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805+ Level|   Geometry|      
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MathRevolution
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The triangle ABC is an acute triangle with AB = 7, AC = 9 and x = BC Updated on: 09 Apr 2020, 22:31
Originally posted by MathRevolution on 09 Apr 2020, 19:12.
Last edited by chetan2u on 09 Apr 2020, 22:31, edited 1 time in total.
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[GMAT math practice question]

The triangle \(ABC\) is an acute triangle with \(AB = 7, AC = 9\) and \(x = BC\) in the figure. How many possible integers of \(x\) do we have?

Attachment:
4.9ps.png
4.9ps.png [ 8.57 KiB | Viewed 5130 times ]

A. \(6\)

B. \(9\)

C. \(11\)

D. \(12\)

E. \(14\)
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A
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The triangle ABC is an acute triangle with AB = 7, AC = 9 and x = BC 09 Apr 2020, 22:30
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[GMAT math practice question]

The triangle \(ABC\) is an acute triangle with \(AB = 7, AC = 9\) and \(x = BC\) in the figure. How many possible integers of \(x\) do we have?

Attachment:
4.9ps.png

A. \(7\)

B. \(9\)

C. \(11\)

D. \(12\)

E. \(14\)
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Firstly when will these be ACUTE angles??
Right angled triangle has the \((L)^2=(S_1)^2+(S_2)^2\)..
So for acute angled triangle ...\((L)^2<(S_1)^2+(S_2)^2\)..

You can distribute these into two cases..

1) when x is the largest
So \(x^2<7^2+9^2....x^2<130....11^2=121..and ...12^2=144\), so MAX value of x is between 11 and 12
Thus, the MAX value of x is 11

2) When 9 is the largest
\(9^2<7^2+x^2....x^2>32....5^2=25..and ...6^2=36\), so MIN value of x is between 5 and 6
Thus, the MIN value of x is 6

Possible values of x from the range \(6\leq{x}\leq{11}\) are 6, 7, 8, 9, 10 and 11.
So 6 values of x possible.

MathRevolution, you have not taken the limitation in MIN value of x, so you are getting wrong answer.
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The triangle ABC is an acute triangle with AB = 7, AC = 9 and x = BC 09 Apr 2020, 21:13
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If angle A is 90 degree the x = sqt of 130 which is 11.2 something, now triangle is acute angle therefore x can take value < = 11.Also difference to two sides should be greater then third side therefore x > 2.
Hence values possible for x are 3,4,5,6,7,8,9,10,11
Nine values are possible.
Option B is the answer.

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The triangle ABC is an acute triangle with AB = 7, AC = 9 and x = BC 09 Apr 2020, 22:25
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MathRevolution
[GMAT math practice question]

The triangle \(ABC\) is an acute triangle with \(AB = 7, AC = 9\) and \(x = BC\) in the figure. How many possible integers of \(x\) do we have?

Attachment:
4.9ps.png

A. \(7\)

B. \(9\)

C. \(11\)

D. \(12\)

E. \(14\)
Show more

In any Acute angled Triangle


\(c^2 < a^2 + b^2\) where a, b and c are sides of triangle and c is the longest side

AB = 7, AC = 9 and BC = x

Case 1: x is longest side,\( x^2 < 7^2 + 9^2\)

i.e. \( x^2 < 130\)

i.e. \( x < 11.3\)

Case 2: 9 is longest side,\( 9^2 < 7^2 + x^2\)

i.e. \( x^2 > 32\)

i.e. \( x > 5.6\)

i.e. 5.6 < x < 11.3

i.e. x = {6, 7, 8, 9, 10, 11}

I am getting 6 values...
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The triangle ABC is an acute triangle with AB = 7, AC = 9 and x = BC 09 Apr 2020, 22:44
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gurmukh
If angle A is 90 degree the x = sqt of 130 which is 11.2 something, now triangle is acute angle therefore x can take value < = 11.Also difference to two sides should be greater then third side therefore x > 2.
Hence values possible for x are 3,4,5,6,7,8,9,10,11
Nine values are possible.

The question should be angle A is acute, I think it's a typo in the question seeing the answer.
If the triangle is acute then 3,4 5 will go out and there will be six choices.

Option B is the answer.

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The triangle ABC is an acute triangle with AB = 7, AC = 9 and x = BC 12 Apr 2020, 04:58
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Since \(7, 9\) and \(x\) are the sides of a triangle, we have \(9 – 7 = 2 < x < 16 < 9 + 7.\)

Case 1: \(x\) is the largest length (\( x ≥ 9\) )

Since the triangle is acute, we have \(x^2 < 7^2 + 9^2 = 49 + 81 = 130\) or \(x < √130 < 12.\)

Case 2: \(9\) is the largest length ( \(x < 9\) )

Since the triangle is acute, we have \(7^2 + x^2 > 9^2\) or \(x^2 > 81 - 49 = 32\) or \(x > √32 > 5.\)

Then we have \(6, 7, …, 11\) as the possible values of \(x.\)

Thus, the number of possible values of \(x\) is \(11 – 6 + 1 = 6.\)

Therefore, A is the answer.
Answer: A
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The triangle ABC is an acute triangle with AB = 7, AC = 9 and x = BC 31 May 2021, 00:48
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For an acute triangle, the following relationship will always hold:

(Longest Length)^2 < (A)^2 + (B)^2


Where A and B are the lengths of the Two shorter sides


The triangle inequality theorem states that X must be:

(9 - 7) < X < (7 + 9)

2 < X < 16


Case 1: X >/= 9 —-making X the longest side

(X)^2 < (9)^2 + (7)^2

X < sqrt(130)

Integer values or X can be: 9, 10, 11


Case 2: 9 is the longest side and X < 9

(9)^2 < (X)^2 + (7)^2

(X)^2 > 32

X > sqrt(32)

X can equal the integer values of: 6, 7, 8


Total possible values for X

6, 7, 8, 9, 10, 11

(A) six values

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The triangle ABC is an acute triangle with AB = 7, AC = 9 and x = BC 31 May 2021, 01:48
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Using the law of triangle, we can say the following three statements:
x + 7 > 9
x + 9 > 7
9 + 7 > x

From these three statements, we get 2 < x < 16.

This gives the following 13 values {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15} —————— (1)

Now, for an acute angled triangle, the following conditions will also hold true:
(AB)^2 (BC)^2 > (AC)^2
(AB)^2 (AC)^2 > (BC)^2
(AB)^2 (AC)^2 > (AB)^2

On substituting the values, we get 32 < (BC)^2 < 130, which gives x = {6, 7, 8, 9, 10, 11} —————— (2)

From 1 and 2, we get 6 values, namely {6, 7, 8, 9, 10, 11}.
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