The triangle ABC is an acute triangle with AB = 7, AC = 9 and x = BC

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? 4.9ps.png A. 6 B. 9 C. 11 D. 12 E. 14

chetan2u · Accepted Answer

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.

gurmukh · Answer

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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GMATinsight · Answer

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...

gurmukh · Answer

MathRevolution · Answer

=> 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

Fdambro294 · Answer

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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AkshaySitlani07 · Answer

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}.

The triangle ABC is an acute triangle with AB = 7, AC = 9 and x = BC

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The triangle ABC is an acute triangle with AB = 7, AC = 9 and x = BC

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