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5.2: Area of a Triangle

Difficulty Level: At Grade Created by: CK-12
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Learning Objectives

  • Apply the area formula to triangles where you know two sides and the included angle.
  • Apply the area formula to triangles where you know all three sides, Heron’s Formula.
  • Use the area formulas in real-world and applied problems.

In this section, we will look at how we can derive a new formula using the area formula that we already know and the sine function. This new formula will allow us to find the area of a triangle when we don’t know the height. We will also look at when we can use this formula and how to apply it to real-world situations.

Deriving an Alternate Formula to the Triangle Area Equation

We can use the area formula from Geometry, \begin{align*}A = \frac{1}{2} bh\end{align*}A=12bh, as well as the sine function, to derive a new formula that can be used when the height, or altitude, is unknown.

In \begin{align*}\triangle{ABC}\end{align*}ABC below, \begin{align*}BD\end{align*}BD is altitude from \begin{align*}B\end{align*}B to \begin{align*}AC\end{align*}AC. We will refer to the length of \begin{align*}BD\end{align*}BD as \begin{align*}h\end{align*}h since it also represents the height of the triangle. Also, we will refer to the area of the triangle as \begin{align*}K\end{align*}K to avoid confusing the area with \begin{align*}\angle{A}\end{align*}A.

\begin{align*}k & = \frac{1}{2}bh && \text{Area of a triangle} \\ k & = \frac{1}{2}b (c \sin A) && \sin A = \frac{h}{c}\ \text{therefore}\ c \sin A = h \\ k & = \frac{1}{2}bc \sin A && \text{Simplify}\end{align*}kkk=12bh=12b(csinA)=12bcsinAArea of a trianglesinA=hc therefore csinA=hSimplify

We can use a similar method to derive all three forms of the area formula, regardless of the angle:

\begin{align*}K & = \frac{1}{2}bc \sin A \\ K & = \frac{1}{2}ac \sin B \\ K & = \frac{1}{2}ab \sin C \end{align*}KKK=12bcsinA=12acsinB=12absinC

The formula \begin{align*}K = \frac{1}{2} \ bc \sin A \end{align*}K=12 bcsinA requires us to know two sides and the included angle (SAS) in a triangle. Once we know these three things, we can easily calculate the area of an oblique triangle.

Example 1: In \begin{align*}\triangle{ABC}, \angle{C} = 62^\circ, b = 23.9\end{align*}ABC,C=62,b=23.9, and \begin{align*}a = 31.6\end{align*}a=31.6. Find the area of the triangle.

Solution: Using our new formula, \begin{align*}K = \frac{1}{2} \ ab \sin C \end{align*}K=12 absinC, plug in what is known and solve for the area.

\begin{align*}K & = \frac{1}{2}(31.6)(23.9)\sin 62 \\ K & \approx 333.4\end{align*}KK=12(31.6)(23.9)sin62333.4

Example 2: The Pyramid Hotel recently installed a triangular pool. One side of the pool is 24 feet, another side is 26 feet, and the angle in between the two sides is \begin{align*}87^\circ\end{align*}87. If the hotel manager needs to order a cover for the pool, and the cost is \begin{align*}\$ 35\end{align*}$35 per square foot, how much can he expect to spend?

Solution: In order to find the cost of the cover, we first need to know the area of the cover. Once we know how many square feet the cover is, we can calculate the cost. In the illustration above, you can see that we know two of the sides and the included angle. This means we can use the formula \begin{align*}K = \frac{1}{2} \ bc \sin A\end{align*}K=12 bcsinA.

\begin{align*}& K = \frac{1}{2} \ (24)(26)\sin 87 \\ & K \approx 311.6 \\ & 311.6 \ sq. ft. \times \$ 35/sq. ft. = \$10,905.03 \end{align*}K=12 (24)(26)sin87K311.6311.6 sq.ft.×$35/sq.ft.=$10,905.03

The cost of the cover will be \begin{align*}\$10,905.03\end{align*}$10,905.03.

Find the Area Using Three Sides: Heron’s Formula

In the last section, we learned how to find the area of an oblique triangle when we know two sides and the included angle using the formula \begin{align*}K = \frac{1}{2} \ bc \sin A\end{align*}K=12 bcsinA. We could also find the area of a triangle in which we know all three sides by first using the Law of Cosines to find one of the angles and then using the formula \begin{align*}K = \frac{1}{2} \ bc \sin A\end{align*}K=12 bcsinA. While this process works, it is time-consuming and requires a lot of calculation. Fortunately, we have another formula, called Heron’s Formula, which allows us to calculate the area of a triangle when we know all three sides. It is derived from \begin{align*}K = \frac{1}{2} \ bc \sin A\end{align*}K=12 bcsinA, the Law of Cosines and the Pythagorean Identity.

