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

points (3,7) (-3,5) (2,3)

Answer 1

To find the area of a triangle given its vertices, you 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 \(A\) can be calculated using the following formula:

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

For the points \((3, 7)\), \((-3, 5)\), and \((2, 3)\), we can assign:

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

Substituting these coordinates into the formula, we get:

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

Calculating each part:

1. \(3(5-3) = 3 \cdot 2 = 6\)
2. \(-3(3-7) = -3 \cdot (-4) = 12\)
3. \(2(7-5) = 2 \cdot 2 = 4\)

Now plug these back into the area formula:

\[ A = \frac{1}{2} \left| 6 + 12 + 4 \right| \]

Calculate \(6 + 12 + 4\):

\[ 6 + 12 + 4 = 22 \]

Thus,

\[ A = \frac{1}{2} \left| 22 \right| = \frac{22}{2} = 11 \]

Therefore, the area of the triangle is \(11\) square units.