11.3: Areas

Note to the Instructor - Why Did We Wait So Long?

I deliberately chose to not cover the included angle formula for the area of a triangle until this section. It was not needed previously, and Heron's formula, by itself, is too short of a topic for its own. Thus, I waited until we had enough mathematical background to do both!

Learning Objectives Find the area of a triangle given two sides and the included angle. Find the area of a triangle using Heron's Formula.

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The Area of a Triangle

Figure \( \PageIndex{ 1 } \) shows part of the map for a new housing development, Pacific Shores. You are interested in the corner lot, number 86, and you would like to know the area of the lot in square feet. The sales representative for Pacific Shores provides you with the dimensions of the lot. Still, you need to know a formula for the area of an irregularly shaped quadrilateral.

Figure \( \PageIndex{ 1 } \): Map of Pacific Shores development

It occurs to you that you can divide the quadrilateral into two triangles and find the area of each.

Figure \( \PageIndex{ 2 } \): Map of Lot 86

You know a formula for the area of a triangle in terms of its base and height, namely, \( A = \frac{1}{2}bh \), but unfortunately, you don’t know the height of either triangle. You can, however, quickly measure the angles at the corners of the lot using the plot map and a protractor (see Figure \( \PageIndex{ 2 } \)). With this information, we can find the area of each triangle via the following theorem.

Theorem: Area of a Triangle

If a triangle has sides of length \(a\) and \(b\), and the angle between those two sides is \(\theta\), then the area of the triangle is given by\[ A=\dfrac{1}{2} a b \sin \left(\theta\right). \nonumber \]

Proof

The figure below shows three possibilities for triangles, depending on whether the angle \(\theta\) is acute, obtuse, or \(90^{\circ}\).

In each case, \(b\) is the base of the triangle, and the altitude is \(h\). In all three triangles\[\sin \left(\theta\right)=\dfrac{h}{a}.\nonumber \](In the third triangle, if \( h = a \), then \( \frac{h}{a} = 1 \). Since the sine of \( 90^{ \circ } \) is 1, \( \sin\left( \theta \right) = 1 = \frac{h}{a} \))

Solving for \(h\) yields \(h=a \sin \left(\theta\right)\). Substituting this expression for \(h\) into our triangle area formula from Geometry, we get\[A=\dfrac{1}{2} b h=\dfrac{1}{2} b a \sin \left(\theta\right) = \dfrac{1}{2} a b \sin\left( \theta \right).\nonumber \]

Example \( \PageIndex{ 1 } \)

Find the area of lot 86.

Solution

For the triangle in the lower portion of lot 86, \(a=120.3\), \(b=141\), and \(\theta=95^{\circ}\). The area of that portion is\[\begin{array}{rcl}
\text{First Area} & = & \dfrac{1}{2} a b \sin \left(\theta\right) \\[6pt] & = & \dfrac{1}{2}(120.3)(141) \sin \left(95^{\circ}\right) \\[6pt] & \approx & 8448.88 \\[6pt] \end{array} \nonumber \]For the triangle in the upper portion of the lot, \(a=161\), \(b=114.8\), and \(\theta=86.1^{\circ}\). The area of that portion is\[\begin{array}{rcl}
\text {Second Area} & = & \dfrac{1}{2} a b \sin \left(\theta\right) \\[6pt] & = & \dfrac{1}{2}(161)(114.8) \sin \left(86.1^{\circ}\right) \\[6pt] & \approx & 9220.00 \\[6pt] \end{array} \nonumber \]The total area of the lot is the sum of the areas of the triangles. Therefore,\[\text{Total area} = \text{First Area} \, + \, \text{Second Area} \, \approx 17668.88. \nonumber \]Lot 86 has an area of approximately 17,669 square feet.

Caution: \( \theta \) is between the given sides

The formula, \(A=\frac{1}{2} a b \sin \left(\theta\right)\), does not mean that we always use the sides labeled \(a\) and \(b\) to find the area of a triangle. The variables \(a\) and \(b\) represent the lengths of the sides that include the known angle. For example, the area of the triangle below is given by \(A=\frac{1}{2}(5 c) \sin \left(\phi\right)\).

