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Challenging Area Problem

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Challenging Area Problem

by mlaboda » Sat May 02, 2009 3:51 pm
Thanks in advance!

One of two ropes equal in length is cut into three segments to form the largest possible triangular area. The other rope is cut into four segments to form the largest possible rectangular area. Which of the following most closely approximates the ratio of the triangle's area to the rectangle's area?

A. 1:2
B. 2:3
C. 3:4
D. 1:1
E. 4:3
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by dumb.doofus » Sat May 02, 2009 4:19 pm
well, you need to remember two things here..

1. Maximum area of a triangle with a given perimeter is when its an equilateral triangle.
2. Maximum area of a rectangle with a given perimeter is when its a square..

so let the length of the rope be 12a (choosing it for simple calculation)


Area of an equilateral triangle = root(3)*x^2/4
here x = 12a/3 = 4a

So area = 4root(3)a^2 ---------- (1)

Area of a square = x^2
here x = 12a/4 = 3a

So area = 9a^2 ----------- (2)

ratio of 1 and 2 = 4root(3)/9 = 0.76

The closest to this is C i.e. 3/4 or 0.75

So answer is C
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by arora007 » Wed Jun 23, 2010 11:30 am
I too got 4root(3)/9 and somehow put the answer as 2:3.

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by Testluv » Wed Jun 23, 2010 9:12 pm
well, you need to remember two things here..

1. Maximum area of a triangle with a given perimeter is when its an equilateral triangle.
2. Maximum area of a rectangle with a given perimeter is when its a square..
We can consolidate these 2 rules into 1:

in general, when we want to maximize the area of a 2-d figure, we need to equalize all of its sides.
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