Heron’s Formula

\begin{align*}K = \sqrt{s(s - a)(s - b)(s - c)}\end{align*} where \begin{align*}s =\frac{1}{2}(a+b+c)\end{align*} or half of the perimeter of the triangle.

Example 3: In \begin{align*}\triangle{ABC}, a = 23, b = 46\end{align*}, and \begin{align*}C = 41\end{align*}. Find the area of the triangle.

Solution: First, you need to find \begin{align*}s\end{align*}: \begin{align*}s = \frac{1}{2}(23 + 41 + 46) = 55\end{align*}. Now, plug s and the three sides into Heron’s Formula and simplify.

\begin{align*}K & = \sqrt{55(55-23)(55-46)(55-41)} \\ K & = \sqrt{55(32)(9)(14)} \\ k & = \sqrt{221760} \\ K & \approx 470.9\end{align*}

Example 4: A handyman is installing a tile floor in a kitchen. Since the corners of the kitchen are not exactly square, he needs to have special triangular shaped tile made for the corners. One side of the tile needs to be 11.3”, the second side needs to be 11.9”,and the third side is 13.6”. If the tile costs \begin{align*}\$4.89\end{align*} per square foot, and he needs four of them, how much will it cost to have the tiles made?

Solution: In order to find the cost of the tiles, we first need to find the area of one tile. Since we know the measurements of all three sides, we can use Heron’s Formula to calculate the area.

\begin{align*}s & = \frac{1}{2}(11.3 + 11.9 + 13.6) = 18.4 \\ K & = \sqrt{18.4(18.4 - 11.3)(18.4 - 11.9)(18.4 - 13.6)} \\ K & = \sqrt{18.4(7.1)(6.5)(4.8)} \\ K & = \sqrt{4075.968}\\ K & \approx63.843\ in^2\end{align*}

The area of one tile is 63.843 square inches. The cost of the tile is given to us in square feet, while the area of the tile is in square inches. In order to find the cost of one tile, we must first convert the area of the tile into square feet.

\begin{align*}1\ \text{square foot} & = 12in \times 12in = 144in^2 \\ \frac{63.843}{144} & = 0.443\ ft^2 && \text{Covert square inches into square feet} \\ 0.443\ ft^2 \times 4.89 & = 2.17 && \text{Multiply by the cost of the tile}.\\ 2.17 \times 4 & = 8.68\end{align*}

The cost for four tiles would be \begin{align*}\$8.68\end{align*}.

Finding a Part of the Triangle, Given the Area

We have already looked at two examples of situations where we can apply the two new area formulas we learned in this section. In this section, we will look at another real-world application where we know the area but need to find another part of the triangle, as well as an application involving a quadrilateral.

Example 5: The jib sail on a sailboat came untied and the rope securing it was lost. If the area of the jib sail is 56.1 square feet, use the figure and information belowto find the length of the rope.

Solution: Since we know the area, one of the sides, and one angle of the jib sail, we can use the formula \begin{align*}K = \frac{1}{2} \ bc \sin A\end{align*} to find the side of the jib sail that is attached to the mast. We will call this side \begin{align*}y\end{align*}.

\begin{align*}56.1 & = \frac{1}{2} \ 28(y)\sin 11 \\ 56.1 & = 2.671325935\ y \\ 21.0 & = y\end{align*}

Now that we know side \begin{align*}y\end{align*}, we know two sides and the included angle in the triangle formed by the mast, the rope, and the jib sail. We can now use the Law of Cosines to calculate the length of the rope.

\begin{align*}x^2 & = 21^2 + 27^2 - 2(21)(27) \cos 18 \\ x^2 & = 91.50191052 \\ x & \approx 9.6\ ft\end{align*}

The length of the rope is approximately 9.6 feet.

Example 6: In quadrilateral \begin{align*}QUAD\end{align*} below, the area of \begin{align*}\triangle{QUD} = 5.64\end{align*}, the area of \begin{align*}\triangle{UAD} = 6.39,\angle{QUD} = 31^\circ, \angle{DUA} = 40^\circ\end{align*}, and \begin{align*}UD = 7.8\end{align*}. Find the perimeter of \begin{align*}QUAD\end{align*}.

