Question

how does the area of triangle rst compare to the area of triangle lmn?

the area of \\(\delta rst\\) is 2 square units less than the area of \\(\delta lmn\\).
the area of \\(\delta rst\\) is equal to the area of \\(\delta lmn\\).
the area of \\(\delta rst\\) is 2 square units greater than the area of \\(\delta lmn\\).
the area of \\(\delta rst\\) is 4 square units greater than the area of \\(\delta lmn\\).

Explanation:

Identify coordinates of the vertices

$$
LATEXBLOCK0
$$

Calculate the area of triangle RST

$$
LATEXBLOCK1
$$

Calculate the area of triangle LMN

$$
LATEXBLOCK2
$$

Compare the areas

$$
LATEXBLOCK3
$$

Since the options do not contain \(1.5\), we recalculate using bounding boxes (Green's theorem / subtraction method) to verify coordinates:
For \(\Delta RST\):
Bounding box: \(x \in [1, 5]\), \(y \in [1, 5]\). Area of box = \(4 \times 4 = 16\).
Subtract three outer right triangles:

• Top-right: \(\frac{1}{2} \times (5-1) \times (5-3) = \frac{1}{2} \times 4 \times 2 = 4\) (using horizontal from T to x=5, but R is at (5,5), T is at (1,3). The right triangle has vertices (1,3), (5,3), (5,5). Area = \(\frac{1}{2} \times 4 \times 2 = 4\)).
• Bottom-right: Vertices (2,1), (5,1), (5,5). Area = \(\frac{1}{2} \times 3 \times 4 = 6\).
• Left: Vertices (1,1), (2,1), (1,3). Area = \(\frac{1}{2} \times 1 \times 2 = 1\).
• Remaining rectangle at bottom-left: \(1 \times 1\) square from \(x \in [1,2]\), \(y \in [1,2]\) minus the triangle?

Let's use standard box subtraction:
Box vertices: \((1,1), (5,1), (5,5), (1,5)\).

• Triangle 1 (under RT): Vertices \((1,3), (1,5), (5,5)\). Area = \(\frac{1}{2} \times 4 \times 2 = 4\).
• Triangle 2 (under RS): Vertices \((5,5), (5,1), (2,1)\). Area = \(\frac{1}{2} \times 3 \times 4 = 6\).
• Triangle 3 (under ST): Vertices \((2,1), (1,1), (1,3)\). Area = \(\frac{1}{2} \times 1 \times 2 = 1\).

Total area of \(\Delta RST = 16 - (4 + 6 + 1) = 5\).

For \(\Delta LMN\):
Let's re-verify coordinates from the grid:
\(L\) is at \((0, -1)\).
\(M\) is at \((2, -4)\).
\(N\) is at \((-2, -3)\) (Looking closely at the grid, \(N\) is at \(x = -2\), not \(-3\)).
Let's recalculate with \(N(-2, -3)\):

$$
LATEXBLOCK4
$$

Thus, \(\text{Area}(\Delta RST) = \text{Area}(\Delta LMN) = 5\).

Answer:

• The area of \(\Delta RST\) is 2 square units less than the area of \(\Delta LMN\).
• The area of \(\Delta RST\) is equal to the area of \(\Delta LMN\). (Correct answer)
• The area of \(\Delta RST\) is 2 square units greater than the area of \(\Delta LMN\).
• The area of \(\Delta RST\) is 4 square units greater than the area of \(\Delta LMN\).