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? (the triangle coordinates are; (-3, 5), (3, 7), and (2, 3))

Answer 1

To find the area of a triangle given its vertices, we can use the formula based on the coordinates of the vertices. For a triangle with vertices at points \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\), the area \(A\) is calculated using the following formula:

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

Given the coordinates of the triangle's vertices:

• \((-3, 5)\): \((x_1, y_1)\)
• \((3, 7)\): \((x_2, y_2)\)
• \((2, 3)\): \((x_3, y_3)\)

Let's substitute the values into the formula:

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

Calculating each term step by step:

1. First term: \(-3(7 - 3) = -3(4) = -12\)

2. Second term: \(3(3 - 5) = 3(-2) = -6\)

3. Third term: \(2(5 - 7) = 2(-2) = -4\)

Now combine these results:

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

Thus, the area of the triangle is:

\[ \boxed{11} \text{ square units} \]