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The sides of a right triangle are consecutive even integers, and the l

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The sides of a right triangle are consecutive even integers, and the l  [#permalink]

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New post 12 Sep 2015, 22:49
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The sides of a right triangle are consecutive even integers, and the length of the longest side is p. Which of the following equations could be used to find p?

A. (p–4)^2=p^2–(p−2)^2
B. (p–2)^2=(p–4)–p^2
C. p^2+4^2+2^2=6^2
D. (p–2)^2=p^2–(p−1)^2
E. (p+2)^2+(p+4)^2=p^2

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Re: The sides of a right triangle are consecutive even integers, and the l  [#permalink]

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New post 13 Sep 2015, 00:41
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The sides are consecutive EVEN integers.
So, if longest side has a length of p, the other two sides will be (p-2) and (p-4)
And using the Pythagorean Theorem, we get: p^2 = (p-2)^2 + (p-4)^2

Ans: A
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Re: The sides of a right triangle are consecutive even integers, and the l  [#permalink]

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New post 12 Aug 2017, 10:11
If sides are three consecutive even integers, the right triangle in talk can have sides 6,8,10 where value of largest side p=10
thus working on all options
A - (10-4)^2 = 10^2 - (10-2)^2 / 6^2 = 10^2 - 8^2 which is true
B - (10-2)^2 = (10-4)^2 - 10^2 / 8^2 = 6^2 - 10^2which is false
C - 10^2 + 4^2 + 2^2 = 6^2 which is false
D - (10-2)^2 = 10^2 - (10-1)^2 / 8^2 = 10^2 - 9^2which is false
E - (10+2)^2 + (10+4)^2 = 10^2 / 12^2 + 14^2 = 10^2which is false
Thus answer is option A
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The sides of a right triangle are consecutive even integers, and the l  [#permalink]

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New post 12 Aug 2017, 21:27
The sides of a right triangle are consecutive even integer and the longest side is P. So, other sides are p-2 & p-4. Using Pythagorean theory p^2=(p-2)^2+(p-4)^2 , => (p-4)^2=p^2-(p-2)^2.
Answer is A.
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Re: The sides of a right triangle are consecutive even integers, and the l  [#permalink]

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New post 06 Sep 2018, 03:48
shasadou wrote:
The sides of a right triangle are consecutive even integers, and the length of the longest side is p. Which of the following equations could be used to find p?

A. (p–4)^2=p^2–(p−2)^2
B. (p–2)^2=(p–4)–p^2
C. p^2+4^2+2^2=6^2
D. (p–2)^2=p^2–(p−1)^2
E. (p+2)^2+(p+4)^2=p^2


One can use the wisdom for GMAT that GMAT uses only specific cases of right angle triangles 3-4-5 being the most frequently used pythagorean triplet

Here three consecutive even integers will be second multiple of 3-4-5 i.e. 6-8-10

As give, p = 10
p-2 = 8
p-4 = 6

Now check options

Option A: \((p–4)^2=p^2–(p−2)^2\) i.e. \(6^2 = 10^2 - 8^2\) fits well hence

Answer: Option A
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Re: The sides of a right triangle are consecutive even integers, and the l  [#permalink]

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New post 06 Sep 2018, 06:15
shasadou wrote:
The sides of a right triangle are consecutive even integers, and the length of the longest side is p. Which of the following equations could be used to find p?

A. (p–4)^2=p^2–(p−2)^2
B. (p–2)^2=(p–4)–p^2
C. p^2+4^2+2^2=6^2
D. (p–2)^2=p^2–(p−1)^2
E. (p+2)^2+(p+4)^2=p^2


\(Base < Altitude < Hypotenuse\)

So, \(p - 4 < p - 2 < p\)

Now, \(p^2 = ( p - 2 )^2 + ( p - 4 )^2\) { Pythagorus Theorem }

Thus, Answer must be (A)
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Re: The sides of a right triangle are consecutive even integers, and the l &nbs [#permalink] 06 Sep 2018, 06:15
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