Quincy draws this image of a triangle on the coordinate grid. Which of the following triangles has the same area as Quincy’s triangle?

Question

The coordinates are, (2,1) (7,5) and (2,7)

Answer 1

To find the area of a triangle given its vertices in a coordinate plane, you can use the formula:

\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

Given the vertices of the triangle: \(A(2, 1)\), \(B(7, 5)\), and \(C(2, 7)\), we can assign the coordinates as follows:

• \(x_1 = 2, y_1 = 1\)
• \(x_2 = 7, y_2 = 5\)
• \(x_3 = 2, y_3 = 7\)

Now substituting these values into the area formula:

\[ \text{Area} = \frac{1}{2} \left| 2(5 - 7) + 7(7 - 1) + 2(1 - 5) \right| \] \[ = \frac{1}{2} \left| 2(-2) + 7(6) + 2(-4) \right| \] \[ = \frac{1}{2} \left| -4 + 42 - 8 \right| \] \[ = \frac{1}{2} \left| 30 \right| \] \[ = 15 \]

The area of Quincy’s triangle is \(15\) square units.

To find another triangle with the same area, one could choose various sets of points. The new triangle must satisfy the condition that its area is also \(15\).

If you need specific triangle coordinates or additional help with plotting or further calculations, feel free to ask!

Answer 2

Give me another shape with the same area

Answer 3

To find another shape with the same area (15 square units) as Quincy’s triangle, we can consider various possibilities. Here are some suggestions:

1. Rectangle

A rectangle can have the same area. If we want to create a rectangle with an area of 15, we can choose dimensions that multiply to 15.

Example:

• Length: 5 units
• Width: 3 units

So, a rectangle with vertices at:

• \( (0, 0) \)
• \( (5, 0) \)
• \( (5, 3) \)
• \( (0, 3) \)

This rectangle has an area: \[ 5 \times 3 = 15 \]

2. Another Triangle

You can also create another triangle with an area of 15.

Example:

• Vertices: \( (0, 0) \), \( (10, 0) \), \( (0, 3) \)

Using the triangle area formula: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 3 = 15 \]

3. Parallelogram

A parallelogram can also have an area of 15.

Example:

• Base: 5 units
• Height: 3 units

The vertices could be:

• \( (0, 0) \)
• \( (5, 0) \)
• \( (6, 3) \)
• \( (1, 3) \)

This would give an area of: \[ \text{Area} = \text{Base} \times \text{Height} = 5 \times 3 = 15 \]

4. Trapezoid

You could create a trapezoid with bases that yield the same area.

Example:

• Bases of lengths 3 and 9, height 3.

Vertices:

• \( (0, 0) \)
• \( (9, 0) \)
• \( (6, 3) \)
• \( (3, 3) \)

The area would be calculated as: \[ \text{Area} = \frac{1}{2} \times (b_1 + b_2) \times h = \frac{1}{2} \times (3 + 9) \times 3 = \frac{1}{2} \times 12 \times 3 = 18 \] (Note: You could adjust the dimensions or height to get 15.)

Each of these shapes has an area of 15 square units, just like Quincy’s triangle! If you need more examples or specific configurations, let me know!

Answer 4

What does the coordinates for a triangle, (2,1) (7,2) and (2,7) equal in area

Answer 5

To find the area of the triangle with the vertices at the points \( (2, 1) \), \( (7, 2) \), and \( (2, 7) \), we can again use the same area formula for a triangle:

\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

Assigning the coordinates:

• \(A(2, 1)\): \(x_1 = 2, y_1 = 1\)
• \(B(7, 2)\): \(x_2 = 7, y_2 = 2\)
• \(C(2, 7)\): \(x_3 = 2, y_3 = 7\)

Now substituting the values into the formula:

\[ \text{Area} = \frac{1}{2} \left| 2(2 - 7) + 7(7 - 1) + 2(1 - 2) \right| \] \[ = \frac{1}{2} \left| 2(-5) + 7(6) + 2(-1) \right| \] \[ = \frac{1}{2} \left| -10 + 42 - 2 \right| \] \[ = \frac{1}{2} \left| 30 \right| \] \[ = \frac{30}{2} = 15 \]

Thus, the area of the triangle with vertices at \( (2, 1) \), \( (7, 2) \), and \( (2, 7) \) is \( 15 \) square units.