Solution: In order to find the perimeter of \begin{align*}QUAD\end{align*}, we need to know sides \begin{align*}QU, QD, UA\end{align*}, and \begin{align*}AD\end{align*}. Since we know the area, one side, and one angle in each of the triangles, we can use \begin{align*}K = \frac{1}{2} \ bc \sin A\end{align*} to figure out \begin{align*}QU\end{align*} and \begin{align*}UA\end{align*}.

\begin{align*}5.64 & = \frac{1}{2}(7.8)(QU)\sin 31 && 6.39 = \frac{1}{2}(7.8)UA\sin 40 \\ 2.8 & \approx QU && 2.5 \approx UA\end{align*}

Now that we know \begin{align*}QU\end{align*} and \begin{align*}UA\end{align*}, we know two sides and the included angle in each triangle (SAS). This means that we can use the Law of Cosines to find the other two sides, \begin{align*}QD\end{align*} and \begin{align*}DA\end{align*}. First we will find \begin{align*}QD\end{align*} and \begin{align*}DA\end{align*}.

\begin{align*}QD^2 & = 2.8^2 + 7.8^2 - 2(2.8)(7.8) \cos 31 && DA^2 = 2.5^2 + 7.8^2 - 2(2.5)(7.8) \cos 40 \\ QD^2 & = 31.23893231 && DA^2 = 37.21426672 \\ QD & \approx 5.6 && DA \approx 6.1\end{align*}

Finally, we can calculate the perimeter since we have found all four sides of the quadrilateral.

\begin{align*}pQUAD = 2.8 + 5.6 + 6.1 + 2.5 = 17 \end{align*}

Points to Consider

  • Why can’t s (half of the perimeter) in Heron’s Formula be smaller than any of the three sides in the triangle?
  • How could we find the area of a triangle is AAS, SSA, and ASA cases?
  • Is it possible to figure out the length of the third side of a triangle if we know the other two sides and the area?

Review Questions

  1. Using the figures and given information below, determine which formula you would need to use in order to find the area of each triangle (\begin{align*}A = \frac{1}{2} \ bh, K = \frac{1}{2} \ bc \sin A\end{align*}, or Heron’s Formula).
Given Figure Formula
a. \begin{align*}CF = 3, FM = 8\end{align*}, and \begin{align*}CO = 5\end{align*}
b. \begin{align*}HC = 4.1, CE = 7.4\end{align*}, and \begin{align*}HE = 9.6\end{align*}
c. \begin{align*}AP = 59.8, PH = 86.3,\angle{APH} = 103^\circ\end{align*}
d. \begin{align*}RX = 11.1, XE = 18.9,\angle{R} = 41^\circ\end{align*}
  1. Find the area of all of the triangles in the chart above to the nearest tenth.
  2. Using the given information and the figures below, decide which area formula you would need to use to find each side, angle, or area.
Given Figure Find Formula
a. Area \begin{align*}= 1618.98, b = 36.3\end{align*} \begin{align*}h\end{align*}
b. Area \begin{align*}= 387.6, b = 25.6, c = 32.9\end{align*} \begin{align*}\angle{A}\end{align*}
c. Area \begin{align*}\triangle{ABD} = 16.96, AD = 3.2, \angle{DBC} = 49.6^\circ\end{align*} Area of \begin{align*}\triangle{ABC}\end{align*}
  1. Using the figures and information from the table above, find the angle, side, or area requested.
  2. The Pyramid Hotel is planning on repainting the exterior of the building. The building has four sides that are isosceles triangles with bases measuring 590 ft and legs measuring 375 ft.
    1. What is the total area that needs to be painted?
    2. If one gallon of paint covers 25 square feet, how many gallons of paint are needed?
  3. A contractor needs to replace a triangular section of roof on the front of a house. The sides of the triangle are 8.2 feet, 14.6 feet, and 16.3 feet. If one bundle of shingles covers \begin{align*}33 \ \frac{1}{3}\end{align*} square feet and costs \begin{align*}\$15.45\end{align*}, how many bundles does he need to purchase? How much will the shingles cost him? How much of the bundle will go to waste?
  4. A farmer needs to replant a triangular section of crops that died unexpectedly. One side of the triangle measures 186 yards, another measures 205 yards, and the angle formed by these two sides is \begin{align*}148^\circ\end{align*}.
    1. What is the area of the section of crops that needs to be replanted?
    2. The farmer goes out a few days later to discover that more crops have died. The side that used to measure 205 yards now measures 288 yards. How much has the area that needs to be replanted increased by?

  5. Find the perimeter of the quadrilateral at the left If the area of \begin{align*}\triangle{DEG} = 56.5\end{align*} and the area of \begin{align*}\triangle{EGF} = 84.7\end{align*}.
  6. In \begin{align*}\triangle{ABC}, BD\end{align*} is an altitude from \begin{align*}B\end{align*} to \begin{align*}AC\end{align*}. The area of \begin{align*}\triangle{ABC} = 232.96, AB = 16.2\end{align*}, and \begin{align*}AD = 14.4\end{align*}. Find \begin{align*}DC\end{align*}.
  7. Show that in any triangle \begin{align*}DEF, d^2 + e^2 + f^2 = 2(ef \cos D + df \cos E + de \cos F)\end{align*}.

