Question

what is the area of this figure? 2 mi 9 mi 14 mi 2 mi 3 mi 2 mi 2 mi 6 mi square miles

Explanation:

Step1: Divide the figure into three rectangles

We can split the L - shaped figure into three rectangles. Let's define the three rectangles as follows:
- Rectangle 1: Width = 2 mi, Height = 14 mi (the left - most vertical rectangle).
- Rectangle 2: Width = 2 mi, Height = 3 + 2=5 mi (the middle rectangle, its height is the sum of 3 mi and 2 mi).
- Rectangle 3: Width = 2 mi, Height = 2 mi (the right - most bottom rectangle). Wait, actually, a better way is to calculate the area by considering the total length and subtracting the missing parts or by dividing into three rectangles with widths 2, 2, 2 (since 2 + 2+2 = 6) and appropriate heights.
Wait, another approach: The figure can be divided into three rectangles:
1. The top - left rectangle: width = 2 mi, height = 9 mi.
2. The middle rectangle: width = 2 mi, height = 9 + 3=12 mi? No, let's do it correctly. Let's use the method of adding the areas of three rectangles:
First rectangle: width = 2 mi, height = 14 mi (the left vertical part). Area of first rectangle $A_1=2\times14 = 28$ square miles.
Second rectangle: width = 2 mi (since 6 - 2 - 2=2? No, 6 mi is the total width. Wait, 2 (first) + 2 (second)+2 (third)=6. So the second rectangle has width = 2 mi, and height = 3 + 2 = 5 mi (because from the bottom, the height of the middle part is 3 + 2, and the top part is 9, but maybe a better way is:
The total height is 14 mi. The bottom part has a height of 2 mi (the lowest rectangle), the middle part has a height of 3 mi, and the top part has a height of 9 mi.
So three rectangles:
1. First rectangle: width = 2 mi, height = 9 mi. Area $A_1 = 2\times9=18$
2. Second rectangle: width = 2 mi, height = 9 + 3=12 mi? No, that's wrong. Wait, let's look at the horizontal lengths. The total width is 6 mi. The left - most rectangle has width 2 mi, the middle rectangle (between the left and right) has width 2 mi, and the right - most rectangle has width 2 mi (2+2 + 2=6).
The heights:
- The left - most rectangle (width 2 mi) has height 14 mi.
- The middle rectangle (width 2 mi) has height 14 - 9=5? No, the figure is L - shaped. Let's use the formula for the area of a composite figure by adding the areas of three rectangles:
First rectangle: width = 2 mi, height = 14 mi. Area $A_1=2\times14 = 28$
Second rectangle: width = 2 mi, height = 3+2 = 5 mi (because the middle part is between the left rectangle and the right rectangle, and its height is 3 + 2, since the bottom rectangle has height 2, the middle has height 3, and the top has height 9). Wait, no, let's calculate the area by considering the three rectangles with widths 2, 2, 2 and heights:
- First rectangle (left): width = 2, height = 14. Area = 2*14 = 28
- Second rectangle (middle): width = 2, height = 3 + 2=5. Area = 2*5 = 10
- Third rectangle (right): width = 2, height = 2. Area = 2*2 = 4
Wait, but that's not correct. Wait, another way: The area of the figure can be calculated as the area of the large rectangle (6 mi width and 14 mi height) minus the area of the two missing rectangles. But the missing rectangles: The first missing rectangle has width 6 - 2=4 mi? No, let's do it by dividing into three rectangles:
1. Bottom rectangle: width = 6 mi, height = 2 mi. Area $A_1=6\times2 = 12$
2. Middle rectangle: width = 6 - 2=4 mi? No, 6 - 2=4? Wait, no. Wait, the middle part (above the bottom rectangle) has a width of 6 - 2=4? No, the figure has a step - like structure. Let's look at the horizontal and vertical dimensions:
The total height is 14 mi. The bottom rectangle has height 2 mi and width 6 mi. Area $A_1 = 6\times2=12$
The middle rectangle (above the bottom rectangle) has height 3 mi and width 6 - 2=4 mi? No, 6 - 2=4? Wait, the right - most part of the middle rectangle is 2 mi less than 6 mi? No, the figure has a width of 6 mi. The bottom rectangle is 6 mi wide and 2 mi tall. The middle rectangle (above the bottom rectangle) is 6 - 2=4 mi wide? No, the middle rectangle (the part with height 3 mi) has width 6 - 2=4? No, the left - most part of the middle rectangle is 2 mi wide? Wait, I think I made a mistake. Let's use the correct division:
The figure can be divided into three rectangles:
1. Rectangle 1: width = 2 mi, height = 14 mi (left - most vertical rectangle). Area $A_1=2\times14 = 28$
2. Rectangle 2: width = 2 mi, height = 3+2 = 5 mi (middle vertical rectangle, between the left and right). Area $A_2=2\times5 = 10$
