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PLEASE HELP!! URGENT

Refer to the diagram of the regular hexagonal nut. What is the area of the hexagonal face to the nearest millimeter?

Please answer this in terms of the questions

1. On the diagram above, what is the area that you need to find?
2. How does this area differ from that of a regular hexagon?
3. What two areas must you find to solve this problem?

Planning the Solution
4. What is the formula for the area of a circle?
5. What is the formula for the area of a regular polygon?
6. What information do you have about the hexagon? What information do you need?
7. What special triangle relationship can you use to find the information you need?


Getting an Answer
8. Use this relationship to find the length of the shorter leg, the perimeter, and the apothem.
9. Use the formula to calculate the area of the hexagon.
10. Use the formula to calculate the area of the circle.
11. What is the area of the hexagonal face?

PLEASE HELP!! URGENT Refer to the diagram of the regular hexagonal nut. What is the-example-1
User Radomeit
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1 Answer

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1. The problem statement tells you to find "the area of the hexagonal face".

2. If we assume the intent is to find the shaded area of the face only, it differs from the area of a regular hexagon in that there is a hole in the middle.

3. You must find the area of the regular hexagon, and subtract the area of the circular hole in the middle.

4. The formula for the area of a circle in terms of its radius is

... A = πr²

5. The formula for the area of a regular hexagon in terms of the radius of the circumcircle is

... A = (3√3)/2·r²

6. The radius of the circumcircle of the regular hexagon is given. No additional information is needed.

7. You can use the trig functions of the angles of an equilateral triangle to find the apothem, but there is no need for that when you use the formula of 5.

8. All this is unnecessary. The apothem is (8 mm)·(√3)/2 = 4√3 mm ≈ 6.9282 mm, the shorter leg is (8 mm)·(1/2) = 4 mm. The perimeter is 6·8 mm = 48 mm.

9. The area of the hexagon is

... A = 3√3/2·(8 mm)² = 96√3 mm² ≈ 166.277 mm²

10. The area of the circle is

... A = π·(4 mm)² = 16π mm² ≈ 50.265 mm²

11. The area of the hexagonal face is approximately ...

... 166.277 mm² - 50.265 mm² = 116.01 mm²

User Multicollinearity
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