Surface Area of a Cuboid: Identify the greater value

To solve this problem, we'll calculate the surface area of each cuboid solar panel and compare them:

• Step 1: Identify dimensions for prototype A: length $l = 70$, width $w = 25$, height $h = 12$.
• Step 2: Calculate the surface area of A using the formula $2(lw + lh + wh)$.
  Surface area $= 2(70 \times 25 + 70 \times 12 + 25 \times 12)$.
• Step 3: Perform the calculations:
  $= 2(1750 + 840 + 300)$
  $= 2(2890)$
  $= 5780$.
• Step 4: Identify dimensions for prototype B: length $l = 82$, width $w = 7$, height $h = 13$.
• Step 5: Calculate the surface area of B using the same formula.
  Surface area $= 2(82 \times 7 + 82 \times 13 + 7 \times 13)$.
• Step 6: Perform the calculations:
  $= 2(574 + 1066 + 91)$
  $= 2(1731)$
  $= 3462$.
• Step 7: Compare the surface areas: $5780$ (A) vs. $3462$ (B).

Therefore, prototype A will absorb more of the sun's energy because it has a larger total surface area.

Thus, the solution to the problem is A.

To solve this problem, we'll follow these steps:

• Step 1: Determine the dimensions of each orthohedron from the diagram.
• Step 2: Calculate the surface area for each cuboid using the formula.
• Step 3: Compare the calculated surface areas.

Now, let's work through each step:

Step 1: Identify the Given Dimensions
From the visual data, we note that the dimensions of the first orthohedron are given as 1, 2, and 3, while the second orthohedron has the same visual size with similar digits 1, 2, and 3 marked, suggesting identical measurements for each dimension.

Step 2: Calculate the Surface Area for Each Cuboid
Utilize the surface area formula for cuboids:

$SA = 2(lw + lh + wh)$

For Both Orthohedra, Given Dimensions:

• Length ($l$): 1 unit
• Width ($w$): 2 units
• Height ($h$): 3 units

The surface area calculation will be:

$SA = 2(1 \cdot 2 + 1 \cdot 3 + 2 \cdot 3)$ $= 2(2 + 3 + 6)$ $= 2 \cdot 11$ $= 22 \text{ square units}$

As both cuboids have the same dimensions, their surface area calculations yield identical results.

Step 3: Compare Surface Areas
Since both orthohedra compute to the same total surface area of 22 square units, we conclude their surface areas are the same.

Therefore, the solution to the problem is The same.

To solve the problem, we'll proceed through these steps:

• Step 1: Identify the dimensions of the orthohedrons from the diagram.
• Step 2: Apply the surface area formula for a cuboid.
• Step 3: Compare the surface areas to determine if they are equal or different.

Step 1: Dimensions from the diagram:

• Orthohedron 1: Length $l = 3$, Width $w = 2$, Height $h = 1$.
• Orthohedron 2: Length $l = 2$, Width $w = 3$, Height $h = 1$.

Step 2: Calculate surface areas using the formula $SA = 2(lw + lh + wh)$.

For Orthohedron 1:
$SA_1 = 2(3 \cdot 2 + 3 \cdot 1 + 2 \cdot 1) = 2(6 + 3 + 2) = 2(11) = 22$.

For Orthohedron 2:
$SA_2 = 2(2 \cdot 3 + 2 \cdot 1 + 3 \cdot 1) = 2(6 + 2 + 3) = 2(11) = 22$.

Step 3: Compare the surface areas:
Both orthohedrons have a surface area of 22; therefore, the surface areas are equal.

Thus, the surface areas of the two orthohedrons are equal.

The correct answer is: $=$.