7.3: The Area of a Triangle

Last updated
Save as PDF

Page ID: 24855

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $ $ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $$\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$ $\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$

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 $\triangle ABC = \dfrac{1}{2} \text{ area of parallelogram } ABCE = \dfrac{1}{2} bh$.
$Screen Shot 2020-12-18 at 4.53.29 PM.png$ Figure $\PageIndex{2}$: 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 $\PageIndex{3}$

$Screen Shot 2020-12-18 at 4.56.39 PM.png$ Figure $\PageIndex{3}$: 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 $\PageIndex{4}$.

$Screen Shot 2020-12-18 at 5.11.27 PM.png$ Figure $\PageIndex{4}$: 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: $A = 25\sqrt{3}, P = 20 + 10 \sqrt{3}$.