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## Quiz of the Month (November 2006)

### Hector C. Parr

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#### SOLUTION TO LAST MONTH'S QUIZ

```  1.  1.950 cm
2.  0.158 cm
3.  0.207 cm2 ```

Notes.

 ``` Angles at centre O: 4q * 7 = 360o So q = 12.857143o = 0.2243995 radians 1. Distance = AD = 2 * OD * cosq = 1.9498559 2. Length of arc DE (centre O) = OD * 4q = 0.8975979 Length of arc DE (centre A) = AD * 2q = 0.8750934 Difference = 0.0225045 So total difference = 0.0225045 * 7 = 0.1575315 3. Area triangle AOD = DP * PO = cosq * sinq = 0.9749279 * 0.2225209 = 0.2169418 Area sector DAE = AD2 * 2q/2 = 0.853153 So area sector DOE of coin = 0.853153 - 2 * 0.2169418 = 0.4192694 But area sector DOE of circle = OD2 * 4q/2 = 2q = 0.4487990 Difference = 0.0295296 So total difference = 0.0295296 * 7 = 0.2067072 ```

#### THIS MONTH'S QUIZ

1. ABCDEFGH is a wooden cube with edges of length 1 ft. The corner E is cut off with a straight sawcut as shown to expose the triangular area AFH (coloured green). Find the volume of the piece removed.

2. Three other corners, B, D and G, are removed in the same way, leaving the regular tetrahedron ACFH. Find the volume of the tetrahedron.

3. Find CP, the perpendicular distance of C from the plane AFH.

[Hint: The volume of any pyramid (and hence of any tetrahedron) is one third the area of its base multiplied by its perpendicular height.]

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