Question

step 1 draw a rectangle around triangle rst. step 2 find the area of the rectangle. step 3 find the area of the three right triangles created by the sides of the rectangle and the intersection with each vertex of the triangle. step 4 subtract the area of the right triangles from the area of the rectangle. 2 units by 4 units 3 units by 4 units 3 units by 5 units 4 units by 5 units

Explanation:

Step1: Determine the length and width of the rectangle

From the graph, the horizontal distance (length) from the left - most to the right - most point of the rectangle is \(4\) units (from \(x=-1\) to \(x = 3\)). The vertical distance (width) from the bottom - most to the top - most point of the rectangle is \(3\) units (from \(y=-1\) to \(y = 2\)). Wait, no, actually, if we consider the rectangle that encloses the triangle \(RST\), by counting the grid units:
The length of the rectangle (horizontal side) is from \(x=-1\) to \(x = 3\), so \(l=4\) units. The width of the rectangle (vertical side) is from \(y=-1\) to \(y = 2\), so \(w = 3\) units. But wait, another way:
The rectangle that encloses the triangle \(RST\) has a length of \(4\) units (from \(x=-1\) to \(x = 3\)) and a width of \(3\) units (from \(y=-1\) to \(y=2\)). Wait, no, looking at the coordinates of the vertices:
Let's use the formula for the distance between two points. But for a rectangle drawn on a grid, we can count the units.
The rectangle has a length of \(4\) units (from \(x=-1\) to \(x = 3\)) and a width of \(3\) units (from \(y=-1\) to \(y = 2\)). Wait, no, actually, if we consider the rectangle that encloses the triangle \(RST\) properly:
The \(x\) - coordinates of the extreme points for the rectangle: the left - most \(x\) value is \(x=-1\) and the right - most \(x\) value is \(x = 3\), so the length \(L=3-(-1)=4\) units. The \(y\) - coordinates of the extreme points: the bottom - most \(y\) value is \(y=-1\) and the top - most \(y\) value is \(y = 2\), so the width \(W=2-(-1)=3\) units. But wait, no, looking at the options, we made a mistake.
Let's re - examine:
The rectangle that encloses the triangle \(RST\):
The \(x\) - range: from \(x=-1\) to \(x = 3\) (length \(l = 4\) units). The \(y\) - range: from \(y=-1\) to \(y=2\) (width \(w=3\) units). But wait, no, if we consider the rectangle that is formed by the perpendiculars from the vertices of the triangle to the \(x\) and \(y\) axes.
The correct way:
The length of the rectangle (horizontal side) is \(4\) units (from \(x=-1\) to \(x = 3\)) and the width (vertical side) is \(3\) units (from \(y=-1\) to \(y = 2\)). But wait, looking at the options, we must have mis - calculated.
Wait, another approach:
The formula for the area of a rectangle \(A = l\times w\).
If we use the coordinate method:
The vertices of the rectangle (assuming it's axis - aligned) will have coordinates such that for the \(x\) values: \(\min(x_R,x_S,x_T)=-1\), \(\max(x_R,x_S,x_T)=3\), so \(l = 3-(-1)=4\). For the \(y\) values: \(\min(y_R,y_S,y_T)=-1\), \(\max(y_R,y_S,y_T)=2\), so \(w=2 - (-1)=3\)

Answer:

3 units by 4 units