Amaury draws an acute triangle on the coordinate grid. He finds the area of the triangle by drawing a rectangle that circumscribes the triangle and forming right triangles. How can Amaury find the area of the triangle? What is the area of the triangle in square units?

Question

A(-3,5) B(3,7) C(2,3)

Answer 1

To find the area of triangle ABC with vertices at points A(-3, 5), B(3, 7), and C(2, 3), we can use the formula for the area of a triangle defined by three points in a Cartesian coordinate system. The formula is given by:

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

We denote the points as follows:

• \( A(x_1, y_1) = (-3, 5) \)
• \( B(x_2, y_2) = (3, 7) \)
• \( C(x_3, y_3) = (2, 3) \)

Now, we substitute the coordinates into the formula:

\[ \text{Area} = \frac{1}{2} \left| (-3)(7 - 3) + 3(3 - 5) + 2(5 - 7) \right| \]

Calculating each term separately:

1. First term: \( -3(7 - 3) = -3 \times 4 = -12 \)
2. Second term: \( 3(3 - 5) = 3 \times (-2) = -6 \)
3. Third term: \( 2(5 - 7) = 2 \times (-2) = -4 \)

Now, we can sum these values:

\[ \text{Area} = \frac{1}{2} \left| -12 - 6 - 4 \right| = \frac{1}{2} \left| -22 \right| = \frac{1}{2} \times 22 = 11 \]

Thus, the area of triangle ABC is \( \boxed{11} \) square units.

Answer 2

Are you sure that's right

Answer 3

Let’s carefully go through the calculations together to ensure correctness.

We have the vertices of the triangle \( A(-3, 5) \), \( B(3, 7) \), and \( C(2, 3) \). To find the area, we will use the formula for the area of a triangle given by three vertices:

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

With our points:

• \( A(x_1, y_1) = (-3, 5) \)
• \( B(x_2, y_2) = (3, 7) \)
• \( C(x_3, y_3) = (2, 3) \)

Substituting into the formula: \[ \text{Area} = \frac{1}{2} \left| (-3)(7 - 3) + 3(3 - 5) + 2(5 - 7) \right| \]

Calculating each component:

1. For \( A \): \[ -3(7 - 3) = -3 \times 4 = -12 \]

2. For \( B \): \[ 3(3 - 5) = 3 \times (-2) = -6 \]

3. For \( C \): \[ 2(5 - 7) = 2 \times (-2) = -4 \]

Now, we sum these results: \[ -12 - 6 - 4 = -22 \]

Thus, we take the absolute value: \[ \left| -22 \right| = 22 \]

Then we apply the final step: \[ \text{Area} = \frac{1}{2} \times 22 = 11 \]

The area of triangle ABC is indeed \( \boxed{11} \) square units. The calculations are confirmed to be correct. Thank you for your patience.