Question

triangle tuv, with vertices t(6,-9), u(7,-2), and v(2,-4), is drawn inside a rectangle, as shown below.

Explanation:

Assuming the problem is to find the area of triangle TUV using the rectangle method (subtracting areas of surrounding right triangles from the rectangle's area), here's the solution:

Step1: Find rectangle dimensions

The rectangle enclosing triangle TUV has length (horizontal) from \( x = 2 \) to \( x = 7 \), so length \( l = 7 - 2 = 5 \).
Height (vertical) from \( y = -9 \) to \( y = -2 \), so height \( h = -2 - (-9) = 7 \).
Area of rectangle: \( A_{\text{rect}} = l \times h = 5 \times 7 = 35 \).

Step2: Find areas of three right triangles

• Triangle 1 (top - left): Vertices V(2, -4), (2, -2), U(7, -2).

Base \( b_1 = 7 - 2 = 5 \), Height \( h_1 = -2 - (-4) = 2 \).
Area \( A_1 = \frac{1}{2} \times 5 \times 2 = 5 \).

• Triangle 2 (top - right): Vertices U(7, -2), (7, -9), T(6, -9).

Base \( b_2 = 7 - 6 = 1 \), Height \( h_2 = -2 - (-9) = 7 \).
Area \( A_2 = \frac{1}{2} \times 1 \times 7 = 3.5 \).

• Triangle 3 (bottom - left): Vertices V(2, -4), T(6, -9), (2, -9).

Base \( b_3 = 6 - 2 = 4 \), Height \( h_3 = -4 - (-9) = 5 \).
Area \( A_3 = \frac{1}{2} \times 4 \times 5 = 10 \).

Step3: Subtract triangle areas from rectangle

Area of \( \triangle TUV = A_{\text{rect}} - (A_1 + A_2 + A_3) \)
\( = 35 - (5 + 3.5 + 10) = 35 - 18.5 = 16.5 \) (or \( \frac{33}{2} \)).

Answer:

The area of triangle TUV is \( \boldsymbol{16.5} \) (or \( \boldsymbol{\frac{33}{2}} \)) square units.