To calculate the total surface area of the given composite figures, we'll break down each question as follows:
Question 1:
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First Rectangular Prism (left):
- Length = 10 mm
- Height = 8 mm
- Width = ?
- Since combined length of both prisms = 16 mm, Width (right's width) = 6 mm.
- So, Width = 6 mm.
Surface Area = 2(lw + lh + wh) \( = 2(10 \times 6 + 10 \times 8 + 6 \times 8) \) \( = 2(60 + 80 + 48) = 2(188) = 376 \text{ mm}^2 \)
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Second Rectangular Prism (right):
- Length = 6 mm
- Width = 6 mm
- Height = 6 mm.
Surface Area = 2(lw + lh + wh) \( = 2(6 \times 6 + 6 \times 6 + 6 \times 6) \) \( = 2(36 + 36 + 36) = 2(108) = 216 \text{ mm}^2 \)
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Combine and adjust for shared faces:
- Shared face area between two prisms = lw = 10 mm (side length) × 6 mm (side width) -> 60 mm^2.
Total Surface Area = Total from both prisms - Shared area \( = 376 + 216 - 60 = 532 \text{ mm}^2 \)
Answer: 532 mm²
Question 2:
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Rectangular Prism:
- Length = 32 m
- Width = 10 m
- Height = 8 m.
Surface Area = 2(lw + lh + wh) \( = 2(32 \times 10 + 32 \times 8 + 10 \times 8) \) \( = 2(320 + 256 + 80) = 2(656) = 1312 \text{ m}^2 \)
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Triangular Prism:
- Base Area of triangular face = \( \frac{1}{2} \times \text{base} \times \text{height} \) -> \( \frac{1}{2} \times 10 \times 12 = 60 \text{ m}^2 \)
- Triangular prism height = 32 m
Surface Area = [2 x base area (triangular) + perimeter x height] \( = 2 \times 60 + (10 + 20 + 20) \times 32 \) \( = 120 + 50 \times 32 \) \( = 120 + 1600 = 1720 \text{ m}^2 \)
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Total Surface Area = Rectangular Prism Area + Triangular Prism Area \( = 1312 + 1720 \) \( = 3032 \text{ m}^2 \)
Answer: 3032 m²
Question 3:
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Total Surface Area of the larger rectangular prism:
- Length = 15 in
- Width = 6 in
- Height = 12 in
Surface Area = 2(lw + lh + wh) \( = 2(15 \times 6 + 15 \times 12 + 6 \times 12) \) \( = 2(90 + 180 + 72) = 2(342) = 684 \text{ in}^2 \)
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Surface Area of the removed section:
- Length = 7 in
- Width = 6 in
- Height = 6 in
Surface Area = 2(lw + lh + wh) \( = 2(7 \times 6 + 7 \times 6 + 6 \times 6) \) \( = 2(42 + 42 + 36) = 2(120) = 240 \text{ in}^2 \)
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Adjust for visibility:
- The top and bottom faces of the removed prism are no longer visible.
- Therefore, the total exposed surfaces will be total large surface minus two faces of the smaller removed area: \( = 684 - (2 \times 42) = 684 - 84 = 600 \text{ in}^2 + 240 \text{ in}^2\)
Total Surface Area = 600 + 240 = 840.
Answer: 840 in²
Question 4:
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Dimensions of L-shaped platform:
- Overall length = 16 in
- Overall width = 10 in
- Cut-out portion’s dimensions:
- Length of removed portion = 12 in
- Width = 4 in (overlap)
- Remaining part's length = 4 in
- Remaining width = 6 in
Two rectangles to calculate:
-
Full rectangle Surface Area: \( A_{\text{full}} = length \times width = 16 \times 10 = 160 \text{ in}^2 \)
-
Removed rectangle Surface Area: \( A_{\text{removed}} = 12 \times 4 = 48 \text{ in}^2 \)
Final Surface Area: \( A_{\text{final}} = 160 - 48 = 112 \)
Thickness addition: \( = 112 \times 3 = 336 \text{ in}^2 \)
Answer: 336 in²
Question 5:
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Each pedestal's dimensions with overall height given:
- Pedestal 1: 15in
- Pedestal 2: 11in
- Pedestal 3: 7in
- Length = 20 inches, Depth = 20 inches (Equal depth for all)
Total Surface Area should consider their different heights. Total surface for each pedestal:
- Base Surface Area: \( 20 \times 20 = 400 \text{ in}^2 \)
Each pedestal's surface: \( SA_1 = 2 \times (20 \times 20) + 2 \times (20 \times 15) = 400 + 600 = 1000 \) \( SA_2 = 2 \times (20 \times 20) + 2 \times (20 \times 11) = 400 + 440 = 840 \) \( SA_3 = 2 \times (20 \times 20) + 2 \times (20 \times 7) = 400 + 280 = 680 \)
We create a model summary: Combine = (base leaves out) so combine each is overlapping.
Total = 1000 + 840 + 680
Area subtraction for overlaps: Overlap area: = \( 2 \times (20*20) \times 3: \Rightarrow = 400 \)
Overall System: \( SA = 1000 + 840 + 680 - 400\) = Total; 2120.
Answer: 2120 in²
Note: Surface area calculations can generally involve careful accounting of overlapping faces. Always check for uniformity in sizes to ensure accurate implementation of the model equations in full 3D calculations.