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If x>0, and two sides of a certain triangle
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30 Aug 2016, 02:20

11

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Question Stats:

If x>0, and two sides of a certain triangle have lengths 2x+1 and 3x+4 respectively, which of the following could be the length of the third side of the triangle?

Indicate all possible lengths.

A) 4x+5

B) x+2

C) 6x+1

D) 5x+6

E) 2x+17

Indicate all possible lengths.

A) 4x+5

B) x+2

C) 6x+1

D) 5x+6

E) 2x+17

Re: If x>0, and two sides of a certain triangle
[#permalink]
30 Aug 2016, 05:10

5

1

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Let the third side be y

then (3x + 4) – (2x + 1) < y < (3x + 4) + (2x + 1)

=> x + 3 < y < 5x + 5

As with option B, x+2 is always less than x+3, therefore B is not possible.

And with option E, if we choose x=5, then 2x+17= 27

=> from inequality, 8<27<30

Hope, the solution is clear!

then (3x + 4) – (2x + 1) < y < (3x + 4) + (2x + 1)

=> x + 3 < y < 5x + 5

As with option B, x+2 is always less than x+3, therefore B is not possible.

And with option E, if we choose x=5, then 2x+17= 27

=> from inequality, 8<27<30

Hope, the solution is clear!

Re: If x>0, and two sides of a certain triangle
[#permalink]
30 Aug 2016, 04:35

2

Expert Reply

Hi,

I think B should be one of the solutions too

Let the triangle have side named A, B and C

A= 2x+1 and B= 3x+4

A+B= 5x+5

Now we know from triangle inequality

C < A +B

Option A

4x+5 < 5x+5

=>x > 0

Option B

x +2 < 5x+5

=> -4x < 3 => x > \(\frac{-3}{4}\).

There is also possible as long as x >0 it is also x > \(\frac{-3}{4}\)

Option C

6x+1 < 5x+5

x < 4

Also a possible solution exits such that 0 < x < 4.

Option D is not possible.

Option E is also similarly possible.

I think B should be one of the solutions too

Let the triangle have side named A, B and C

A= 2x+1 and B= 3x+4

A+B= 5x+5

Now we know from triangle inequality

C < A +B

Option A

4x+5 < 5x+5

=>x > 0

Option B

x +2 < 5x+5

=> -4x < 3 => x > \(\frac{-3}{4}\).

There is also possible as long as x >0 it is also x > \(\frac{-3}{4}\)

Option C

6x+1 < 5x+5

x < 4

Also a possible solution exits such that 0 < x < 4.

Option D is not possible.

Option E is also similarly possible.

Re: If x>0, and two sides of a certain triangle
[#permalink]
11 Nov 2018, 08:19

Hey I got the answer as A, C ,E..I took x as 1,3 and 8... But my workbook still shows the answer as incorrect. Can somebody help me with this question.

Re: If x>0, and two sides of a certain triangle
[#permalink]
11 Nov 2018, 09:24

1

Expert Reply

Reetika1990 wrote:

Hey I got the answer as A, C ,E..I took x as 1,3 and 8... But my workbook still shows the answer as incorrect. Can somebody help me with this question.

The answer is correct as A, C and E.

the equations can be formed in following way..

(I) the third side is less than the sum of the other two sides...

so third side < (2x+1) + (3x+4) or third side < 5x+5

(II) the third side is greater than the difference of the other two sides...

so third side > |(2x+1) - (3x+4)| or > |x+3|

Therefore, equation becomes \(x+3 < third side < 5x+5\)

Re: If x>0, and two sides of a certain triangle
[#permalink]
12 Sep 2019, 11:49

Hi All ,

While 4x+5 , 6x+1 , and 2x +17 are mentioned as correct answers for x>0 there might e challenge

X tending to 0 e.g. 0.01

2x+1 = 1.02

3x+4 = 4.03

here 6x+1 as third side will become 1.06 thus three sides as 1.02,1.06, 4.03 which is not possible

here 2x+17 = 17.02 which again is not possible

Again if x is big as x=1000

2x+1 =2001

3x+4 =3004

6x+1 becomes 6001

Not so sure about the answers now ?

While 4x+5 , 6x+1 , and 2x +17 are mentioned as correct answers for x>0 there might e challenge

X tending to 0 e.g. 0.01

2x+1 = 1.02

3x+4 = 4.03

here 6x+1 as third side will become 1.06 thus three sides as 1.02,1.06, 4.03 which is not possible

here 2x+17 = 17.02 which again is not possible

Again if x is big as x=1000

2x+1 =2001

3x+4 =3004

6x+1 becomes 6001

Not so sure about the answers now ?

