Mr. Thompson drew a rectangle on the chalkboard with a length to width ratio of 5 to 2. He asked students to draw a rectangle wi
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Answer:
1480 in^2
Step-by-step explanation:
We can number the surfaces so we can talk about them. Starting with the very top horizontal surface, call it #1. Then the vertical surface to its right (clockwise) is #2; the lower horizontal surface you can see is #3, and the rightmost end vertical surface is #4. The bottom horizontal surface on which the figure rests is #5, and the left vertical surface you can't see is #6. Call the front vertical L-shaped surface #7, and the back vertical L-shaped surface #8.
These can all be "unfolded" into the figure shown in the attachment. (Each grid square in the attached figure is 5 inches.)
Surfaces 1–6 constitute a rectangle that is 88 inches long and 10 inches wide. The length (88 inches) is the perimeter of the L-shaped front (and back) surfaces (7 and 8). The width is the front-to-back width of the shape, marked as 10 inches at the lower right of the given figure.
The areas of the two L-shaped surfaces can be calculated several ways. One way is to recognize the L-shape as a 20 in × 24 in rectangle with a 12 in × 15 in rectangle removed from it. Another way to calculate the area is to consider it to be two trapezoids (as cut by the dashed line JV in the attached figure).
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The discussion above and the attached figure (net) give the information we need to calculate the surface area.
The area of the central pink net is ...
(10 in)×(88 in) = 880 in².
The area of one golden L-shape is ...
(20 in)×(24 in) - (12 in)×(15 in) = 480 in² -180 in² = 300 in²
Then the total surface area is that of the central (pink) rectangle and two (2) (golden) L-shapes, or ...
total area = 880 in² + 2×300 in²
total area = 1480 in²
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<em>Comment on the L as trapezoids</em>
The dimensions of the edges of the L-shapes are shown in the attachment. They are computed using the information in the given figure and by subtracting heights or lengths to find the unknown dimensions.
The upper left trapezoid has a height of 9 units and bases of 12 and 20. Its area is given by the formula
A = (1/2)(b1 +b2)h = (1/2)(12 +20)·9 = 144 . . . . in²
The lower right trapezoid has a height of 8 units and bases of 24 and 15. Its area is given using the same formula
A = (1/2)(24 +15)·8 = 156 . . . . in²
Then the total area of one L-shape is the sum of these areas, or ...
L-shape area = 144 in² + 156 in² = 300 in² . . . . . same as above
