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

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

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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

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

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

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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

$$Base < Altitude < Hypotenuse$$

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  [#permalink]

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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

GIVEN: p = length of the longest side

Since the side lengths are consecutive even integers, we can say:
p - 2 = length of the 2nd longest side
p - 4 = length of the shortest side

NOTE: The longest side is the HYPOTENUSE.
So, p = length of the HYPOTENUSE
And p - 2 = length of one leg of the right triangle
And p - 4 = length of the other leg of the right triangle

By the Pythagorean Theorem, we can write: (p - 2)² + (p - 4)² = p²
Check the answers . . . not there! Looks like we need to fiddle with the equation to make it look like on of the answer choices.

Take: (p - 2)² + (p - 4)² = p²
Subtract (p - 2)² from both sides to get: (p - 4)² = p² - (p - 2)² . . . BINGO!!

Cheers,
Brent
_________________ Re: The sides of a right triangle are consecutive even integers, and the l   [#permalink] 07 Feb 2019, 12:38
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