The Pythagorean Theorem for right triangles states a relationship between the three sides. Let \(a\) and \(b\) be the lengths of the two legs, and let \(c\) be the length of the hypotenuse. Then,
$$a^2 + b^2 = c^2$$
However, if a triangle is not a right triangle, then this equation is not true. There are certain inequalities that will hold for acute and obtuse triangles that are very similar to the Pythagorean Theorem. Those inequalities will be introduced here, along with how to use them.
Also make sure that the following inequalities are true:
$$a + b > c$$ $$a + c > b$$ $$b + c > a$$
Otherwise you have a degenerate triangle where the sides are not all connected. This is the statement provided by the Triangle Inequality and applies to all triangles.
Example 1: Consider a right triangle with legs of lengths \(a\) and \(b\) and a hypotenuse of length \(c\). Then
$$a^2 + b^2 = c^2$$
If you try to cut the length of the hypotenuse at one end of the hypotenuse, you will form a degenerate triangle unless you pull the two legs closer together, thus shrinking the angle in between them. Therefore the triangle now has three acute angles and is an acute triangle. Since only the original hypotenuse decreased in length, and the other two sides stayed the same length,
$$a^2 + b^2 > c^2$$
This is the Pythagorean Theorem for Acute Triangles.
Be warned that there is no definitive hypotenuse for a triangle that is acute or obtuse. Therefore we can assign the side lengths of these triangles to the three variable slots in the formula (\(a\), \(b\), and \(c\)) in any configuration we wish.
Example 2: Show that a triangle with sides of lengths \(6\), \(7\), and \(8\) is an acute triangle.
Solution: The sum of the squares of two of the sides must be greater than the square of the third. However, we must test all three distinct arrangements of the side lengths into the inequality for acute triangles. Plugging the side lengths into the formula in the three different configurations gives:
$$6^2 + 7^2 > 8^2$$ $$6^2 + 8^2 > 7^2$$ $$7^2 + 8^2 > 6^2$$ $$\Downarrow$$ $$36 + 49 > 64$$ $$36 + 64 > 49$$ $$49 + 64 > 36$$ $$\Downarrow$$ $$85 > 64$$ $$100 > 49$$ $$113 > 36$$
So all three conditions of the inequality are satisfied, and thus this triangle appears to be an acute triangle. Lastly we must check the conditions of the Triangle Inequality. Our side lengths are \(6\), \(7\), and \(8\), so we will rotate those values in for \(a\), \(b\), and \(c\) in the Triangle Inequality (this is how we check to ensure a triangle is non-degenerate whenever we know all the side lengths):
$$6 + 7 > 8$$ $$6 + 8 > 7$$ $$7 + 8 > 6$$
Simplify these inequalities:
$$13 > 8$$ $$14 > 7$$ $$15 > 6$$
These are all true, so the triangle is non-degenerate.
Example 3: An acute non-degenerate triangle has two sides of lengths \(11\) and \(13\). Use the Pythagorean Inequality for acute triangles to find a range of possible side lengths for the third side.
Solution: We know that
$$a^2 + b^2 > c^2$$
for all three distinct combinations of setting up the inequality. Let the unknown side be in the slot of \(a\) in the formula in one case, \(b\) in the second, and \(c\) in the third. However we will call this unknown side \(c\) every time. Thus we get
$$c^2 + 11^2 > 13^2$$ $$13^2 + c^2 > 11^2$$ $$11^2 + 13^2 > c^2$$
Simplify these inequalities:
$$c^2 + 121 > 169$$ $$169 + c^2 > 121$$ $$121 + 169 > c^2$$ $$\Downarrow$$ $$c^2 > 48$$ $$c^2 > -48$$ $$290 > c^2$$
We can ignore the second inequality because we cannot take the square root of a negative number. Therefore, \(c\) fits these restrictions:
$$48 < c^2 < 290$$
Take the square root of this inequality:
$$\sqrt{48} < c^2 < \sqrt{290}$$
We do not have to worry about the \(\pm\) in front of the square roots because side lengths cannot be negative. Also notice that even if \(c^2\) were to equal \(\sqrt{290}\), the other two sides combined would still be longer than it. Still, we don't need to check all the Triangle Inequality conditions because the triangle is assumed to be non-degenerate.
The procedure for finding an inequality for obtuse triangles is similar to that for acute triangles. It is demonstrated in Example 4 below.
Example 4: For a right triangle with sides of lengths \(a\), \(b\), and \(c\):
$$a^2 + b^2 = c^2$$
Obviously \(c\) is the longest side and is the hypotenuse. However, if we make it even longer while keeping the other two sides the same length, the angle between the two shorter sides will become larger as those two sides are separated. This angle between the two shorter sides was the right angle of the triangle, so it is now an obtuse angle, and we have an obtuse triangle. Since the length of \(c\) increased and \(a\) and \(b\) stayed the same:
$$a^2 + b^2 < c^2$$
This applies only for obtuse triangles. You must be able to interchange the three side lengths in the different variable slots in this formula, and the inequality must be satisfied every time to identify a triangle as obtuse.
