# 7.3: Triangle CIassifications

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Triangles may be classified according to the relative lengths of their sides:

• An equilateral triangle has three equal sides,
• An isosceles triangle has two equal sides.
• A scalene triangle has no equal sides.

Triangles may also be classified according to the measure of their angles:

• An acute triangle is a triangle with three acute angles.
• An obtuse triangle is a triangle with one obtuse angle.
• An equiangular triangle is a triangle with three equal angles,

Each angle of an equiangular triangle must be $$60^{\circ}$$, We will show in section 2, 5 that equiangular triangle are the same as equilateral triangles,

A right triangle is a triangle with one right angle, The sides of the right angle are called the legs of the triangle and the remaining side is called the hypotenuse,

##### Example $$\PageIndex{1}$$

Find $$x$$ if $$\triangle ABC$$ is isosceles with $$AC = BC$$:

Solution

$\begin{array} {rcl} {x + \dfrac{1}{2}} & = & {2x - \dfrac{1}{2}} \\ {(2)(x + \dfrac{1}{2})} & = & {(2) (2x - \dfrac{1}{2})} \\ {2x + 2 (\dfrac{1}{2})} & = & {(2)(2x) - (2) (\dfrac{1}{2})} \\ {2x + 1} & = & {4x - 1} \\ {1 + 1} & = & {4x - 2x} \\ {2} & = & {2x} \\ {1} & = & {x} \end{array}$

Check:

Answer: $$x = 1$$.

##### Example $$\PageIndex{2}$$

$$\triangle ABC$$ is equilateral. Find $$x$$:

Solution

$\begin{array} {rcl} {\dfrac{x}{2} + 5} & = & {\dfrac{x}{3} + 7} \\ {(6)(\dfrac{x}{2} + 5)} & = & {(6) (\dfrac{x}{3} + 7)} \\ {(6) (\dfrac{x}{2} + (6)(5)} & = & {(6)(\dfrac{x}{3} + (6)(7)} \\ {3x + 30} & = & {2x + 42} \\ {3x - 2x} & = & {42 - 30} \\ {x} & = & {12} \end{array}$

Check:

Answer: $$x = 12$$.

An altitude of a triangle is a line segment from a vertex perpendicular to the opposite·side, In Figure 4, $$CD$$ and $$GH$$ are altitudes, Note that altitude $$GH$$ lies outside $$\triangle EFG$$ and side $$EF$$ must be extended to meet it.

A median of a triangle is a line segment from a vertex to the midpoint of the opposite side, In Figure 5, CD is a median,

An angle bisector is a ray which divides an angle into two eaual angles. In Figure $$\PageIndex{6}$$, $$\overrightarrow{CD}$$ is an angle bisector.

##### Example $$\PageIndex{3}$$

Find $$AB$$ if $$CD$$ is a median:

Solution

$\begin{array} {rcl} {AD} & = & {DB} \\ {x^2} & = & {10x} \\ {x^2 - 10x} & = & {0} \\ {(x)(x - 10)} & = & {0} \end{array}$

$$\begin{array} {rclcrcl} {x} & = & {0} & \text{ or } & {x - 10} & = & {0} \\ {} & & {} & \text{ } & {x} & = & {10} \end{array}$$

Check, $$x = 0$$:

Check, $$x = 10$$:

We reject the answer $$x = 0$$ because the length of a line segment must be greater than 0, Therefore $$AB = AD + DB = 100 + 100 = 200$$.

Answer: $$AB = 200$$.

##### Example $$\PageIndex{4}$$

Find $$\angle ACB$$ if $$\overrightarrow{CD}$$ is an angle bisector:

Solution

$\begin{array} {rcl} {\anlge ACD} & = & {\angle BCD} \\ {x^2 + x} & = & {6x} \\ {x^2 + x - 6x} & = & {0} \\ {x^2 - 5x} & = & {0} \\ {(x)(x - 5)} & = & {0} \end{array}$

$$\begin{array} {rclcrcl} {x} & = & {0} & \text{ or } & {x - 5} & = & {0} \\ {} & & {} & \text{ } & {x} & = & {5} \end{array}$$

Check, $$x = 0$$:

Check, $$x = 5$$:

We reject the answer $$x = 0$$ because the measures of $$\angle ACD$$ and $$\angle BCD$$ must be greater than $$0^{\circ}$$. Therefore $$\angle ACB = \angle ACD + \angle BCD = 30^{\circ} + 30^{\circ} = 60^{\circ}$$.

Answer: $$\angle ACB = 60^{\circ}$$.

The perimeter of a triangle is the sum of the lengths of the sides. The perimeter of $$\triangle ABC$$ in Figure $$\PageIndex{7}$$ is $$3 + 4 + 5 = 12$$.

##### Theorem $$\PageIndex{1}$$

The sum of any two sides of a triangle is greater than the remaining side.

For example, in Figure $$\PageIndex{7}$$, $$AC + BC = 3 + 4 > AB = 5$$.

Proof

This follows from the postulate that the shortest distance between two noints is along a straight line, For example, in Figure $$\PageIndex{7}$$, the length $$AB$$ (a straight line segment) must be less than the combined lengths of $$AC$$ and $$CB$$ (not on a straight line from $$A$$ to $$B$$),

##### Example $$\PageIndex{5}$$

Find the perimeter of the triangle in terms of $$x$$, Then find the perimeter if $$x = 1$$:

Solution

$\begin{array} {rcl} {} & = & {} \\ {} & = & {} \\ {} & = & {} \\ {} & = & {} \\ {} & = & {} \end{array}$

If $$x = 1$$, $$\dfrac{24 - x}{12} = \dfrac{24 - 1}{12} = \dfrac{23}{12}$$.

Check:

Answer: $$\dfrac{24 - x}{12}$$, $$\dfrac{23}{12}$$.

## Problems

1 - 2. Find $$x$$ if $$\triangle ABC$$ is isosceles with $$AC = BC$$:

1. 2.

3 - 4. Find $$x$$ if $$\triangle ABC$$ is equilateral:

3. 4.

5 - 6. Find $$AB$$ if $$CD$$ is a median:

5. 6.

7 - 8. Find $$\angle ACB$$ if $$\overrightarrow{CD}$$ is an angle bisector:

7. 8.

9 - 10. Find the perimeter of the triangle in terms of $$x$$, Then find the perimeter if $$x = 4$$:

9. 10.

11. Find $$x$$ if the perimeter of $$\triangle ABC$$ is 33.

12. Find $$x$$ if the perimeter of $$\triangle ABC$$ is 11.

This page titled 7.3: Triangle CIassifications is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Henry Africk (New York City College of Technology at CUNY Academic Works) via source content that was edited to the style and standards of the LibreTexts platform.