1.6: Triangle CIassifications

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.

Figure $\PageIndex{1}$: Triangles classified according to their 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,

Figure $\PageIndex{2}$: Triangles classified according to their angles,

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,

Figure $\PageIndex{3}$: Right triangles.

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.

Figure $\PageIndex{4}$. $CD$ and $GH$ are altitudes.

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.

Figure $\PageIndex{5}$: $CD$ is a median.

Figure $\PageIndex{6}$. $\overrightarrow{CD}$ is an angle bisector of $\angle ACB$.

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 $3 + 4 + 5 = 12$.

Figure $\PageIndex{7}$. The perimeter of $\triangle ABC$ is 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.