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7.3: The Area of a Triangle

  • Page ID
    24855
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    For each of the triangles in Figure \(\PageIndex{1}\), side \(AB\) is called the base and \(CD\) is called the height or altitude drawn to this base. The base can be any state of the triangle though it is usually chosen to be the side on which the triangle appears to be resting. The height is the line drawn perpendicular to the base from the opposite vertex. Note that the height may fall outside the triangle, If the triangle is obtuse, and that the height may be one of the legs, if the triangle is a right triangle.

    clipboard_efd670e805c28238bdc4efc37dd102001.png clipboard_e6101f70eb26d7d2e6dd656f6e0849fc5.png clipboard_e7539c3e97e505b76ecea452531f7aa9b.png
    Figure \(\PageIndex{1}\): Triangles with base \(b\) and height \(h\).
    Theorem \(\PageIndex{1}\)

    The area of a triangle is equal to one-half of its base times its height.

    \[A = \dfrac{1}{2} bh\]

    Proof

    For each of the triangles illustrated in Figure \(\PageIndex{1}\), draw \(AE\) and \(CE\) so that \(ABCE\) is a parallelogram (Figure \(PageIndex{2}\)). \(\triangle ABC \cong \triangle CEA\) so area of \(\triangle ABC = \text{ area of } \triangle CEA\). Therefore area of \(\triangle ABC = \dfrac{1}{2} \text{ area of parallelogram } ABCE = \dfrac{1}{2} bh\).

    Screen Shot 2020-12-18 at 4.53.29 PM.pngFigure \(\PageIndex{2}\): Draw \(AE\) and \(CE\) so that \(ABCE\) is a parallelogram.
    Example \(\PageIndex{1}\)

    Find the area:

    Screen Shot 2020-12-18 at 4.54.27 PM.png

    Solution

    \(A = \dfrac{1}{2} bh = \dfrac{1}{2} (9) (4) = \dfrac{1}{2} (36) = 18.\)

    Answer: 18.

    Example \(\PageIndex{2}\)

    Find the area to the nearest tenth:

    Screen Shot 2020-12-18 at 4.55.58 PM.png

    Solution

    Draw the height \(h\) as shown in Figure \(\PageIndex{3}\)

    Screen Shot 2020-12-18 at 4.56.39 PM.pngFigure \(\PageIndex{3}\): Draw height \(h\).

    \[\begin{array} {rcl} {\sin 40^{\circ}} & = & {\dfrac{h}{10}} \\ {.6428} & = & {\dfrac{h}{10}} \\ {(10)(.6428)} & = & {\dfrac{h}{10}(10)} \\ {6.428} & = & {h} \end{array}\]

    Area = \(\dfrac{1}{2} bh = \dfrac{1}{2} (15)(6.428) = \dfrac{1}{2} (96.420) = 48.21 = 48.2\)

    Answer: \(A = 48.2\)

    Example \(\PageIndex{3}\)

    Find the area and perimeter:

    Screen Shot 2020-12-18 at 5.00.27 PM.png

    Solution

    \(A = \dfrac{1}{2} bh = \dfrac{1}{2} (5)(6) = \dfrac{1}{2} (30) = 15.\)

    To find \(AB\) and \(BC\) we use the Pythagorean theorem:

    \(\begin{array} {rcl} {\text{AD}^2 + \text{BD}^2} & = & {\text{AB}^2} \\ {8^2 + 6^2} & = & {\text{AB}^2} \\ {64 + 36} & = & {\text{AB}^2} \\ {100} & = & {\text{AB}^2} \\ {10} & = & {\text{AB}} \end{array}\) \(\begin{array} {rcl} {\text{CD}^2 + \text{BD}^2} & = & {\text{BC}^2} \\ {3^2 + 6^2} & = & {\text{BC}^2} \\ {9 + 36} & = & {\text{BC}^2} \\ {45} & = & {\text{BC}^2} \\ {\text{BC} = \sqrt{45}} & = & {\sqrt{9} \sqrt{5} = 3\sqrt{5}} \end{array}\)

    Perimeter = \(AB + AC + BC = 10 + 5 + 3\sqrt{5} = 15 + 3 \sqrt{5}\)

    Answer: \(A = 15, P = 15 + 3\sqrt{5}\).

    Example \(\PageIndex{4}\)

    Find the area and perimeter:

    Screen Shot 2020-12-18 at 5.08.58 PM.png

    Solution

    \(\angle A = \angle B = 30^{\circ}\) so \(\triangle ABC\) is isosceles with \(BC = AC = 10\). Draw height \(h\) as in Figure \(\PageIndex{4}\).

    Screen Shot 2020-12-18 at 5.11.27 PM.pngFigure \(\PageIndex{4}\): Draw height \(h\).

    \(\triangle ACD\) is a \(30^{\circ} - 60^{\circ} -90^{\circ}\) triangle hence

    \[\begin{array} {rcl} {\text{hypotenuse}} & = & {2 (\text{short leg})} \\ {10} & = & {2h} \\ {5} & = & {h} \\ {\text{long leg}} & = & {(\text{short leg}) (\sqrt{3})} \\ {AD} & = & {h\sqrt{3} = 5\sqrt{3}.} \end{array} \]

    Similarly \(BD = 5\sqrt{3}\).

    Area = \(\dfrac{1}{2} bh = \dfrac{1}{2} (5\sqrt{3} + 5\sqrt{3})(5) = \dfrac{1}{2} (10\sqrt{3})(5) = \dfrac{1}{2}(50\sqrt{3}) = 25\sqrt{3}\).

    Perimeter = \(10 + 10 + 5\sqrt{3} + 5 \sqrt{3} = 20 + 10 \sqrt{3}\).

    Answer: \(A = 25\sqrt{3}, P = 20 + 10 \sqrt{3}\).

    Problems

    1 - 4. Find the area of \(\triangle ABC\):

    1.

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    2.

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    3.

    Screen Shot 2020-12-18 at 5.23.48 PM.png

    4.

    Screen Shot 2020-12-18 at 5.24.04 PM.png

    5 - 6. Find the area to the nearest tenth:

    5.

    Screen Shot 2020-12-18 at 5.24.20 PM.png

    6.

    Screen Shot 2020-12-18 at 5.26.18 PM.png

    7 - 20. Find the area and perimeter of \(\triangle ABC\):

    7.

    Screen Shot 2020-12-18 at 5.26.39 PM.png

    8.

    Screen Shot 2020-12-18 at 5.27.38 PM.png

    9.

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    10.

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    11.

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    12.

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    13.

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    14.

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    15.

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    16.

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    17.

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    18.

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    19 - 20. Find the area and perimeter to the nearest tenth:

    19.

    Screen Shot 2020-12-18 at 5.33.35 PM.png

    20.

    Screen Shot 2020-12-18 at 5.34.09 PM.png

    21. Find \(x\) if the area of \(\triangle ABC\) is 35:

    Screen Shot 2020-12-18 at 5.36.02 PM.png

    22. Find \(x\) if the area of \(\triangle ABC\) is 24.

    Screen Shot 2020-12-18 at 5.36.23 PM.png

    23. Find \(x\) if the area of \(\triangle ABC\) is 12:

    Screen Shot 2020-12-18 at 5.36.50 PM.png

    24. Find \(x\) if the area of \(\triangle ABC\) is 108:

    Screen Shot 2020-12-18 at 5.37.17 PM.png


    This page titled 7.3: The Area of a Triangle is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Henry Africk (New York City College of Technology at CUNY Academic Works) .