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The sides of a right triangle are consecutive even integers,

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The sides of a right triangle are consecutive even integers,

by BTGmoderatorDC » Thu Feb 07, 2019 4:57 am

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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

OA A

Source: Veritas Prep
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by [email protected] » Thu Feb 07, 2019 10:21 am
Hi All,

We're told that the sides of a right triangle are CONSECUTIVE EVEN INTEGERS, and the length of the LONGEST side is P. We're asked which of the following equations could be used to find P. This question can be solved in a couple of different ways, including by TESTing VALUES.

To start, you might recognize that a right triangle with sides that are CONSECUTIVE EVEN INTEGERS would be a 6/8/10 right triangle - meaning that P = 10 and the two legs are 6 and 8. Using those values, there's only one answer that matches. In addition, a number of the answers can be quickly eliminated for logical reasons (without doing much math at all). For example, Answer C adds three values (including 10^2)... but there's no way those 3 values would equal 6^2 and Answer E squares two values that are each GREATER than P, so the sum of those values couldn't possibly equal P^2.

Final Answer: A

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by Brent@GMATPrepNow » Thu Feb 07, 2019 11:24 am
BTGmoderatorDC 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 = p² - (p−2)²
B. (p-2)² = (p-4)² - p²
C. p² + 4² + 2² = 6²
D. (p-2)² = p² - (p−1)²
E. (p+2)² + (p+4)² = p²

OA A

Source: Veritas Prep
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!!

Answer: A

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Scott@TargetTestPrep » Sun Feb 10, 2019 7:36 am
BTGmoderatorDC 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

OA A

Source: Veritas Prep
Many of us know the 3-4-5 right triangle. Though it is not the only right triangle with sides of integer lengths, it is the only one with side lengths that are consecutive integers. And if we double each of these numbers, we get the 6-8-10 right triangle, which is the only right triangle with side lengths that are consecutive even integers. Therefore, we see that that p = 10, which makes 6 = p - 4 and 8 = p - 2. By the Pythagorean theorem, we have:

6^2 + 8^2 = 10^2

That is,

(p - 4)^2 + (p - 2)^2 = p^2

(p - 4)^2 = p^2 - (p - 2)^2

Alternate Solution:

Since the sides are consecutive even integers and since the longest side is p, the remaining sides are p - 2 and p - 4.

By the Pythagorean Theorem, we have:

(p - 4)^2 + (p - 2)^2 = p^2,

which is equivalent to

(p - 4)^2 = p^2 - (p - 2)^2

Answer: A

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