# Math Help - Acute Angled Triangle Question

1. ## Acute Angled Triangle Question

Hi All,
In the following question and answer explanation. 1. Can anyone explain why for the Rule 2 part of the explanation below we start at a value of 3, (why not 1 or 2)? 2. How is it known that the values 3,4,5,6,16,17,18,19,20,21 will result in an obtuse angled triangle? I.E. How do we know these values will produce an angle greater than 90degrees?

Question....
If 10, 12 and 'x' are sides of an acute angled triangle, how many integer values of 'x' are possible?
1. 7
2. 12
3. 9
4. 13
5. 11

Finding the answer to this question requires one to know two rules in geometry.

Rule 1: For an acute angled triangle, the square of the LONGEST side MUST BE LESS than the sum of squares of the other two sides.

Rule 2: For any triangle, sum of any two sides must be greater than the third side.

The sides are 10, 12 and 'x'.

From Rule 2, x can take the following values: 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21 – A total of 19 values.

When x = 3 or x = 4 or x = 5 or x = 6, the triangle is an OBTUSE angled triangle (Rule 1 is NOT satisfied).

The smallest value of x that satisfies BOTH conditions is 7. (102 + 72 > 122).

The highest value of x that satisfies BOTH conditions is 15. (102 + 122 > 152).

When x = 16 or x = 17 or x = 18 or x = 19 or x = 20 or x = 21, the triangle is an OBTUSE angled triangle (Rule 1 is NOT satisfied).

Hence, the values of x that satisfy both the rules are x = 7, 8, 9, 10, 11, 12, 13, 14, 15. A total of 9 values.

2. To answer your first question, you can't choose side lengths of 1 or 2 because $\displaystyle 10 + 1 = 11 < 12$, so rule 2 isn't satisfied, and $\displaystyle 10 + 2 = 12 = 12$, so rule 2 isn't satisfied.

$\displaystyle c^2 = a^2 + b^2 - 2ab\cos{C}$.

You should know that when $\displaystyle \angle{C} < 90^{\circ}, \cos{C} > 0$ and therefore $\displaystyle 2ab\cos{C} > 0$. So that means that $\displaystyle c^2 < a^2 + b^2$, since you're subtracting something positive.

However, when $\displaystyle \angle{C} = 90^{\circ}$, this is a right-angle triangle and an application of Pythagoras' Theorem, and when $\displaystyle \angle{C} > 90^{\circ}, \cos{C} < 0$ and therefore $\displaystyle 2ab\cos{C} < 0$. So that means that $\displaystyle c^2 > a^2 + b^2$, since you are subtracting something negative (i.e. adding something).

3. Hello, dumluck!

$\text{If }10, 12\text{ and }x\text{ are sides of an acute triangle,}$
. . $\text{how many integer values of }x\text{ are possible?}$

. . $(a)\;7 \qquad (b)\;12 \qquad (c)\;9 \qquad (d)\;13 \qquad (e)\;11$

$\text{Rule 1: For an acute triangle, the square of the longest side must}$
. . . . . . . . . $\text{be }less\:than\text{ the sum of squares of the other two sides.}$

$\text{Rule 2: For any triangle, sum of any two sides must be}$
. . . . . . . . . $\text{greater than the third side.}$
Code:

*
*  *
*     *   12
10 *        *
*           *
*              *
*  *  *  *  *  *  *
x

From Rule 1: . $x^2 \:<\:10^2 + 12^2 \quad\Rightarrow\quad x^2 \:<\<244 \quad\Rightarrow\quad x \:<\:16$

From Rule 2: . $10 + x\:>\:12 \quad\Rightarrow\quad x \:>\:2$

Hence: . $x \,\in\, \{3,4,5,6,7,8,9,10,11,12,13,14,15\}$

$\,x$ can have 13 integer values . . . answer (d).

4. Originally Posted by Soroban
Hello, dumluck!

Code:
           *
*  *
*     *   12
10 *        *
*           *
*              *
*  *  *  *  *  *  *
x
From Rule 1: . $x^2 \:<\:10^2 + 12^2 \quad\Rightarrow\quad x^2 \:<\<244 \quad\Rightarrow\quad x \:<\:16$

From Rule 2: . $10 + x\:>\:12 \quad\Rightarrow\quad x \:>\:2$

Hence: . $x \,\in\, \{3,4,5,6,7,8,9,10,11,12,13,14,15\}$

$\,x$ can have 13 integer values . . . answer (d).

Spot on, thanks for that. Is their any good resources for learning to translating these questions into expressions like that? The above looks obvious when you have the answer.

5. Originally Posted by Soroban
Hello, dumluck!

Code:

*
*  *
*     *   12
10 *        *
*           *
*              *
*  *  *  *  *  *  *
x

From Rule 1: . $x^2 \:<\:10^2 + 12^2 \quad\Rightarrow\quad x^2 \:<\<244 \quad\Rightarrow\quad x \:<\:16$

From Rule 2: . $10 + x\:>\:12 \quad\Rightarrow\quad x \:>\:2$

Hence: . $x \,\in\, \{3,4,5,6,7,8,9,10,11,12,13,14,15\}$

$\,x$ can have 13 integer values . . . answer (d).

But notice that if x = 3, 4, 5 or 6 then $10^2+x^2<12^2$. So those also give obtuse-angled triangles and must be ruled out by Rule 1, leaving a total of 9 possible values for x, as indicated in the original post.

6. Originally Posted by Opalg
But notice that if x = 3, 4, 5 or 6 then $10^2+x^2<12^2$. So those also give obtuse-angled triangles and must be ruled out by Rule 1, leaving a total of 9 possible values for x, as indicated in the original post.
Thanks Opalg. Just for my own clarificiation though, using the formula above and substituting for 3,4,5,6 = 10 + 3 < 12 : 13 < 12. Is it because 13 is NOT less than twelve we know that these are obtuse?

7. Originally Posted by dumluck
Thanks Opalg. Just for my own clarificiation though, using the formula above and substiruting for 3,4,5,6 = 10 + 3 < 12 : 13 < 12. Is it because 13 is NOT less than twelve we know that these are obtuse?
No, you have to square the numbers. If x=3, for example, then $10^2+x^2 = 10^2+3^2 = 100+9 = 109$. But $12^2=144$. So $10^2+3^2<12^2$. The same thing works for x=4, 5 and 6. When x=6, for example, $x^2=36$, and then we get $10^2+x^2 = 136$, which is still less than 144. But x=7 is okay, because $7^2=49$, and 149 is greater than 144.

8. Originally Posted by Opalg
No, you have to square the numbers. If x=3, for example, then $10^2+x^2 = 10^2+3^2 = 100+9 = 109$. But $12^2=144$. So $10^2+3^2<12^2$. The same thing works for x=4, 5 and 6. When x=6, for example, $x^2=36$, and then we get $10^2+x^2 = 136$, which is still less than 144. But x=7 is okay, because $7^2=49$, and 149 is greater than 144.
Thanks Opalg. I get it , I'm refering to the inequality rule. Which is that given side (12) must be the less than the sum of the other two sides. Your formula is for evaluating a vertices. What's the mathematical term for THAT formula?

Thanks,