use base $\overline{DC}$: length from $x=1$ to $x=5$ is 4? No, look at coordinates: $A(-4,5)$, $B(1,5)$, $D(1,0)$, $C(5,0)$? No, $C$ is at $(5,0)$, $D$ at $(1,0)$, $A(-4,5)$, $B(1,5)$. So base $\overline{AB}$ length is $1 - (-4) = 5$ units. Height is vertical distance from $y=0$ to $y=5$? No, correction: the vertical height between the two horizontal lines $y=5$ and $y=0$ is 5? No, wait the side $BC$ goes from $(1,5)$ to $(5,0)$, so the height corresponding to base $AB$ is the vertical drop, but actually, base $AB$ is 5 units, and the horizontal distance? No, correct formula: area = base × perpendicular height. Base $AB = 5$ units, the perpendicular height is the horizontal distance? No, no: the vertical distance between the line $AB$ (y=5) and line $DC$ (y=0) is 5? But that would be 5×5=25, which is wrong. Wait no, coordinates: $A(-4,5)$, $B(1,5)$, $C(5,0)$, $D(1,0)$. So this is a parallelogram with base $AD$: length of $AD$ is $\sqrt{(-4-1)^2 + (5-0)^2} = 5\sqrt{2}$, no, better to use the shoelace formula.

Step1: List coordinates in order

$A(-4,5)$, $B(1,5)$, $C(5,0)$, $D(1,0)$, back to $A(-4,5)$

Step2: Apply shoelace formula

Area = $\frac{1}{2} |x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)|$
Substitute values:
$\frac{1}{2} |(-4)(5) + 1(0) + 5(0) + 1(5) - (5(1) + 5(5) + 0(1) + 0(-4))|$

Step3: Calculate each part

First part: $(-20) + 0 + 0 + 5 = -15$
Second part: $5 + 25 + 0 + 0 = 30$

Step4: Compute absolute difference

$\frac{1}{2} |-15 - 30| = \frac{1}{2} |-45| = \frac{1}{2}×45 = 22.5$? No, wrong, correction: coordinates are $A(-4,5)$, $B(1,5)$, $C(5,0)$, $D(1,0)$? No, $D$ is at $(1,0)$, $A(-4,5)$, so $AD$ is from $(-4,5)$ to $(1,0)$, $BC$ is from $(1,5)$ to $(5,0)$. So correct base: length of $AB$ is $1 - (-4) = 5$ units. The height is the vertical distance? No, the perpendicular height to base $AB$ is the horizontal distance? No, the slope of $AB$ is 0 (horizontal), so the perpendicular height is the vertical distance from $C$ to line $AB$: line $AB$ is $y=5$, $C$ is at $(5,0)$, so vertical distance is $5-0=5$? No, that would be 5×5=25, but that's a rectangle? No, this is a parallelogram, actually, the correct base is $AB=5$, and the horizontal component? No, use the formula for area of parallelogram with vectors: vector $\overrightarrow{AB}=(5,0)$, vector $\overrightarrow{AD}=(5,-5)$. Area is $|5×(-5) - 0×5| = |-25|=25$? No, that's wrong, because the figure is a parallelogram with base 5 and height 4? Wait, no, count the grid squares: the parallelogram can be seen as a rectangle minus two triangles. The rectangle from $x=-4$ to $x=5$, $y=0$ to $y=5$ has area $9×5=45$. Subtract the two triangles on the sides: each triangle has base 4, height 5, area $\frac{1}{2}×4×5=10$, two triangles total 20. So $45-20=25$? No, no, the left triangle is from $x=-4$ to $x=1$, $y=0$ to $y=5$, area $\frac{1}{2}×5×5=12.5$, right triangle from $x=1$ to $x=5$, $y=0$ to $y=5$, area $\frac{1}{2}×4×5=10$. No, that's not right. Wait, correct coordinates: $A(-4,5)$, $B(1,5)$, $C(5,0)$, $D(1,0)$. So this is a parallelogram with base $AB=5$, and the height is the vertical distance? No, the perpendicular height to $AB$ is 5, but that would make area 25. But wait, the side $BC$ has length $\sqrt{(5-1)^2 + (0-5)^2}=\sqrt{16+25}=\sqrt{41}$. The area can also be calculated as base $DC$ (length 4) times height: the height corresponding to base $DC$ is $\sqrt{(-4-1)^2 + (5-0)^2}$? No, no, the correct height for base $DC$ (length 4) is 5, so 4×5=20? Wait, I made a mistake in coordinates. $D$ is at $(1,0)$, $C$ at $(5,0)$, so $DC$ length is 4. The height is the vertical distance from $A$ to line $DC$ (which is $y=0$), so height is 5? No, 4×5=20. Ah, here's the mistake: $A$ is at $(-4,5)$, so the horizontal distance from $A$ to $D$ is 5, vertical distance 5. The parallelogram has base $DC=4$, and the slant side $AD$ has horizontal component 5, so the area is base $AB$ (5) times the vertical height? No, no, use the shoelace formula correctly:
$x_1=-4, y_1=5; x_2=1, y_2=5; x_3=5, y_3=0; x_4=1, y_4=0$
Area = $\frac{1}{2} |(-4)(5) + 1(0) + 5(0) + 1(5) - (5(1) + 5(5) + 0(1) + 0(-4))|$
= $\frac{1}{2} |-20 + 0 + 0 + 5 - (5 + 25 + 0 + 0)|$
= $\frac{1}{2} |-15 - 30|$
= $\frac{1}{2} |-45| = 22.5$? That can't be, the grid has integer units. Oh! I misread the coordinates: $C$ is at $(5,0)$? No, $C$ is at $(5,0)$, $D$ is at $(1,0)$, $B$ is at $(1,5)$, $A$ is at $(-3,5)$? No, the grid: $A$ is at $(-4,5)$, $B$ at $(1,5)$, $D$ at $(0,0)$? No, the x-axis: $D$ is at $(1,0)$, $C$ at $(5,0)$, $A$ at $(-3,5)$? Wait, the horizontal distance between $A$ and $B$ is 4 units? From $x=-3$ to $x=1$ is 4, then height is 5, area 20. Yes, that makes sense. I misread $A$ as $(-4,5)$ but it's $(-3,5)$. So correction:

Step1: Correct coordinates

$A(-3,5)$, $B(1,5)$, $D(1,0)$, $C(5,0)$

Step2: Calculate base length

Base $\overline{AB} = 1 - (-3) = 4$ units

Step3: Calculate height

Height is vertical distance between $y=5$ and $y=0$, which is 5 units

Step4: Compute area

Area = base × height = $4×5 = 20$ square units

Answer:

20 square units (Option B)