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1.6: Triangle CIassifications

  • Page ID
    34122
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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.
    屏幕快照 2020-10-29 下午7.58.38.png
    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,

    屏幕快照 2020-10-29 下午8.00.49.png
    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,

    屏幕快照 2020-10-29 下午8.02.49.png
    Figure \(\PageIndex{3}\): Right triangles.
    Example \(\PageIndex{1}\)

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

    屏幕快照 2020-10-29 下午8.04.55.png

    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:

    屏幕快照 2020-10-29 下午8.06.17.png

    Answer: \(x = 1\).

    Example \(\PageIndex{2}\)

    \(\triangle ABC\) is equilateral. Find \(x\):

    屏幕快照 2020-10-29 下午8.09.41.png

    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:

    屏幕快照 2020-10-29 下午8.13.21.png

    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.

    屏幕快照 2020-10-29 下午8.15.47.png
    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.

    屏幕快照 2020-10-29 下午8.18.06.png
    Figure \(\PageIndex{5}\): \(CD\) is a median.
    屏幕快照 2020-10-29 下午8.18.51.png
    Figure \(\PageIndex{6}\). \(\overrightarrow{CD}\) is an angle bisector of \(\angle ACB\).
    Example \(\PageIndex{3}\)

    Find \(AB\) if \(CD\) is a median:

    屏幕快照 2020-10-29 下午8.20.56.png

    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, \(x = 0\):

    屏幕快照 2020-10-29 下午8.25.43.png

    Check, \(x = 10\):

    屏幕快照 2020-10-29 下午8.29.40.png

    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:

    屏幕快照 2020-10-29 下午8.30.06.png

    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, \(x = 0\):

    屏幕快照 2020-10-29 下午8.31.41.png

    Check, \(x = 5\):

    屏幕快照 2020-10-29 下午8.32.36.png

    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 \(3 + 4 + 5 = 12\).

    屏幕快照 2020-10-29 下午8.34.53.png
    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 \(x = 1\):

    屏幕快照 2020-10-29 下午8.41.38.png

    Solution

    \[\begin{array} {rcl} {} & = & {} \\ {} & = & {} \\ {} & = & {} \\ {} & = & {} \\ {} & = & {} \end{array}\]

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

    Check:

    屏幕快照 2020-10-29 下午8.42.23.png

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

    Problems

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

    1. Screen Shot 2020-10-29 at 8.51.03 PM.png 2. Screen Shot 2020-10-29 at 8.51.34 PM.png

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

    3. Screen Shot 2020-10-29 at 8.52.11 PM.png 4. Screen Shot 2020-10-29 at 8.52.28 PM.png

    5 - 6. Find \(AB\) if \(CD\) is a median:

    5. Screen Shot 2020-10-29 at 8.52.54 PM.png 6. Screen Shot 2020-10-29 at 8.53.11 PM.png

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

    7. Screen Shot 2020-10-29 at 8.53.31 PM.png 8. Screen Shot 2020-10-29 at 8.53.47 PM.png

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

    9. Screen Shot 2020-10-29 at 8.54.07 PM.png 10. Screen Shot 2020-10-29 at 8.54.23 PM.png

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

    Screen Shot 2020-10-29 at 8.54.48 PM.png

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

    Screen Shot 2020-10-29 at 8.55.02 PM.png


    This page titled 1.6: 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.

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