6.5: The Area of a Trapezoid

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