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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # The right triangle △ABC has side lengths 10 – x, 9 – x, and 8 – x. Giv

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Joined: 16 Jun 2018
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GMAT 1: 640 Q49 V29 GPA: 4
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The right triangle △ABC has side lengths 10 – x, 9 – x, and 8 – x. Giv  [#permalink]

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Question Stats: 51% (02:17) correct 49% (02:05) wrong based on 57 sessions

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The right triangle △ABC has side lengths 10 – x, 9 – x, and 8 – x. Given that only one value of x is possible, which of the following statements is true?

a) x is not an integer
b) The perimeter of △ABC must be greater than 10.
c) △ABC is isosceles
d) The area of △ABC cannot be determined from the information given
e) None of the above statements is true

Originally posted by amresh09 on 12 Feb 2019, 01:00.
Last edited by Bunuel on 12 Feb 2019, 01:04, edited 1 time in total.
Renamed the topic.
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Re: The right triangle △ABC has side lengths 10 – x, 9 – x, and 8 – x. Giv  [#permalink]

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2
amresh09 wrote:
The right triangle △ABC has side lengths 10 – x, 9 – x, and 8 – x. Given that only one value of x is possible, which of the following statements is true?

a) x is not an integer
b) The perimeter of △ABC must be greater than 10.
c) △ABC is isosceles
d) The area of △ABC cannot be determined from the information given
e) None of the above statements is true

10-x, 9-x and 8-x are sequential numbers. They immediately remind one of 3-4-5 pythagorean triplet. If x = 5, we get the sides as 3-4-5.

The perimeter will be 3+4+5 = 12 which is greater than 10.

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Karishma
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The right triangle △ABC has side lengths 10 – x, 9 – x, and 8 – x. Giv  [#permalink]

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2
amresh09 wrote:
The right triangle △ABC has side lengths 10 – x, 9 – x, and 8 – x. Given that only one value of x is possible, which of the following statements is true?

a) x is not an integer
b) The perimeter of △ABC must be greater than 10.
c) △ABC is isosceles
d) The area of △ABC cannot be determined from the information given
e) None of the above statements is true

My thoughts on this, first thing which i thought after reading right triangle △ABC, the first Pythagorean triplet 3,4,5

Given that only one value of x is possible

Now Question is which of the following statements is true.

a) x is not an integer
As per my initial assumption this is not true.

b) The perimeter of △ABC must be greater than 10.
perimeter, sum of all sides => 27-3x > 10
we can take x as 5
12> 10
This is true

c) △ABC is isosceles
It can be possible, it cannot be possible

d) The area of △ABC cannot be determined from the information given
This for sure was false

e) None of the above statements is true
Irrelevant
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If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.

Originally posted by KanishkM on 12 Feb 2019, 01:29.
Last edited by KanishkM on 12 Feb 2019, 01:30, edited 1 time in total.
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Re: The right triangle △ABC has side lengths 10 – x, 9 – x, and 8 – x. Giv  [#permalink]

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1
Total perimeter will be 27-3x
Take x = 5 --- p = 12 . Construction of triangle is valid it's a 3 4 5 triplet . Let's inrease x, so that we can come to know the maximum perimeter that can be made.
B

Let's say x is 6 ---> p= 9.
This is not a valid triangle construction , there can be no triangle with legs 2 3 and 4 because this violated the side rule.
So the perimeter will always be greater than 10.

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Re: The right triangle △ABC has side lengths 10 – x, 9 – x, and 8 – x. Giv  [#permalink]

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1
amresh09 wrote:
The right triangle △ABC has side lengths 10 – x, 9 – x, and 8 – x. Given that only one value of x is possible, which of the following statements is true?

a) x is not an integer
b) The perimeter of △ABC must be greater than 10.
c) △ABC is isosceles
d) The area of △ABC cannot be determined from the information given
e) None of the above statements is true

From the three sides given we can apply Pythagoras theorem:

(8-x)^2 + (9-x)^2 = (10-x)^2
=> (8-x)^2 = (10-x)^2 - (9-x)^2
=> (8-x)^2 = (10 - x + 9 - x) * (10 - x - 9 + x)
=> (8 - x)^2 = (19 - 2x) * (1)
=> x^2 + 64 - 16x = 19 - 2x
=> x^2 - 14x + 45 = 0
=> x = 9, 5
X cant be 9 as it would make one of the sides as zero

=> x = 5
Hence the three sides as 3, 4, & 5

Therefore the correct answer is B. Re: The right triangle △ABC has side lengths 10 – x, 9 – x, and 8 – x. Giv   [#permalink] 17 Feb 2019, 01:00
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