If 10, 12 and 'x' are sides of an acute angled triangle, how many integer values of 'x' are possible?	Choose the correct option:
A) 7
B) 12
C) 9
D) 13
E) 11
Explanatory Answer   Key property about sides of an acute triangle   Finding the answer to this question requires that you know this key property about sides of an acute triangle.   For an acute angled triangle, the square of the LONGEST side MUST BE LESS than the sum of squares of the other two sides.   If 'a', 'b', and 'l' are the 3 sides of an acute triangle where 'l' is the longest side then l2 &lt; a2 + b2   The sides are 10, 12, and 'x'.   Scenario 1: Among the 3 sides 10, 12, and x, for values of x ≤ 12, 12 is the longest side.   Scenario 2: For values of x &gt; 12, x is the longest side   Possibilities in scenario 1: When x ≤ 12, let us find the number of values for x that will satisfy the inequality 122 &lt; 102 + x2   i.e., 144 &lt; 100 + x2   The least integer value of x that satisfies this inequality is 7.   The inequality will hold true for x = 7, 8, 9, 10, 11, and 12. i.e., 6 values.   Possibilities in scenario 1: When x &gt; 12, x is the longest side.   Let us count the number of values of x that will satisfy the inequality x2 &lt; 102 + 122   i.e., x2 &lt; 244   x = 13, 14, and 15 satisfy the inequality. That is 3 more values.   Hence, the values of x for which 10, 12, and x will form sides of an acute triangle are x = 7, 8, 9, 10, 11, 12, 13, 14, 15. A total of 9 values.   9hoice 9
<result>C</result>
If 10, 12 and 'x' are sides of an acute angled triangle, how many integer values of 'x' are possible?