Question

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?
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.