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# Geometry

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Two sides of an acute triangle are 8 and 15. How many possible lengths are there for the third side, if it is a positive integer?

eileenthecoolbean  Aug 15, 2017
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#1
+6794
+1

Let x be the length of the 3rd side.

From triangle inequality, we know:

8 + 15 > x

x < 23

and

8 + x > 15

x > 7

combine the 2 results, we get:

7 < x < 23.

The question now becomes:

Find the number of positive integers that satisfies 7 < x < 23.

We can count the answer, which is 15.

Therefore there are 15 possible lengths for the 3rd side, if it is a +ve int.

MaxWong  Aug 15, 2017
#2
+1114
+1

Maxwong has the correct idea, but I think missed the criterion that states that the triangle must be acute. With this restriction, there are not as many possibilities.

Have you ever heard of the Pythagorean Theorem? You probably have. But have you heard of the Pythagorean Inequalities Theorem? Maybe you have and maybe you haven't. I think my textbook summarizes the concept nicely and in a way that is fairly straightforward:

Of course, we seek triangles that only abide to acute triangles. We know 2 side lengths: 8 and 15. The remaining side I have labeled as s. Of course, there are 2 cases that are possible with s. Either it is the largest side or it is not. We will have to take both into account to solve this prolem:

For the first example, I will assume that s is the longest side, which corresponds to c in the picture above. Now, let's solve:

$$c^2 s is c. Whichever side you plug in for a and b does not matter. \(s^2<8^2+15^2$$Simplify the right hand side of the inequality to solve for s
$$s^2<64+225$$
$$s^2<289$$Take the square root of both sides.
$$|s|<17$$In an inequality, the absolute value has the following rule$$|a| . Now, let's apply it.  \(s<17$$ $$-s<17$$

One inequality is already done. Divide both sides by -1 to get the other answer. Don't forget to flip the inequality sign!
 $$s<17$$ $$s>-17$$

We can clean up this solution by recognizing that we can represent the solutions as a compound inequality.
$$-17 Now, let's solve for s when is not the longest side: \(c^2 This time, when we substitute, b will be in either a or b. \(15^2 Simplify both sides. \(225 SUbtract 64 on both sides. \(161 Take the square root of both sides. \(\sqrt{161}<|s|\hspace{1mm}\text{or}\hspace{1mm}|s|>\sqrt{161}$$The same logic applies for greater than symbols, too. In other words, $$|a|>b\rightarrow a>b\hspace{1mm}\text{and}\hspace{1mm} -a>b$$
 $$s>\sqrt{161}$$ $$-s>\sqrt{161}$$

Just like above, divide by -1 on the right inequality.
 $$s>\sqrt{161}$$ $$s<-\sqrt{161}$$

These inequalities cannot be combined into a compound inequality, unfortunately.

The triangle must adhere to both conditions of $$-17 and ( \(s>\sqrt{161}$$ or $$s<-\sqrt{161}$$ )

Let's think about this logicall before our head explodes, though! If s is less than 17 and greater than √161, that means the following compound inequality arises:

$$\sqrt{161} We don't care about the other inequality as it contains negative numbers, and negative numbers are nonsensical in the context of this problem, so I have excluded them. We aren't done yet! We have to figure out the first integer that is greater that 161. Well, \(\sqrt{161}$$ is in between 12 and 13, so 13 is the first integer that is allowable under our given conditions.

Therefore, the only possible integer solutions are 13, 14, 15, and 16.

TheXSquaredFactor  Aug 15, 2017

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