Re: If x>0, and two sides of a certain triangle
[#permalink]
15 Oct 2019, 06:57

3

x+3<3rd side length<5x+5

let's say x=1000

1003<3rd side<5005

option E: 2(1000)+17= 2017 which falls right into the region. N.B if you consider small values for x, ie 2/3 then option E is invalid. but the ques asks for which of the following could be** the length. not which of the following must be**. you have to go for every possible way out there.

let's say x=1000

1003<3rd side<5005

option E: 2(1000)+17= 2017 which falls right into the region. N.B if you consider small values for x, ie 2/3 then option E is invalid. but the ques asks for which of the following could be** the length. not which of the following must be**. you have to go for every possible way out there.

Re: If x>0, and two sides of a certain triangle
[#permalink]
04 Jun 2020, 18:31

4

phoenixio wrote:

If x>0, and two sides of a certain triangle have lengths 2x+1 and 3x+4 respectively, which of the following could be the length of the third side of the triangle?

Indicate all possible lengths.

A) 4x+5

B) x+2

C) 6x+1

D) 5x+6

E) 2x+17

Indicate all possible lengths.

A) 4x+5

B) x+2

C) 6x+1

D) 5x+6

E) 2x+17

solution:

of course 3x+4 > 2x+1 so, third side of the triangle is 3X+4-(2x+1)<third_side<3x+4+2x+1.

which gives x+3<third_side<5x+5.

note: question asked which "could be" the length of third side.

x>0:

A) x+3<4x+5<5x+5 which is true.

B) x+3<x+2<5x+5 never true.

C) x+3<6X+1<5x+5 which is not true for all value of x, but is true for x<4. so, could be true.

D) x+3<5x+6<5x+5 never true.

E) x+3<2x+17<5x+5 which is not true for all value of x, but is true for x>=5.

A,C,E

Re: If x>0, and two sides of a certain triangle
[#permalink]
28 Oct 2021, 14:29

1

C and E should be out of the answers because we don't know the real value of X

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Joined: **10 Apr 2015 **

Posts: **6218**

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Re: If x>0, and two sides of a certain triangle
[#permalink]
23 Oct 2022, 07:17

2

phoenixio wrote:

If x>0, and two sides of a certain triangle have lengths 2x+1 and 3x+4 respectively, which of the following could be the length of the third side of the triangle?

Indicate all possible lengths.

A) 4x+5

B) x+2

C) 6x+1

D) 5x+6

E) 2x+17

Indicate all possible lengths.

A) 4x+5

B) x+2

C) 6x+1

D) 5x+6

E) 2x+17

IMPORTANT PROPERTY: If two sides of a triangle have lengths A and B, then . . .

DIFFERENCE between A and B < length of third side < SUM of A and B

ASIDE: If x > 0, then we can be certain that 3x+4 is greater than 2x+1

Let's plug the two given lengths into the above property to get: (3x+4) - (2x+1) < length of 3rd side < (3x+4) + (2x+1)

Simplify to get: x + 3 < length of 3rd side < 5x + 5

So any length that could satisfy the above inequality will be a possible length.

So let's start testing!!!

A) 4x+5

Plug this value into our inequality: x+3 < 4x+5 < 5x+5

If x = 1, the inequality holds.

So answer choice A COULD be the length of the third side.

B) x+2

Plug this value into our inequality: x+3 < x+2 < 5x+5

Let's focus on this part of the inequality: x+3 < x+2

Subtract x from both sides to get: 3 < 2

At this point, we can conclude that there are no possible values of x that will satisfy the inequality.

So answer choice B CANNOT be the length of the third side.

C) 6x+1

Plug this value into our inequality: x+3 < 6x+1 < 5x+5

If x = 1, the inequality holds.

So answer choice C COULD be the length of the third side.

D) 5x+6

Plug this value into our inequality: x+3 < 5x+6 < 5x+5

Let's focus on this part of the inequality: 5x+6 < 5x+5

Subtract 5x from both sides to get: 6 < 5

At this point, we can conclude that there are no possible values of x that will satisfy the inequality.

So answer choice D CANNOT be the length of the third side.

E) 2x+17

Plug this value into our inequality: x+3 < 2x+17 < 5x+5

Don't forget that all we need is one value of x that satisfies the above any quality in order to show that the answer choice COULD be the length of the third side.

After a bit of testing I found that when x = 10, the inequality holds.

So answer choice E COULD be the length of the third side.

Answer: A, C, E

Re: If x>0, and two sides of a certain triangle
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23 Jun 2024, 06:28

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