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6.5: The Area of a Trapezoid

  • Page ID
    34150
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    In Figure \(\PageIndex{1}\), \(b_1\) and \(b_2\) are the bases of trapezoid \(ABCD\) and \(h\) is the height or altitude. The formula for the area is given in the following theorem:

    Screen Shot 2020-12-28 at 11.40.42 AM.png
    Figure \(\PageIndex{1}\): Trapezoid \(ABCD\) with bases \(b_1\) and \(b_2\) and height \(h\).
    Theorem \(\PageIndex{1}\)

    The area of a trapezoid is equal to one-half the product of its height and the sum of its bases.

    \[A = \dfrac{1}{2} h(b_1 + b_2)\]

    Proof

    In Figure \(\PageIndex{1}\) draw \(BD\) (see Figure \(\PageIndex{2}\)). Note that \(CD = b_2\) is the base and \(BF = h\) is the height of \(\triangle BCD\). Area of trapezoid \(ABCD =\) Area of \(\triangle ABD\) + Area of \(\triangle BCD\) = \(\dfrac{1}{2} b_1h + \dfrac{1}{2} b_2 h = \dfrac{1}{2} h(b_1 + b_2)\).

    Screen Shot 2020-12-28 at 11.46.50 AM.png
    Figure \(\PageIndex{2}\): Draw \(BD\). \(CD\) is the base and \(BF\) is the height of \(\triangle BCD\).
    Example \(\PageIndex{1}\)

    Find the area:

    Screen Shot 2020-12-28 at 11.42.45 AM.png

    Solution

    \(A = \dfrac{1}{2} h(b_1 + b_2)= \dfrac{1}{2} (6)(28 + 16) = \dfrac{1}{2} (6)(44) = 132\).

    Answer: \(A = 132\).

    Example \(\PageIndex{2}\)

    Find the area and perimeter:

    Screen Shot 2020-12-28 at 11.48.05 AM.png

    Solution

    Draw heights \(DE\) and \(CF\) (Figure \(\PageIndex{3}\)). \(\triangle ADE\) is \(30^{\circ}-60^{\circ}-90^{\circ}\) triangle. So \(AE\) = short leg = \(\dfrac{1}{2}\) hypotenuse = \(\dfrac{1}{2}(10) = 5\), and \(DE\) = long leg = (short leg)(\(\sqrt{3}\)) = \(5\sqrt{3}\). \(CDEF\) is a rectangle so \(EF = CD = 10\). Therefore \(BF = AB - EF = 22 -10 - 5 = 7\). Let \(x = BC\).

    Screen Shot 2020-12-28 at 11.50.59 AM.png
    Figure \(\PageIndex{3}\): Draw heights \(DE\) and \(CF\).

    \[\begin{array} {rcl} {\text{CF}^2 + \text{BF}^2} & = & {\text{BC}^2} \\ {(5\sqrt{3})^2 + 7^2} & = & {x^2} \\ {75 + 49} & = & {x^2} \\ {124} & = & {x^2} \\ {x} & = & {\sqrt{124} = \sqrt{4} \sqrt{31} = 2\sqrt{31}} \end{array}\]

    Area = \(\dfrac{1}{2} h(b_1 + b_2) = \dfrac{1}{2} (5\sqrt{3}) (22 + 10) = \dfrac{1}{2} (5\sqrt{3})(32) = 80\sqrt{3}\).

    Perimeter = \(22 + 10 + 10 + 2\sqrt{31} = 42 + 2\sqrt{31}\).

    Answer: \(A = 80\sqrt{3}\), \(P = 42 + 2\sqrt{31}\).

    Problems

    1 - 2. Find the area of \(ABCD\):

    1.

    Screen Shot 2020-12-28 at 11.57.18 AM.png

    2.

    Screen Shot 2020-12-28 at 11.57.49 AM.png

    3 - 12. Find the area and perimeter of \(ABCD\):

    3.

    Screen Shot 2020-12-28 at 11.58.07 AM.png

    4.

    Screen Shot 2020-12-28 at 11.58.23 AM.png

    5.

    Screen Shot 2020-12-28 at 11.58.42 AM.png

    6.

    Screen Shot 2020-12-28 at 11.58.55 AM.png

    7.

    Screen Shot 2020-12-28 at 11.59.10 AM.png

    8.

    Screen Shot 2020-12-28 at 11.59.28 AM.png

    9.

    Screen Shot 2020-12-28 at 11.59.44 AM.png

    10.

    Screen Shot 2020-12-28 at 12.00.10 PM.png

    11.

    Screen Shot 2020-12-28 at 12.00.28 PM.png

    12.

    Screen Shot 2020-12-28 at 12.00.42 PM.png

    13 - 14. Find the area and perimeter to the nearest tenth of \(ABCD\):

    13.

    Screen Shot 2020-12-28 at 12.00.58 PM.png

    14.

    Screen Shot 2020-12-28 at 12.01.15 PM.png

    15. Find \(x\) if the area of \(ABCD\) is 50:

    Screen Shot 2020-12-28 at 12.01.30 PM.png

    16. Find \(x\) if the area of \(ABCD\) is 30:

    Screen Shot 2020-12-28 at 12.01.53 PM.png


    This page titled 6.5: The Area of a Trapezoid 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.