3. Rectangle 3: width = 2 mi, height = 2 mi (right - most vertical rectangle). Area $A_3=2\times2 = 4$
Wait, but 2+2 + 2=6 (the total width). And 14 mi is the total height. But let's check the heights: The left - most rectangle has height 14 mi, the middle rectangle has height 3 + 2=5 mi (because the bottom of the middle rectangle is at the same level as the bottom of the right - most rectangle's top), and the right - most rectangle has height 2 mi. Wait, no, the total height of the left - most rectangle is 14 mi, which is 9+3 + 2. So:
- The top rectangle (height 9 mi, width 2 mi): Area $A_1=2\times9 = 18$
- The middle rectangle (height 3 mi, width 2 + 2=4 mi? No, 2+2 = 4? Wait, 2 (left) + 2 (middle)=4. Area $A_2=4\times3 = 12$
- The bottom rectangle (height 2 mi, width 6 mi): Area $A_3=6\times2 = 12$
Now, sum these areas: $18 + 12+12=42$? No, that's not right. Wait, let's use the correct method. Let's calculate the area by adding the areas of three rectangles:
1. The left - most rectangle: width = 2 mi, height = 14 mi. Area = 2*14 = 28
2. The middle rectangle: width = 2 mi, height = 3 + 2=5 mi (because the middle part is between the left and right, and its height is the sum of the 3 mi and 2 mi sections). Area = 2*5 = 10
3. The right - most rectangle: width = 2 mi, height = 2 mi. Area = 2*2 = 4
Now, sum them up: 28+10 + 4=42? Wait, no, that can't be. Wait, let's check the total width: 2+2 + 2=6, which matches the given width of 6 mi. The total height: 14 mi. But let's check with another method. Let's calculate the area as the area of the large rectangle (6*14) minus the area of the two missing rectangles. The large rectangle area is 6*14 = 84. The first missing rectangle: width = 6 - 2=4 mi, height = 9 mi (the top missing part). Area = 4*9 = 36. The second missing rectangle: width = 6 - 2 - 2=2 mi, height = 3 mi (the middle missing part). Area = 2*3 = 6. Then the area of the figure is 84-(36 + 6)=42. Wait, no, that's not correct. Wait, I think I messed up the missing parts. Let's look at the figure again:
The figure has a height of 14 mi and width of 6 mi. The top part has a height of 9 mi, and the width of the top part is 2 mi (the left - most 2 mi). Then, below the top part, there is a section with height 3 mi, and the width of this section is 2 + 2=4 mi (because the left - most 2 mi is already used in the top part). Then, below that, there is a section with height 2 mi, and width 6 mi.
So area of top part: 2*9 = 18
Area of middle part: 4*3 = 12
Area of bottom part: 6*2 = 12
Total area: 18+12 + 12=42. Wait, but let's calculate using the three - rectangle method with correct dimensions:
First rectangle (left - top): width = 2, height = 9. Area = 2*9 = 18
Second rectangle (middle): width = 2, height = 9 + 3=12? No, that's wrong. Wait, the correct way is:
The figure can be divided into three rectangles:
1. Rectangle 1: width = 2 mi, height = 14 mi. Area = 2×14 = 28
2. Rectangle 2: width = 2 mi, height = 3 + 2 = 5 mi. Area = 2×5 = 10
3. Rectangle 3: width = 2 mi, height = 2 mi. Area = 2×2 = 4
Sum: 28+10 + 4 = 42. Wait, but let's check the vertical dimensions. The total height of the left - most rectangle is 14 mi, which is 9+3 + 2. The middle rectangle has height 3 + 2=5 mi (since it starts from the bottom of the top rectangle (height 9 mi) down to the bottom of the middle section), and the right - most rectangle has height 2 mi.
Yes, because 9 (top) + 3 (middle) + 2 (bottom)=14. And the widths: 2+2 + 2=6. So the areas are calculated correctly.
Wait, another way: Let's use the formula for the area of a composite figure by adding the areas of all the rectangles:
- The bottom rectangle: width = 6 mi, height = 2 mi. Area = 6×2 = 12
- The middle rectangle (above the bottom rectangle): width = 6 - 2=4 mi? No, 6 - 2=4? Wait, no, the middle rectangle (height 3 mi) has width 6 - 2=4 mi? No, the left - most 2 mi is part of the top rectangle. Wait, I think I was overcomplicating. Let's use the following approach:
The figure is made up of three rectangles:
1. Left rectangle: 2 mi (width) ×14 mi (height) = 28
2. Middle rectangle: 2 mi (width) ×(3 + 2) mi (height) = 2×5 = 10
3. Right rectangle: 2 mi (width) ×2 mi (height) = 4
Now, sum these areas: 28+10 + 4 = 42.

Step2: Verify the calculation

Let's check the total width: 2+2 + 2=6 mi (which matches the given width of 6 mi). The total height: 14 mi (which matches the given height). The sum of the areas of the three rectangles is 28 + 10+4 = 42 square miles.

Answer:

42