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Quiz of the Month (May 2003)

Hector C. Parr

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SOLUTION TO LAST MONTH'S QUIZ

   1.  82.84 cm2
   2.  6 cm, 8 cm, 9.5 cm
   3.  4.55 m
Notes 
1. AB = 10 - 2x
   AC = sqr(2) * x
    So 10 - 2x = sqr(2) * x 
    Solving, x = 2.929
      (rejecting the other solution).

    Then area = 100 - 2x2
              = 82.843
 
2.   xy = 76
     yz = 57
     zx = 48
     Multiplying all three equations together,
        x2y2z2 = 207936
     So   xyz = sqr(207936) = 456
     Then   x = xyz/yz = 456/57 = 8
            y = xyz/zx = 456/48 = 9.5
            z = xyz/xy = 456/76 = 6


  

3. Triangle PQX is similar to SRX
     So QX:RX = 13:7
     So QX:QR = 13:20
     So x:7   = 13:20
       and  x = 91/20 = 4.55

THIS MONTH'S QUIZ

The illustration represents a perspective drawing (i.e. projection from the eye position onto a vertical plane) of a straight row of vertical fence posts standing on level ground. The posts are of equal height and equally spaced. On the drawing the length of the first post, AB, is 10 cm, the third post, EF, is 5 cm, and the distance along the ground between the first two posts, AC, is 8 cm. Find the lengths on the drawing of the following lines:

  1. CD
  2. GH
  3. CE

[Hint:  Complete the quadrilateral ABFE and draw its diagonals.]

***

(c) Hector C. Parr (2003)


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