Figure \( \PageIndex{ 3 } \)

Checkpoint \( \PageIndex{ 1 } \)

A triangle has sides of length 6 and 7, and the angle between those sides is \(150^{\circ}\). Find the area of the triangle.

Answer

\(\dfrac{21}{2}\)

Area of a Triangle - Heron's Formula

To use the previous theorem when finding the area of a triangle, we must be given two sides and an included angle; however, is it possible to find the area of a triangle given all three sides but no angles? The answer to this question is yes, and the following theorem, Heron’s Formula, explains how to find such an area.

Theorem: Heron’s Formula

Suppose \(a\), \(b\), and \(c\) denote the lengths of the three sides of a triangle. Let \(s\) be the semiperimeter of the triangle. That is, let\[s = \dfrac{1}{2}\left(a+b+c\right). \nonumber \]Then the area, \(A\), enclosed by the triangle is given by\[A=\sqrt{s(s-a)(s-b)(s-c)}.\nonumber\]

Proof

Using the convention that the angle \(C\) is opposite the side \(c\), and borrowing the area theorem from earlier in this section, we have\[A=\dfrac{1}{2} a b \sin \left( C \right).\nonumber \]To simplify computations, we start by manipulating the expression for \(A^{2}\).\[\begin{array}{rrclcl}
& A^2 & = & \left(\frac{1}{2} a b \sin \left( C \right)\right)^{2} & \quad & \left( \text{squaring both sides of the area formula} \right) \\[6pt] \implies & A^2 & = & \dfrac{1}{4} a^{2} b^{2} \sin ^{2}\left( C \right) & \quad & \left( \text{Laws of Exponents} \right) \\[6pt] \implies & A^2 & = & \dfrac{a^{2} b^{2}}{4}\left(1-\cos^{2}\left( C \right)\right) & \quad & \left( \text{Pythagorean Identity} \right) \\[6pt] \implies & A^2 & = & \dfrac{a^{2} b^{2}}{4}\left[1-\left(\cos\left( C \right)\right)^2\right] & \quad & \\[6pt] \implies & A^2 & = & \dfrac{a^{2} b^{2}}{4}\left[1-\left(\dfrac{c^2 - a^2 - b^2}{-2ab}\right)^2\right] & \quad & \left( \text{Law of Cosines} \right) \\[6pt] \implies & A^2 & = & \dfrac{a^{2} b^{2}}{4}\left[1-\left(\dfrac{a^2 + b^2 - c^2}{2ab}\right)^2\right] & \quad & \left( \text{multiplying numerator and denominator by }-1 \right) \\[6pt] \implies & A^2 & = & \dfrac{a^{2} b^{2}}{4}\left[1-\dfrac{\left(a^2 + b^2 - c^2\right)^2}{4a^2b^2}\right] & \quad & \left( \text{Laws of Exponents} \right) \\[6pt] \implies & A^2 & = & \dfrac{a^{2} b^{2}}{4}\left[\dfrac{4a^2b^2}{4a^2b^2}-\dfrac{\left(a^2 + b^2 - c^2\right)^2}{4a^2b^2}\right] & \quad & \left( \text{making equivalent fractions} \right) \\[6pt] \implies & A^2 & = & \dfrac{a^{2} b^{2}}{4}\left[\dfrac{4a^2b^2 -\left(a^2 + b^2 - c^2\right)^2}{4a^2b^2}\right] & \quad & \left( \text{combining like fractions} \right) \\[6pt] \implies & A^2 & = & \dfrac{\cancelto{1}{a^{2}} \cancelto{1}{b^{2}}}{4}\left[\dfrac{4a^2b^2 -\left(a^2 + b^2 - c^2\right)^2}{4\cancelto{1}{a^2}\cancelto{1}{b^2}}\right] & \quad & \left( \text{cancel like factors} \right) \\[6pt] \implies & A^2 & = & \dfrac{4a^2b^2 -\left(a^2 + b^2 - c^2\right)^2}{16} & \quad & \left( \text{arithmetic} \right) \\[6pt] \implies & A^2 & = & \dfrac{\left( 2ab - (a^2 + b^2 - c^2) \right)\left(2ab + (a^2 + b^2 - c^2)\right)}{16} & \quad & \left( \text{factoring via Difference of Squares} \right) \\[6pt] \implies & A^2 & = & \dfrac{\left( 2ab - a^2 - b^2 + c^2 \right)\left(2ab + a^2 + b^2 - c^2\right)}{16} & \quad & \left( \text{distribution} \right) \\[6pt] \implies & A^2 & = & \dfrac{\left(c^2 - a^2 + 2ab - b^2 \right)\left(a^2 + 2ab + b^2 - c^2\right)}{16} & \quad & \left( \text{Commutative Property of Addition} \right) \\[6pt] \implies & A^2 & = & \dfrac{\left[c^2 - \left(a^2 - 2ab + b^2 \right)\right]\left[\left(a^2 + 2ab + b^2\right) - c^2\right]}{16} & \quad & \left( \text{factoring out a negative and grouping} \right) \\[6pt] \implies & A^2 & = & \dfrac{\left[c^2 - \left(a - b \right)^2\right]\left[\left(a + b\right)^2 - c^2\right]}{16} & \quad & \left( \text{factoring trinomials} \right) \\[6pt] \implies & A^2 & = & \dfrac{\left[c - \left(a - b \right)\right]\left[c + \left(a - b \right)\right]\left[\left(a + b\right) - c\right]\left[\left(a + b\right) + c\right]}{16} & \quad & \left( \text{factoring via Difference of Squares} \right) \\[6pt] \implies & A^2 & = & \dfrac{\left(c - a + b \right)\left(c + a - b \right)\left(a + b - c\right)\left(a + b + c\right)}{16} & \quad & \left( \text{distribution} \right) \\[6pt] \implies & A^2 & = & \dfrac{c - a + b}{2} \cdot \dfrac{c + a - b}{2} \cdot \dfrac{a + b - c}{2} \cdot \dfrac{a + b + c}{2} & \quad & \left( \text{arithmetic} \right) \\[6pt] \implies & A^2 & = & \dfrac{c - a + b}{2} \cdot \dfrac{c + a - b}{2} \cdot \dfrac{a + b - c}{2} \cdot s & \quad & \left( \text{let }s = \dfrac{a + b + c}{2} \right) \\[6pt] \implies & A^2 & = & (s - a) \cdot \dfrac{c + a - b}{2} \cdot \dfrac{a + b - c}{2} \cdot s & \quad & \left( \text{then }s - a = \dfrac{a + b + c}{2} - a = \dfrac{b + c - a}{2} \right) \\[6pt] \implies & A^2 & = & (s - a) \cdot (s - b) \cdot \dfrac{a + b - c}{2} \cdot s & \quad & \left( \text{and }s - b = \dfrac{a + b + c}{2} - b = \dfrac{a + c - b}{2} \right) \\[6pt] \implies & A^2 & = & (s - a) \cdot (s - b) \cdot (s - c) \cdot s & \quad & \left( \text{and }s - c = \dfrac{a + b + c}{2} - c = \dfrac{a + b - c}{2} \right) \\[6pt] \implies & A & = & \sqrt{s(s - a)(s - b)(s - c)} & \quad & \left( \text{Extraction of Roots} \right) \\[6pt] \end{array}\nonumber\]

Example \( \PageIndex{3} \)

Find the area of the given triangle.

Figure \( \PageIndex{ 4 } \)

Solution

Let \( s = \frac{1}{2}\left( a + b + c \right) = \frac{1}{2}\left( 10 + 15 + 7 \right) = \frac{1}{2}\left( 32 \right) = 16 \). Then, by Heron's Formula,\[ A = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{16(16 - 10)(16 - 15)(16 - 7)} = \sqrt{16(6)(1)(9)} = 12 \sqrt{6}\approx 29.4 \, \text{square units}. \nonumber \]

Checkpoint \(\PageIndex{3}\)

Find the area of a triangle with sides of lengths \(a = 29.7\) ft, \(b = 42.3\) ft, and \( c = 38.4\) ft.

Answer

\(\approx 552.3\)