If 10, 12 and ‘x’ are sides of an acute angled triangle, ho

Actually, chix475ntu, jeetesh and gurpreet have all given the correct answer either using diagrams or formulas.. 6 < x < 16 is the same as values from 7 to 15 because only integral values are allowed.

Anyway, I am putting down the explanation.
The question asks you for an acute triangle i.e. a triangle with all angles less than 90. An obtuse angled triangle has one angle more than 90. So the logic is that before one of the angles reaches 90, find out all the values that x can take.
Starting from the first diagram where x is minimum and the angle is very close to 90, to the 2nd diagram where all angles are much less than 90 to the third diagram where the other angle is going towards 90.



Note: The remaining angle cannot be 90 because that will make 10 the hypotenuse but hypotenuse is always the greatest side.

In the leftmost diagram \(x = \sqrt{(12^2 - 10^2)}\)
x = root 44 which is 6.something
x should be greater than 6.something because the angle cannot be 90.

In the rightmost diagram, \(x = \sqrt{(12^2 + 10^2)}\)
x = root 244 which is 15.something
x should be less than 15.something so that the angle is not 90.

Values that x can take range from 7 to 15 which is 15 - 7 = 8 + 1 = 9 values.

Check out this video for an explanation on this question: https://youtu.be/y05XIAWgAT0