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In triangle ABC, angle ABC is 90 degrees. What the perimeter [#permalink]
26 Dec 2007, 10:49

In triangle ABC, angle ABC is 90 degrees.
What the perimeter of triangle ABC?

1. AB + BC = 7
2. AB/BC = 3/4

Can someone tell me why statement 1 is insuff. if we shift the values between AB and BC, the total of these two sides will still be 7. how does it affect the perimeter?

A^2+B^2=C^2 for the sides of a right triangle with the legs being A and B and the hypotenuse being C.

1. tells us the two legs add up to 7.

3^3+4^2=25 so the legs are 3-4-5 for a perimeter of 12

BUT they could also be 5 and 2 since 5+2 = 7 (or they could be 6 and 1)

2^2+5^2=29 so the legs would be 2-5-sqrt(29) for a different perimeter altogether.

2. AB/BC tells us the ratio of the two legs is 3/4. This is insufficient on it's own as the legs could be 3-4, 9-12, 30-40...any two numbers in the 3/4 ratio.

HOWEVER, when taken together...we know the ratio between the two legs is 3/4 and they must add up to 7. This means the only possible values for the legs are 3 and 4. Giving the entire perimeter 3+4+5 or 12

Re: perimeter of a triangle [#permalink]
26 Dec 2007, 11:26

1

This post received KUDOS

bmwhype2 wrote:

In triangle ABC, angle ABC is 90 degrees. What the perimeter of triangle ABC?

1. AB + BC = 7 2. AB/BC = 3/4

Can someone tell me why statement 1 is insuff. if we shift the values between AB and BC, the total of these two sides will still be 7. how does it affect the perimeter?

if you shift values , total will remain same but the hypotenuse will change which affects the perimeter.

Re: perimeter of a triangle [#permalink]
21 Jun 2011, 02:25

Expert's post

Let me point out here that Pythagorean triplets (3,4,5 etc) give you the sides of a right angled triangle such that each side is an integer. But the sides of a right angled triangle needn't necessarily be integers. If it is given that hypotenuse is 5, the sides could be 3,4 or they could be 1, \sqrt{24} or 2, \sqrt{21} etc... Similarly, if sum of the legs is 7, the sides could be 3,4 or 1,6 or 2,5 etc... For each of these, the hypotenuse will be different.

What helps in visualizing whether there are multiple solutions is drawing. Make a hypotenuse of length 5 cm. As you keep changing the incline of the hypotenuse, you keep getting different values for the other two sides. _________________