6.2: The Area of a Parallelogram

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In parallelogram $ABCD$ of Figure $\PageIndex{1}$, side $AB$ is called the base and the line segment $DE$ is called the height or altitude. The base may be any side of the parallelogram, though it is usually chosen to be the side on which the parallelogram appears to be resting. The height is a line drawn perpendicular to the base from the opposite side.

Figure $\PageIndex{1}$: Parallelogram $ABCD$ with base $b$ and height $h$.

Theorem $\PageIndex{1}$

The are of a parallelogram is equal to its base times its height.

\[A = bh\]

Proof

Draw $BF$ and $CF$ as shown in Figure $\PageIndex{2}$. $\angle A=\angle CBF$, $\angle AED=\angle F=90^{\circ},$ and $AD=BC$. Therefore $\triangle ADE \cong \triangle BCF$ and the area of $\triangle ADE$ equals the area of $\triangle BCF$. We have:

\[\begin{array} {rcl} {\text{Area of parallelogram } ABCD} & = & {\text{Area of } \triangle ADE + \text{ Area of trapezoid } BCDE} \\ {} & = & {\text{Area of } \triangle BCF + \text{ Area of trapezoid } BCDE} \\ {} & = & {\text{Area of rectangle } CDEF} \\ {} & = & {bh.} \end{array}\]

Figure $\PageIndex{2}$: Draw $BF$ and $CF$.

Example $\PageIndex{1}$

Find the area and perimeter of $ABCD$:

Solution

$b = AB = CD = 8$, $h = 3$. $\text{Area } = bh = (8)(3) = 24$. $AB = CD=8$. $BC = AD =5$. Perimeter = 8 + 8 + 5 + 5 = 26.

Answer:

Area = 24, Perimeter = 26.

Example $\PageIndex{2}$

Find the area and perimeter of $ABCD$:

Solution

Apply the Pythagorean theorem to right triangle $ADE$:

\[\begin{array} {rcl} {\text{AE}^2 + \text{DE}^2} & = & {\text{AD}^2} \\ {2^2 + h^2} & = & {3^2} \\ {4 + h^2} & = & {9} \\ {h^2} & = & {5} \\ {h} & = & {\sqrt{5}} \end{array}\]

Area = $bh = (8)(\sqrt{5}) = 8\sqrt{5}$

Perimeter $=8+8+3+3=22$

Answer: $A = 8 \sqrt{5}, P = 22$.

Example $\PageIndex{3}$

Find the area and perimeter to the nearest tenth

$clipboard_e1ef2b57a9f0e8c09097be7affb1a9648.png$

Solution

To find the area we must first find the height $h$ (Figure $\PageIndex{3}$), Using trigonometry

Screen Shot 2020-12-18 at 4.16.50 PM.png — Figure $\PageIndex{3}$: Draw in height $h$.

$\begin{array} {rcl} {\sin 40^{\circ}} & = & {\dfrac{h}{4}} \\ {(4) .6428} & = & {\dfrac{h}{\cancel{4}} (\cancel{4})} \\ {2.5712} & = & {h} \end{array}$

Area = $bh = (10)(2.5712) = 25.712 - 25.7$

Perimeter = 10 + 10 + 4 + 4 = 28.

Answer

$A = 25.7$, $P = 28$.

Example $\PageIndex{4}$

Find $x$ if the area is 21.

$clipboard_e127af6096e673fe0dff3d3913cbe85d3.png$

Solution

\[\begin{array} {rcl} {A} & = & {bh} \\ {21} & = & {(x + 3)(\dfrac{12}{x})} \\ {(x)21} & = & {(x + 3)(\dfrac{12}{\cancel{x}})(\cancel{x})} \\ {21x} & = & {12x + 36} \\ {9x} & = & {36} \\ {x} & = & {4} \end{array}\]

Check:

$Screen Shot 2020-12-18 at 4.25.44 PM.png$

Answer

$x = 4$.

Example $\PageIndex{5}$

The area of parallelogram $ABCD$ is 48 and the perimeter is 34. Find $x$ and $y$:

$clipboard_e5df41c57031286a531de30ceb33fbd5e.png$

Solution

\[\begin{array} {rcl} {\text{Perimeter}} & = & {AB +BC+CD+DA} \\ {34} & = & {x + 5 + x+5} \\ {34} & = & {2x+10} \\ {24} & = & {2x} \\ {12} & = & {x} \\ {\text{Area}} & = & {xy} \\ {48} & = & {12y} \\ {4} & = & {y} \end{array}\]

Check:

$Screen Shot 2020-12-18 at 4.29.55 PM.png$