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Sagot :
Sure, let's solve the problem step-by-step to find the surface area of the nut after a hole is drilled through it.
### Step 1: Understand the Problem
You have a square metal nut with a length of 6 cm and a thickness of 4 cm. A cylindrical hole with a diameter of 3.5 cm is drilled through this nut.
### Step 2: Calculate the Surface Area of the Nut Without the Hole
The nut is a rectangular prism (or cuboid). To find the surface area, we need to calculate the surface area of each face of the cube.
- Each side of the cube has:
- Two square faces of 6 cm x 6 cm.
- Two rectangular faces of 6 cm x 4 cm.
- Two rectangular faces of 6 cm x 4 cm (since thickness appears twice in rectangular faces)
The formula used for cuboid surface area is:
[tex]\[ \text{Surface Area} = 2 \times (\text{length} \times \text{length} + \text{length} \times \text{thickness} + \text{thickness} \times \text{length}) \][/tex]
So, for our nut:
[tex]\[ \text{Surface Area} = 2 \times (6 \times 6 + 6 \times 4 + 4 \times 6) \][/tex]
[tex]\[ = 2 \times (36 + 24 + 24) \][/tex]
[tex]\[ = 2 \times 84 \][/tex]
[tex]\[ = 168 \text{ cm}^2 \][/tex]
### Step 3: Calculate the Surface Area of the Cylindrical Hole
First, we need to find the radius of the hole:
[tex]\[ \text{radius} = \frac{\text{diameter}}{2} = \frac{3.5}{2} = 1.75 \text{ cm} \][/tex]
Next, we calculate the surface area of the cylindrical hole.
- The lateral (curved) surface area of the cylinder:
[tex]\[ \text{Lateral Surface Area} = 2 \times \pi \times \text{radius} \times \text{thickness} \][/tex]
[tex]\[ = 2 \times \pi \times 1.75 \times 4 \][/tex]
[tex]\[ = 14 \pi \][/tex]
- Surface area of the two circular faces (which are not included in the final total, but are part of the calculations):
[tex]\[ \text{Area of the circles} = 2 \times \pi \times (\text{radius})^2 \][/tex]
[tex]\[ = 2 \times \pi \times (1.75)^2 \][/tex]
[tex]\[ = 2 \times \pi \times 3.0625 \][/tex]
[tex]\[ = 6.125 \pi \][/tex]
Even though the circular faces need to be considered for accurate calculation steps, they are part of both the metal nut and cylindrical hole.
### Step 4: Combining Both Areas for the Surface Area of the Nut with the Hole
The total surface area of the nut with the hole should now be:
[tex]\[ \text{Surface Area} = \text{Surface Area of the Nut Without the Hole} + \text{Lateral Surface Area of the Hole} - \text{Surface Area of the two Circular Faces}\][/tex]
[tex]\[ = 168 + 14\pi - 6.125\pi\][/tex]
[tex]\[ \approx 168 + 2 \pi (i.e. additional surface area impact)\][/tex]
[tex]\[ = 168 + 43.982297 = 211.982297 \][/tex]
### Conclusion:
The surface area of the nut with the hole is approximately 211.98 cm².
### Step 1: Understand the Problem
You have a square metal nut with a length of 6 cm and a thickness of 4 cm. A cylindrical hole with a diameter of 3.5 cm is drilled through this nut.
### Step 2: Calculate the Surface Area of the Nut Without the Hole
The nut is a rectangular prism (or cuboid). To find the surface area, we need to calculate the surface area of each face of the cube.
- Each side of the cube has:
- Two square faces of 6 cm x 6 cm.
- Two rectangular faces of 6 cm x 4 cm.
- Two rectangular faces of 6 cm x 4 cm (since thickness appears twice in rectangular faces)
The formula used for cuboid surface area is:
[tex]\[ \text{Surface Area} = 2 \times (\text{length} \times \text{length} + \text{length} \times \text{thickness} + \text{thickness} \times \text{length}) \][/tex]
So, for our nut:
[tex]\[ \text{Surface Area} = 2 \times (6 \times 6 + 6 \times 4 + 4 \times 6) \][/tex]
[tex]\[ = 2 \times (36 + 24 + 24) \][/tex]
[tex]\[ = 2 \times 84 \][/tex]
[tex]\[ = 168 \text{ cm}^2 \][/tex]
### Step 3: Calculate the Surface Area of the Cylindrical Hole
First, we need to find the radius of the hole:
[tex]\[ \text{radius} = \frac{\text{diameter}}{2} = \frac{3.5}{2} = 1.75 \text{ cm} \][/tex]
Next, we calculate the surface area of the cylindrical hole.
- The lateral (curved) surface area of the cylinder:
[tex]\[ \text{Lateral Surface Area} = 2 \times \pi \times \text{radius} \times \text{thickness} \][/tex]
[tex]\[ = 2 \times \pi \times 1.75 \times 4 \][/tex]
[tex]\[ = 14 \pi \][/tex]
- Surface area of the two circular faces (which are not included in the final total, but are part of the calculations):
[tex]\[ \text{Area of the circles} = 2 \times \pi \times (\text{radius})^2 \][/tex]
[tex]\[ = 2 \times \pi \times (1.75)^2 \][/tex]
[tex]\[ = 2 \times \pi \times 3.0625 \][/tex]
[tex]\[ = 6.125 \pi \][/tex]
Even though the circular faces need to be considered for accurate calculation steps, they are part of both the metal nut and cylindrical hole.
### Step 4: Combining Both Areas for the Surface Area of the Nut with the Hole
The total surface area of the nut with the hole should now be:
[tex]\[ \text{Surface Area} = \text{Surface Area of the Nut Without the Hole} + \text{Lateral Surface Area of the Hole} - \text{Surface Area of the two Circular Faces}\][/tex]
[tex]\[ = 168 + 14\pi - 6.125\pi\][/tex]
[tex]\[ \approx 168 + 2 \pi (i.e. additional surface area impact)\][/tex]
[tex]\[ = 168 + 43.982297 = 211.982297 \][/tex]
### Conclusion:
The surface area of the nut with the hole is approximately 211.98 cm².
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