Ninth Grade Mathematics: Triangle Inequality Theorems
While congruence criteria determine when two triangles are identical, geometric inequalities govern the comparative sizes of sides and angles within a triangle, establishing the boundary conditions required for a closed polygon to exist. In this chapter, you will master the fundamental Triangle Inequality Theorems that dictate which side lengths can form a triangle and how angle sizes correlate with opposite side lengths.
The Side-Angle Inequality Theorem
In any triangle, the size of an interior angle is directly proportional to the length of the side facing it. A wider angle opening requires a longer opposite side to connect the arms.
Theorem 1 (Angle-Side Inequality):
If two sides of a triangle are unequal, the angle opposite to the longer side
is strictly greater than the angle opposite to the shorter side.
In Triangle ABC:
If side AC > side AB, then Angle B > Angle C.
A
*
/ \
/ \
c / \ b (b > c)
/ \
/ \
*-----------*
B C
Angle B > Angle C
Conversely, if two angles of a triangle are unequal, the side opposite to the larger angle is strictly longer than the side opposite to the smaller angle.
Corollary for Right and Obtuse Triangles:
1. In a right-angled triangle, the right angle (90 deg) is the largest angle.
Therefore, the hypotenuse is strictly the longest side of a right triangle.
2. In an obtuse triangle, the obtuse angle (> 90 deg) is the largest angle.
Therefore, the side opposite the obtuse angle is the longest side.
The Fundamental Triangle Inequality Theorem
Can any three arbitrary sticks form a triangle? If you have sticks of length 2 cm, 3 cm, and 10 cm, the two shorter sticks cannot meet because their combined length is far shorter than the 10 cm stick.
The Fundamental Triangle Inequality Theorem:
The sum of the lengths of any two sides of a triangle is strictly greater
than the length of the third side.
For any triangle with sides a, b, and c:
1. a + b > c
2. b + c > a
3. c + a > b
Sticks 2, 3, 10 (IMPOSSIBLE) Valid Triangle (5, 7, 10)
*
2 3 / \
*-----* *-------* 5 / \ 7
------------------------- / \
10 *-------*
(Cannot meet! 2 + 3 < 10) 10 (5 + 7 > 10)
If the sum of the two shorter sides equals the third side (a + b = c), the three points collapse into a straight line segment (degenerate triangle). For a non-degenerate triangle to exist, the sum must be strictly greater.
Range of Possible Lengths for the Third Side
When two side lengths a and b of a triangle are fixed (with a >= b), the Triangle Inequality Theorem establishes a strict double inequality for the allowed length of the third side x:
Range of the Third Side:
Difference of sides < Third Side < Sum of sides
|a - b| < x < a + b
Example: Two sides of a triangle have lengths 7 cm and 11 cm.
What are the possible lengths for the third side x?
Lower bound: 11 - 7 = 4 cm.
Upper bound: 11 + 7 = 18 cm.
Allowed range: 4 cm < x < 18 cm.
Any length between 4 cm and 18 cm can form a valid triangle; lengths less than or equal to 4 cm, or greater than or equal to 18 cm, cannot.
The Perpendicular as Shortest Distance
A critical consequence of the triangle inequality is that the shortest path from a point to a line is the perpendicular segment:
Theorem:
Of all line segments that can be drawn to a given line from a point not lying on it,
the perpendicular line segment is the shortest.
P (External Point)
*
/|\
/ | \
/ | \
/ | \
-------------*----+----*------------- Line L
A N B
Perpendicular PN is shortest!
In right triangle PNA, hypotenuse PA is opposite the 90-degree angle N. Since the hypotenuse is the longest side, PA > PN. Therefore, the perpendicular PN is strictly shorter than any oblique segment.
Chapter Exercises
Exercise 1: Determine whether it is possible to construct a triangle with the following sets of side lengths: (a) 4 cm, 6 cm, 9 cm; (b) 3 cm, 5 cm, 8 cm; (c) 7 cm, 2 cm, 11 cm.
Exercise 2: The lengths of two sides of a triangle are 9 cm and 14 cm. Between what two numbers must the length of the third side fall?
Exercise 3: Point O is any point inside a triangle ABC. Prove that OA + OB + OC is strictly greater than half the perimeter of triangle ABC: (1/2) * (AB + BC + CA).
Solutions and Explanations
Solution 1: For set (a): the sum of the two shorter sides is 4 + 6 = 10 cm. Since 10 > 9, all three inequality conditions hold (4+6 > 9, 4+9 > 6, 6+9 > 4), so a triangle can be formed. For set (b): the sum of the two shorter sides is 3 + 5 = 8 cm. Since 8 is not strictly greater than 8 (8 = 8), the points lie on a straight line and cannot form a triangle. For set (c): the sum of the two shorter sides is 7 + 2 = 9 cm. Since 9 < 11, the segments cannot meet, so no triangle can be formed.
Solution 2: Let the two known sides be a = 14 cm and b = 9 cm, and let the third side be x. By the Triangle Inequality Theorem, the third side must be strictly less than the sum of the two sides: x < 14 + 9, so x < 23 cm. Furthermore, the sum of the third side and the smaller side must exceed the larger side: x + 9 > 14, which means x > 14 - 9, so x > 5 cm. Therefore, the length of the third side must fall strictly between 5 cm and 23 cm: 5 cm < x < 23 cm.
Solution 3: Consider the three smaller triangles formed by connecting interior point O to the vertices: triangle OAB, triangle OBC, and triangle OCA. In triangle OAB, by the Triangle Inequality Theorem: OA + OB > AB (Inequality 1). In triangle OBC: OB + OC > BC (Inequality 2). In triangle OCA: OC + OA > CA (Inequality 3). Add all three inequalities together: (OA + OB) + (OB + OC) + (OC + OA) > AB + BC + CA. Combine like terms on the left side: 2OA + 2OB + 2*OC > AB + BC + CA, which is 2 * (OA + OB + OC) > AB + BC + CA. Dividing both sides by 2 yields OA + OB + OC > (1/2) * (AB + BC + CA). The sum of distances from any interior point to the vertices is strictly greater than the semi-perimeter.