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Heron's Formula

Area formula based on lengths of sides of a triangle and half its perimeter.

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Heron's Formula

You are in History class learning about different artifacts from other cultures, when the subject of pyramids is presented by your teacher. He informs the class that pyramids come in a variety of sizes, designs, and styles, and are found not only in Egypt, but in many other countries around the world. He tells everyone that a typical pyramid might be approximately 200 meters long at the base and 175 meters up each of the diagonal sides.

Your mind wanders back to your math class from that morning, and you find yourself wondering if there is a straightforward way to determine the area of one of the faces of the pyramid from the information you have been given.

Do you think there is a way to do this?

Heron's Formula

One way to find the area of an oblique triangle when we know two sides and the included angle is by 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.

\begin{align*}K = \sqrt{s(s - a)(s - b)(s - c)}\end{align*}K=s(sa)(sb)(sc) where \begin{align*}s =\frac{1}{2}(a+b+c)\end{align*}s=12(a+b+c) or half of the perimeter of the triangle.


Finding the Area of a Triangle 

1. In \begin{align*}\triangle{ABC}, a = 23, b = 46\end{align*}ABC,a=23,b=46, and \begin{align*}c = 41\end{align*}c=41. Find the area of the triangle.

First, you need to find \begin{align*}s\end{align*}s: \begin{align*}s = \frac{1}{2}(23 + 41 + 46) = 55\end{align*}s=12(23+41+46)=55. 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*}KKkK=55(5523)(5546)(5541)=55(32)(9)(14)=221760470.9

2. 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*}$4.89 per square foot, and he needs four of them, how much will it cost to have the tiles made?

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*}sKKKK=12(11.3+11.9+13.6)=18.4=18.4(18.411.3)(18.411.9)(18.413.6)=18.4(7.1)(6.5)(4.8)=4075.96863.843 in2

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{Convert 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*}1 square foot63.8431440.443 ft2×4.892.17×4=12in×12in=144in2=0.443 ft2=2.17=8.68Convert square inches into square feetMultiply by the cost of the tile.

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

3. In \begin{align*}\triangle{GHI}, g = 11, h = 24\end{align*}GHI,g=11,h=24, and \begin{align*}i = 18\end{align*}i=18. Find the area of the triangle.

First, you need to find \begin{align*}s\end{align*}s: \begin{align*}s = \frac{1}{2}(11 + 24 + 18) = 26.5\end{align*}s=12(11+24+18)=26.5. Now, plug s and the three sides into Heron’s Formula and simplify.

\begin{align*}K & = \sqrt{26.5(26.5-11)(26.5-24)(26.5-18)} \\ K & = \sqrt{26.5(15.5)(2.5)(8.5)} \\ k & = \sqrt{8728.44} \\ K & \approx 93.43\end{align*}KKkK=26.5(26.511)(26.524)(26.518)=26.5(15.5)(2.5)(8.5)=8728.4493.43


Example 1

Earlier, you were given a problem about the pyramid. 

You can use Heron's Formula to find the area of one of the faces of the pyramid.

The equation for the area of the triangle is: \begin{align*}K = \sqrt{s(s - a)(s - b)(s - c)}\end{align*}K=s(sa)(sb)(sc) where \begin{align*}s =\frac{1}{2}(a+b+c)\end{align*}s=12(a+b+c) or half of the perimeter of the triangle.

so, in this case,

\begin{align*}s =\frac{1}{2}(200+175+175) = \frac{550}{2} = 225\end{align*}s=12(200+175+175)=5502=225


\begin{align*}K = \sqrt{225(225-200)(225-175)(225-175)} = 3,750\end{align*}K=225(225200)(225175)(225175)=3,750 square meters.

Example 2

Use Heron's formula to find the area of a triangle with the following sides: \begin{align*}HC = 4.1, CE = 7.4\end{align*}HC=4.1,CE=7.4, and \begin{align*}HE = 9.6\end{align*}HE=9.6

\begin{align*}A = 14.3\end{align*}A=14.3

Example 3

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.

What is the total area that needs to be painted?

Use Heron’s Formula, then multiply your answer by 4, for the 4 sides.

\begin{align*}s = \frac{1}{2}(375 + 375 + 590) = 670\end{align*}s=12(375+375+590)=670

\begin{align*}A = \sqrt{670(670 - 375)(670-375)(670-590)} = 68,297.4\end{align*}A=670(670375)(670375)(670590)=68,297.4

The area multiplied by 4: \begin{align*}68,297.4\cdot 4 = 273,189.8\end{align*}68,297.44=273,189.8 total square feet.

If one gallon of paint covers 25 square feet, how many gallons of paint are needed?

\begin{align*}\frac{273,189.8}{25} \approx 10,928\end{align*}273,189.82510,928 gallons of paint are needed.

Example 4

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*}33 13 square feet and costs \begin{align*}\$15.45\end{align*}$15.45, how many bundles does he need to purchase? How much will the shingles cost him? How much of the bundle will go to waste?

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*}s=12(8.2+14.6+16.3)=19.55 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*}A=19.55(19.558.2)(19.5514.6)(19.5516.3)=59.75 sq.ft. He will need 2 bundles \begin{align*}\left (\frac{59.75}{33.3} = 1.8 \right )\end{align*}(59.7533.3=1.8). The shingles will cost him \begin{align*}2\cdot \$15.45 = \$30.90\end{align*}2$15.45=$30.90 and 6.92 square feet will go to waste \begin{align*}(66.67-59.75 = 6.92)\end{align*}(66.6759.75=6.92).


Find the area of each triangle with the three given side lengths.

  1. 2, 14, 15
  2. 6, 8, 9
  3. 10, 14, 20
  4. 11, 15, 6
  5. 4, 4, 4
  6. 4, 5, 3
  7. 32, 40, 50
  8. 20, 18, 22
  9. 20, 20, 20
  10. 18, 17, 12
  11. 9, 12, 10
  12. 11, 18, 8
  13. Describe when it makes the most sense to use Heron's formula to find the area of a triangle.
  14. A tiling is made of 30 congruent triangles. The lengths of the sides of each triangle are 3 inches, 5 inches, and 7 inches. What is the area of the tiling?
  15. What type of triangle with have the maximum area for a given perimeter? Show or explain your reasoning.

Review (Answers)

To see the Review answers, open this PDF file and look for section 5.5. 

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Heron's Formula

Heron's formula is used to calculate the area of a triangle when the lengths of all three sides are known. Heron's formula states that for a triangle with sides a, b, and c, the area of the triangle is A=\sqrt{s(s-a)(s-b)(s-c)} where s=\frac{a+b+c}{2}.

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