Worksheet for plotting polygons on a coordinate plane with four math challenges.

Here are the step-by-step solutions for each clue on your worksheet.

Clue 1: The Rectangle

Goal: Find the other two vertices of a rectangle with an area of 18 square units.
* Given Vertices: $(-1, 6)$ and $(5, 3)$.
* Step 1: Analyze the coordinates.
Notice that the x-coordinates are different ($-1$ and $5$) and the y-coordinates are different ($6$ and $3$). This means these two points are opposite corners (diagonal) of the rectangle. The sides of the rectangle must be horizontal and vertical.
* Step 2: Calculate side lengths.
* Horizontal distance (width): From $x = -1$ to $x = 5$ is $5 - (-1) = 6$ units.
* Vertical distance (height): From $y = 3$ to $y = 6$ is $6 - 3 = 3$ units.
* Step 3: Check the Area.
$\text{Area} = \text{width} \times \text{height} = 6 \times 3 = 18$ square units. This matches the clue!
* Step 4: Find the missing vertices.
The other two corners share an x-coordinate with one point and a y-coordinate with the other.
* Vertex 3: Take the x from $(-1, 6)$ and the y from $(5, 3)$. Result: $(-1, 3)$.
* Vertex 4: Take the x from $(5, 3)$ and the y from $(-1, 6)$. Result: $(5, 6)$.

Clue 2: The Triangle Classification

Goal: Plot the triangle and classify it by its angles.
* Vertices: $(3, -2)$, $(-4, 6)$, and $(-1, -4)$.
* Step 1: Visualize or calculate slopes.
To classify by angles, we check if any lines are perpendicular (form a $90^\circ$ right angle). Perpendicular lines have slopes that multiply to equal $-1$.
* Slope between $(3, -2)$ and $(-4, 6)$: $\frac{6 - (-2)}{-4 - 3} = \frac{8}{-7}$.
* Slope between $(-4, 6)$ and $(-1, -4)$: $\frac{-4 - 6}{-1 - (-4)} = \frac{-10}{3}$.
* Slope between $(3, -2)$ and $(-1, -4)$: $\frac{-4 - (-2)}{-1 - 3} = \frac{-2}{-4} = \frac{1}{2}$.
* Step 2: Check for Right Angles.
None of these slopes are negative reciprocals of each other (e.g., $2$ and $-\frac{1}{2}$). So, there are no right angles.
* Step 3: Check for Obtuse vs. Acute.
Let's look at the side lengths using the distance formula ($d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$):
* Side A (longest?): Distance between $(-4,6)$ and $(3,-2)$ is $\sqrt{(-7)^2 + 8^2} = \sqrt{49+64} = \sqrt{113} \approx 10.6$. Square is $113$.
* Side B: Distance between $(-4,6)$ and $(-1,-4)$ is $\sqrt{3^2 + (-10)^2} = \sqrt{9+100} = \sqrt{109} \approx 10.4$. Square is $109$.
* Side C: Distance between $(3,-2)$ and $(-1,-4)$ is $\sqrt{(-4)^2 + (-2)^2} = \sqrt{16+4} = \sqrt{20} \approx 4.5$. Square is $20$.

Using the Pythagorean theorem check ($a^2 + b^2$ vs $c^2$):
$109 + 20 = 129$.
Since $129 > 113$ (the sum of the squares of the two shorter sides is greater than the square of the longest side), all angles are less than $90^\circ$.

Clue 3: The Square

Goal: Find the other three vertices of a square with perimeter 16.
* Given Vertex: $(-1, 5)$.
* Step 1: Determine side length.
$\text{Perimeter} = 4 \times \text{side}$.
$16 = 4 \times s$, so the side length is 4 units.
* Step 2: Plot the square.
Assuming the sides are parallel to the axes (standard for this level unless rotated):
Start at $(-1, 5)$. We need to move 4 units in different directions to form a square.
* Move Right 4: $x$ becomes $-1 + 4 = 3$. Point: $(3, 5)$.
* Move Down 4: $y$ becomes $5 - 4 = 1$. Point: $(-1, 1)$.
* Move Right 4 AND Down 4: Point: $(3, 1)$.

*(Note: You could also go Up or Left, but typically we fill the quadrant closest to the origin or follow a standard pattern. Any combination of $\pm 4$ works as long as they form a square. The set below is the most common solution).*

Clue 4: The Triangle Area

Goal: Find the area of the triangle.
* Vertices: $(-4, -6)$, $(-1, -6)$, and $(-1, -8)$.
* Step 1: Identify the Base and Height.
* Look at $(-4, -6)$ and $(-1, -6)$. They have the same y-coordinate ($-6$). This is a horizontal line.
Length (Base) = distance from $-4$ to $-1$ = 3 units.
* Look at $(-1, -6)$ and $(-1, -8)$. They have the same x-coordinate ($-1$). This is a vertical line.
Length (Height) = distance from $-6$ to $-8$ = 2 units.
* Since one side is horizontal and the other is vertical, they meet at a right angle. This is a right triangle.
* Step 2: Calculate Area.
$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$
$\text{Area} = \frac{1}{2} \times 3 \times 2$
$\text{Area} = 3$ square units.

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Final Answer:

Clue 1: The other two vertices are $(-1, 3)$ and $(5, 6)$.

Clue 2: The triangle is an Acute Triangle.

Clue 3: The other three vertices are $(3, 5)$, $(-1, 1)$, and $(3, 1)$.
*(Note: Other valid answers exist if the square is drawn in a different direction, such as $(-5, 5), (-1, 9), (-5, 9)$, but the ones listed above are the standard solution moving toward the positive axes).*

Clue 4: The area of the triangle is 3 square units.