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January 2nd, 2008, 04:16 PM   #1
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From: dasmariƱas, cavite

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Please Help me!!!

1. If the width of a rectangle is 2cm more than one-half its length and its perimeter iis 40 cm, what are the dimensions?

2. The longest side of a triangle is twice a long as the shotest side and 2 cm longer than the third side. If the perimeter of the triangle is 33 cm, what is the length of each side?
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January 3rd, 2008, 08:49 AM   #2
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Please put any future questions into the proper forums.... And be specific (i.e. find length of side)

That being said:

These problems are not difficult; it will make math infinitely easier to be able to solve these fully by yourself sometime soon.
Are you having trouble setting up the problem from the words, or are you having trouble substituting? Or is the algebra itself the problem?
Please let me know so I can try to help you understand whatever is slipping by you.

#1
we will call perimeter P, width w, and length l

w=l/2+2 (2 more than half of l)

P=2w+2l (4 sides) = 40

Substitute and solve.

#2
we will call the three sides A, B, and C. Following convention, C will be the longest.

C=B+2 (2 more than B)
A=B-2 (2 less than B)

P=A+B+C (3 sides) = 33

Again, substitute and solve.
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January 18th, 2008, 09:11 PM   #3
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Quote:
Originally Posted by ansfer02
2. The longest side of a triangle is twice a long as the shortest side and 2 cm longer than the third side. If the perimeter of the triangle is 33 cm, what is the length of each side?
If x is the length of the shortest side, the length of the longest side is 2x and the length of the third side is 2x - 2 cm.

Hence 2x + x + 2x - 2 cm = 33 cm, so x = (35/5) cm = 7 cm, and so the three sides have lengths 14 cm, 12 cm, and 7 cm.
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