Example 5: Show that the triangle with sides of lengths \(14\), \(15\), and \(17\) is not an obtuse triangle.
Solution: For this triangle to not be obtuse, the inequality
$$a^2 + b^2 < c^2$$
must not be satisfied for some placement of the side lengths into this inequality. First, let \(a = 14\), \(b = 15\), and \(c = 17\):
$$14^2 + 15^2 < 17^2 \Rightarrow$$ $$196 + 225 < 289 \Rightarrow$$ $$421 < 289$$
This inequality is not true. Despite the possibility of other arrangements of the side lengths into this inequality being true, this one false statement singlehandedly proves that the triangle is not an obtuse triangle.
Example 6: Two sides of an obtuse triangle have lengths \(11\) and \(17\). If the third side is longer than both of these sides, what is the range of possible side lengths for the third side?
Solution: Since this is an obtuse triangle, the following inequality must be satisfied:
$$11^2 + 17^2 < c^2$$
Here, we let \(c\) equal the length of the third side. We can simplify this inequality:
$$121 + 289 < c^2 \Rightarrow$$ $$410 < c^2$$
Take the square root of both sides of this inequality:
$$\sqrt{410} < c$$
However, this third side cannot be infinitely long. The Triangle Inequality comes into play. It states that for a triangle to not be degenerate,
$$a + b > c$$
Since the shorter sides have lengths of \(11\) and \(13\), the third side's length must be less than that of the shorter two combined, so \(c\) is less than \(24\), giving a range of
$$\sqrt{410} < c < 24$$
Here we will use the side lengths of a triangle to classify it as right, acute, or obtuse.
Example 7: In Example 5, we showed that a triangle with sides of lengths \(14\), \(15\), and \(17\) cannot be obtuse. Determine whether the triangle is acute or right.
Solution: The triangle must be either a right triangle or an acute triangle. It is much easier to test if the triangle is a right triangle because we only need to plug variables in one time. If it is a right triangle, the side of length \(17\) is the hypotenuse, so in the Pythagorean Theorem \(c = 17\). It does not matter which variable we assign to which of the other two variables because the result will be the same. So just let \(a = 14\) and \(b = 15\). If this is a right triangle, then by the Pythagorean Theorem this equation is true:
$$a^2 + b^2 = c^2 \Rightarrow$$ $$14^2 + 15^2 = 17^2 \Rightarrow$$ $$196 + 225 = 289 \Rightarrow$$ $$421 = 289$$
This is not true. In fact, \(a^2 + b^2 > c^2\), so by Pythagoras's Inequalities this triangle must be acute.
Example 8: Is a triangle with sides of lengths \(3\), \(5\), and \(7\) acute, right, or obtuse?
Solution: It is easiest to determine whether the triangle is a right triangle with the Pythagorean Theorem. The hypotenuse would have a length of \(7\). Therefore \(c = 7\). By the Pythagorean Theorem, this is a right triangle if
$$3^2 + 5^2 = 7^2$$
Simplify this equation:
$$9 + 25 = 49 \Rightarrow$$ $$34 = 49$$
This is not true. So the triangle is not a right triangle.
However, this tells us something else. It also shows us that \(a^2 + b^2 < c^2\), so this cannot be an acute triangle either. We conclude that this is an obtuse triangle.
However, we must use the Triangle Inequality to check that this triangle is not degenerate:
$$3 + 5 > 7$$ $$3 + 7 > 5$$ $$5 + 7 > 3$$
Simplify these inequalities to verify that they are all true:
$$8 > 7$$ $$10 > 5$$ $$12 > 3$$
These inequalities are all true, so the Triangle Inequality is satisfied. Therefore this triangle is not degenerate, and it is an obtuse triangle.
Example 9: A triangle has sides of lengths \(7\), \(24\), and \(25\). Show that this triangle is not degenerate, and then determine whether it is right, acute, or obtuse.
Solution: To show that this triangle is not degenerate, we must show that
$$a + b > c$$ $$b + c > a$$ $$a + c > b$$
In a right triangle, usually the hypotenuse is labeled \(c\), so let's just let \(a = 7\), \(b = 24\), and \(c = 25\). So plug those values into the three displayed conditions for the Triangle Inequality:
$$7 + 24 > 25$$ $$24 + 25 > 7$$ $$7 + 25 > 24$$
Simplify these inequalities:
$$31 > 25$$ $$49 > 7$$ $$32 > 24$$
These inequalities are all true, so the triangle is not degenerate.
We next determine if this is a right triangle. The hypotenuse would be the side with a length of \(25\), so by the Pythagorean Theorem:
$$7^2 + 24^2 = 25^2 \Rightarrow$$ $$49 + 576 = 625 \Rightarrow$$ $$625 = 625$$
This equation is true, so the triangle is a right triangle.
It is important to be able to use the sides of a triangle to classify the triangle as acute, obtuse, or right. It is also important to be able to construct triangles with given restrictions (i.e. it must be a certain type of triangle, etc.). Lastly, be sure to check that the triangle is not degenerate, because if it is, neither Pythagoras's Inequalities nor the Pythagorean Theorem will apply.