Challenging Area Problem

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mlaboda: Junior | Next Rank: 30 Posts; Posts: 25; Joined: Sun Jan 25, 2009 2:34 pm; Thanked: 1 times

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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Source: Beat The GMAT — Problem Solving | Sat May 02, 2009 3:51 pm

dumb.doofus: Master | Next Rank: 500 Posts; Posts: 435; Joined: Sat Sep 27, 2008 2:02 pmFollowed by: 1 member; Location: San Jose, CA; Thanked: 43 times; GMAT Score:720

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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arora007: Community Manager; Posts: 1048; Joined: Mon Aug 17, 2009 3:26 amFollowed by: 27 members; Location: India; Thanked: 51 times; GMAT Score:670

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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Testluv: GMAT Instructor; Posts: 1302; Joined: Mon Oct 19, 2009 2:13 pmFollowed by: 164 members; Location: Toronto; Thanked: 539 times; GMAT Score:800

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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