Math Problem Statement

The cross- section of a piece of conduit is an annulus. It has an outer diameter of 60mm and an inner diameter of 50mm. A- find the area of the cross-section to the nearest square mm. B- what is the increase in cross- section area if the outer diameter is increased 65mm? Answer to nearest square mm.

Solution

Let's go step by step through the problem.

Part A: Find the area of the cross-section of the annulus.

The area of an annulus is given by the formula:

A=π(R2r2)A = \pi \left( R^2 - r^2 \right)

Where:

  • RR is the outer radius,
  • rr is the inner radius.

We are given:

  • The outer diameter is 60 mm, so R=602=30R = \frac{60}{2} = 30 mm.
  • The inner diameter is 50 mm, so r=502=25r = \frac{50}{2} = 25 mm.

Substituting these values into the formula:

A=π(302252)=π(900625)=π×275A = \pi \left( 30^2 - 25^2 \right) = \pi \left( 900 - 625 \right) = \pi \times 275

Now, let's calculate the area to the nearest square millimeter.

A3.1416×275864mm2A \approx 3.1416 \times 275 \approx 864 \, \text{mm}^2

So, the area of the cross-section is approximately 864 mm².


Part B: What is the increase in cross-section area if the outer diameter is increased to 65 mm?

For this, we calculate the new area with the outer diameter of 65 mm and find the difference.

  1. The new outer radius is R=652=32.5R = \frac{65}{2} = 32.5 mm.
  2. The inner radius remains r=25r = 25 mm.

The new area is:

Anew=π(32.52252)=π(1056.25625)=π×431.25A_{\text{new}} = \pi \left( 32.5^2 - 25^2 \right) = \pi \left( 1056.25 - 625 \right) = \pi \times 431.25

Now, let's calculate the new area:

Anew3.1416×431.251355mm2A_{\text{new}} \approx 3.1416 \times 431.25 \approx 1355 \, \text{mm}^2

To find the increase in area, subtract the original area from the new area:

ΔA=AnewAoriginal=1355864=491mm2\Delta A = A_{\text{new}} - A_{\text{original}} = 1355 - 864 = 491 \, \text{mm}^2

So, the increase in the cross-section area is approximately 491 mm².


Final Answers:

  • Part A: The area of the cross-section is approximately 864 mm².
  • Part B: The increase in cross-section area is approximately 491 mm².

Would you like more details on any part of this? Here are some related questions to explore:

  1. How would the area change if the inner diameter also increased?
  2. What is the significance of the annular shape in conduit design?
  3. How can we model the volume of such a conduit if the length is known?
  4. How does the change in radius affect the area growth in general?
  5. How would you calculate the material cost for the conduit based on cross-sectional area?

Tip: When working with diameters and radii, always remember to halve the diameter to find the radius before applying area formulas!

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Math Problem Analysis

Mathematical Concepts

Geometry
Annulus Area Calculation
Area Difference

Formulas

Area of annulus: A = π(R^2 - r^2)
Outer radius: R = diameter / 2
Inner radius: r = diameter / 2

Theorems

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Suitable Grade Level

Grades 9-12