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The right triangle △ABC has side lengths 10 – x, 9 – x, and 8 – x. Giv

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

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New post Updated on: 12 Feb 2019, 01:04
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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.
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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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New post 12 Feb 2019, 01:18
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.

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

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New post Updated on: 12 Feb 2019, 01:30
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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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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New post 12 Feb 2019, 01:29
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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New post 17 Feb 2019, 01:00
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.
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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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