6.5: The Area of a Trapezoid
- Page ID
- 34150
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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:

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 \(\triangle BCD\) = \(\dfrac{1}{2} b_1h + \dfrac{1}{2} b_2 h = \dfrac{1}{2} h(b_1 + b_2)\).
Figure \(\PageIndex{2}\): Draw \(BD\). \(CD\) is the base and \(BF\) is the height of \(\triangle BCD\).
Find the area:
Solution
\(A = \dfrac{1}{2} h(b_1 + b_2)= \dfrac{1}{2} (6)(28 + 16) = \dfrac{1}{2} (6)(44) = 132\).
Answer: \(A = 132\).
Find the area and perimeter:
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 \(x = BC\).

\[\begin{array} {rcl} {\text{CF}^2 + \text{BF}^2} & = & {\text{BC}^2} \\ {(5\sqrt{3})^2 + 7^2} & = & {x^2} \\ {75 + 49} & = & {x^2} \\ {124} & = & {x^2} \\ {x} & = & {\sqrt{124} = \sqrt{4} \sqrt{31} = 2\sqrt{31}} \end{array}\]
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 \(ABCD\):
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3 - 12. Find the area and perimeter of \(ABCD\):
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13 - 14. Find the area and perimeter to the nearest tenth of \(ABCD\):
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15. Find \(x\) if the area of \(ABCD\) is 50:
16. Find \(x\) if the area of \(ABCD\) is 30: