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If $10$, $12$ and '$x$' are sides of an acute angled triangle, how many integer values of '$x$' are possible ?

- $7$
- $12$
- $9$
- $13$

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**Option C. 9**

Acute Angle : between 0 and 89 degrees.

*Case 1: X is the longest side*

then X < $\sqrt{10^{2}+12^{2}}$ i.e. X < 16

*Case 2: 12 is the longest side.*

then X > $\sqrt{12^{2}-10^{2}}$ i.e. X > 6

Therefore x ranges from 7 to 15 = **9 Values**

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Angles can be classified into five groups, based on their measure in degrees.

**Acute**: angles with measure $< 90^\circ$**Right**: angles with measure $= 90^\circ$**Obtuse**: angles with measure $> 90^\circ$ and $< 180 ^ \circ$**Straight**: angles with measure $= 180^ \circ$**Reflex**: angles with measure $> 180^ \circ$> and $< 360 ^ \circ$

If $a, b,$ and $c$ are the $3$ sides of an acute triangle where $c$ is the longest side then $c^{2} < a^{2} + b^{2}$

The sides are $10, 12,$ and $'x'.$

Case$1:$ Among the $3$ sides $10, 12,$ and $x,$ for values of $x \leq 12,\; 12 $ is the longest side.

When $x \leq 12,$ let us find the number of values for $x$ that will satisfy the inequality $12^{2} < 10^{2} + x^{2}$

$\implies144 < 100 + x^{2}$

$\implies 44<x^{2}$

$\implies x^{2}>44$

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.

Case$2:$ For values of $x > 12, x$ is the longest side.

Let us count the number of values of $x$ that will satisfy the inequality $x^{2} < 10^{2} + 12^{2}$

i.e., $x^{2} < 244$

$x = 13, 14,$ 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 $x = 7, 8, 9, 10, 11, 12, 13, 14, 15.$

So, the correct answer is $(C).$

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