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 Quiz of the Month (May 2003)
Hector C. Parr
***
SOLUTION TO LAST MONTH'S QUIZ
1. 82.84 cm^{2}
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  2x^{2}
= 82.843 

2. xy = 76
yz = 57
zx = 48
Multiplying all three equations together,
x^{2}y^{2}z^{2} = 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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