Question

error analysis describe the error in finding the perimeter of the rectangle.
image: a rectangle on a coordinate grid (x-axis: -2 to 4, y-axis with a vertical segment labeled 2). calculation: ( p = 2l + 2w = 2(4) + 2(3) = 14 ). the perimeter is 14 units. then: \the (square) is not calculated correctly. correct the error. the correct measure is (square) units, so the correct perimeter is (square) units.\

Explanation:

Step1: Find correct length and width

From the graph, the length (horizontal side) spans from \( x = -2 \) to \( x = 4 \), so length \( l=4 - (-2)=6 \)? Wait, no, wait. Wait, the rectangle's vertices: looking at the x - coordinates, the left vertex is at \( x=-1 \)? Wait, no, the grid: the vertical lines are at -2, -1, 0, 1, 2, 3, 4? Wait, the rectangle has left x at -1? No, the blue rectangle: the left side is at x = -1? Wait, no, the coordinates: the top - left corner is at (-1, 3)? Wait, no, the y - axis: the vertical side is from y = 1 to y = 3? Wait, no, the original mistake: the length and width were miscalculated. Let's re - examine the graph. The rectangle: the horizontal distance (length) is from x=-1 to x = 3? No, wait, the x - axis: the left end is at x=-1? No, the given calculation used l = 4 and w = 3, which is wrong. Let's find the correct length and width. From the graph, the horizontal side (length) is from x=-1 to x = 3? Wait, no, the distance between x=-1 and x = 3 is 4? No, wait, the x - coordinates of the right and left sides: the left x is -1, right x is 3? No, the grid lines: the x - axis has marks at -2, -1, 0, 1, 2, 3, 4. The rectangle's left side is at x=-1, right side at x = 3? No, the blue rectangle: the left vertex is at (-1, 3) and right at (3, 3), bottom left at (-1, 1), bottom right at (3, 1). Wait, no, the y - axis: the vertical side is from y = 1 to y = 3, so height (width) is \( 3 - 1=2 \). The horizontal side: from x=-1 to x = 3, so length is \( 3-(-1)=4 \)? No, that can't be. Wait, the original error: the length and width were misidentified. Wait, the formula for the perimeter of a rectangle is \( P = 2l+2w \), where \( l \) is length and \( w \) is width. In the graph, the horizontal length (from x=-1 to x = 3) is 4? No, wait, the x - coordinates: if the left is at x=-1 and right at x = 3, the distance is \( 3-(-1)=4 \), and the vertical width (from y = 1 to y = 3) is \( 3 - 1 = 2 \). Wait, but the original calculation used l = 4 and w = 3, which is wrong. Wait, maybe the x - coordinates are from x=-2 to x = 4? No, the blue rectangle: the left vertex is at (-1, 3), right at (3, 3), bottom left at (-1, 1), bottom right at (3, 1). So length \( l=3-(-1)=4 \), width \( w = 3 - 1=2 \). Wait, the original calculation used w = 3, which is wrong. So the error is in calculating the width (or length). Let's do it correctly.

Step2: Correct the perimeter calculation

First, find the correct length and width. From the graph, the horizontal side (length) is \( 4 - (-1)=4 \)? No, wait, the x - coordinates of the two horizontal vertices: let's take the bottom side, from x=-1 to x = 3, so length \( l = 3-(-1)=4 \)? Wait, no, the distance between -1 and 3 is 4? Yes, \( 3-(-1)=4 \). The vertical side (width) is from y = 1 to y = 3, so \( w=3 - 1 = 2 \). Wait, but the original calculation used w = 3. So the correct length and width: let's re - check the graph. The rectangle is drawn with left at x=-1, right at x = 3 (so length \( l = 4 \)), bottom at y = 1, top at y = 3 (so width \( w = 2 \)). Wait, no, maybe the x - coordinates are from x=-2 to x = 4? No, the blue rectangle's left vertex is at (-1, 3), right at (3, 3), bottom left at (-1, 1), bottom right at (3, 1). So length \( l=4 \), width \( w = 2 \). Then the perimeter formula is \( P=2l + 2w \).

Wait, maybe I misread the graph. Let's look at the x - axis: the marks are at -2, -1, 0, 1, 2, 3, 4. The rectangle's left side is at x=-1, right side at x = 3 (so the length is \( 3-(-1)=4 \)), and the bottom side is at y = 1, top at y = 3 (so width is \( 3 - 1=2 \)). The original calculation used \( l = 4 \) and \( w = 3 \), which is wrong because the width is 2, not 3.

So the error is in the calculation of the width (or the side lengths). Let's correct it.

Correct length \( l \): from x=-1 to x = 3, \( l=3-(-1)=4 \)? Wait, no, if the left is at x=-2 and right at x = 4, then \( l=4-(-2)=6 \), and the vertical side from y = 1 to y = 3, \( w = 2 \). Wait, maybe the graph has the rectangle with left x=-2, right x = 4 (so length \( l=4-(-2)=6 \)) and bottom y = 1, top y = 3 (width \( w = 2 \)). Let's check the original calculation: \( 2(4)+2(3)=14 \). So they took length as 4 and width as 3. But if the left is at x=-2 and right at x = 4, length is \( 4-(-2)=6 \), and top at y = 3, bottom at y = 1, width is \( 3 - 1=2 \). Then the correct perimeter is \( 2\times6+2\times2=12 + 4=16 \). Wait, that makes more sense. So the error is in calculating the length (or width). The length should be \( 4-(-2)=6 \) (distance between x=-2 and x = 4) and width is \( 3 - 1=2 \) (distance between y = 1 and y = 3). So the original calculation used length = 4 (wrong, should be 6) and width = 3 (wrong, should be 2).

So step by step:

1. Find correct length and width:
- Length \( l \): The horizontal distance between \( x=-2 \) and \( x = 4 \) is \( l=4-(-2)=6 \).
- Width \( w \): The vertical distance between \( y = 1 \) and \( y = 3 \) is \( w=3 - 1=2 \).
2. Calculate the correct perimeter:
- Using the formula \( P = 2l+2w \), substitute \( l = 6 \) and \( w = 2 \).
- \( P=2\times6+2\times2=12 + 4 = 16 \).

Answer:

The error is that the length and width of the rectangle were calculated incorrectly. The correct length is 6 units and the correct width is 2 units. The correct perimeter is \( 2\times6 + 2\times2=16 \) units. So the correct measure for the perimeter is 16 units.