Review Answers

    1. \begin{align*}A = \frac{1}{2} \ bh\end{align*}
    2. Heron’s formula
    3. \begin{align*}K = \frac{1}{2} \ bc \sin A\end{align*}
    4. \begin{align*}A = \frac{1}{2} \ bh\end{align*}
    1. \begin{align*}A = 22\end{align*}
    2. \begin{align*}A = 14.3\end{align*}
    3. \begin{align*}A = 2514.2\end{align*}
    4. \begin{align*}A = 144.7\end{align*}
    1. \begin{align*}A = \frac{1}{2} \ bh\end{align*}
    2. \begin{align*}K = \frac{1}{2} \ bc \sin A\end{align*}
    3. \begin{align*}A = \frac{1}{2} \ bh\end{align*}
    1. \begin{align*}h = 89.2\end{align*}
    2. \begin{align*}\angle{A} = 67^\circ\end{align*}
    3. Area of \begin{align*}\triangle{ABC} = 83.0\end{align*}
  1. (a) Use Heron’s Formula, then multiply your answer by 4, for the 4 sides. \begin{align*}s = \frac{1}{2}(375 + 375 + 590) = 670, A = \sqrt{670(670 - 375)(670-375)(670-590)} = 68,297.4\end{align*}, multiplied by \begin{align*}4 = 273,189.8\end{align*} total square feet. (b) \begin{align*}\frac{273,189.8}{25} \approx 10,928\end{align*} gallons of paint are needed.
  2. Using Heron’s Formula, s and the area are: \begin{align*}s = \frac{1}{2}(8.2 + 14.6+16.3) = 19.55\end{align*} and \begin{align*}A = \sqrt{19.55(19.55 - 8.2)(19.55-14.6)(19.55 - 16.3)} = 59.75\ sq. ft\end{align*}. He will need 2 bundles \begin{align*}\left (\frac{59.75}{33.3} = 1.8 \right )\end{align*}. The shingles will cost him \begin{align*}2^*\$15.45 = \$30.90\end{align*} and 6.92 square feet will go to waste \begin{align*}(66.67-59.75 = 6.92)\end{align*}.
    1. Use \begin{align*}K = \frac{1}{2} \ bc \sin A, K= \frac{1}{2}(186)(205)\sin 148^\circ\end{align*}. So, the area that needs to be replaced is 10102.9 square yards.
    2. \begin{align*}K = \frac{1}{2}(186)(288)\sin 148^\circ = 14193.4\end{align*}, the area has increased by 4090.5 yards.
  3. You need to use the \begin{align*}K = \frac{1}{2} \ bc \sin A\end{align*} formula to find \begin{align*}DE\end{align*} and \begin{align*}GF\end{align*}. \begin{align*}56.5 = \frac{1}{2}(13.6)DE \sin 39^\circ \rightarrow DE = 13.2 && 84.7 = \frac{1}{2}(13.6)EF \sin 60^\circ \rightarrow EF = 14.4\end{align*} Second, you need to find sides \begin{align*}DG\end{align*} and \begin{align*}GF\end{align*} using the Law of Cosines. \begin{align*}DG^2 & = 13.2^2 + 13.6^2 - 2 \cdot 13.2 \cdot 13.6 \cdot \cos 39^\circ \rightarrow DG = 8.95 \\ GF^2 & = 14.4^2 + 13.6^2 - 2 \cdot 14.4 \cdot 13.6 \cdot \cos 60^\circ \rightarrow GF = 14.0\end{align*} The perimeter of the quadrilateral is 50.55.
  4. First, find \begin{align*}BD\end{align*} by using the Pythagorean Theorem. \begin{align*}BD = \sqrt{16.2^2 - 14.4^2} = 7.42\end{align*}. Then, using the area and formula \begin{align*}(A = \frac{1}{2}bh)\end{align*}, you can find \begin{align*}AC\end{align*}. \begin{align*}232.96 = \frac{1}{2}(7.42)AC \rightarrow AC = 62.78\end{align*}. \begin{align*}DC = 62.78 - 14.4 = 48.38\end{align*}.
  5. \begin{align*}d^2 & = e^2 + f^2 - 2ef \cos D \\ e^2 & = d^2 + f^2 - 2df \cos E && \text{All three versions of the Law of Cosines}\\ f^2 & = d^2 + e^2 - 2de \cos F\end{align*} Add the three formulas together, we get: \begin{align*}d^2 + e^2 + f^2 & = e^2 + f^2 - 2ef \cos D + d^2 + f^2 - 2df \cos E + d^2 + e^2 - 2de \cos F \\ d^2 + e^2 + f^2 & = 2(d^2 + e^2 + f^2) - 2(ef \cos D + df \cos E + de \cos F) \\ - (d^2 + e^2 + f^2) & = -2(ef \cos D + df \cos E + de \cos F) \\ d^2 + e^2 + f^2 & = 2(ef \cos D + df \cos E + de \cos F)\end{align*}

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