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Powerprep Geometry

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by caramel3536 » Sat May 03, 2008 1:18 pm
May someone help me with this question:

A rectangular box is 10 inches wide, 10 inches long, and 5 inches high. What is the greatest possible (straight-line) distance in inches between any two points on the box (ans 15)
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Source: — Problem Solving |

by sripardhu » Sat May 03, 2008 2:46 pm
= sqrt((sqrt(10 ^ 2 + 10 ^ 2)) ^ 2 + 5 ^ 2)

= 15
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by AleksandrM » Sat May 03, 2008 4:47 pm
I think that everyone should use the visualization principle, and draw first [see my attachement - the dotted line in the second box, is one of the sides that you would view when you turn the box to see the inside triangle as it would appear in a two-dimensional space].

Realize that you need to get the length of the line from one angle to the furthest opposite angle of the box.

You now have a triangle with a known height, but an unknown base and hypotenuse. The base and the two sides measuring 10, form a lying right triangle. By calculating the hypotenuse of the lying triangle, you will calculate the base of the bigger triangle inside the box.

10^2 + 10^2 = h^2

h = 10sqroot2 -------> this is the base of the bigger triangle.

Note: you don't have to go through this calculation if you memorize the common triangles, which will lead you to recognize that this equilateral triangle has sides in the ratio of x : x : xsqroot2

Now, go on to get your answer:

(10sqroot2)^2 + 5^2 = x^2 [the new hypotenuse]

200 + 25 = x^2 take a sqroot of both sides

x = 15
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two triangles.doc
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Thanks guys!

by caramel3536 » Sun May 04, 2008 1:37 pm
.........
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