Lesson Explainer: Heron’s Formula | Nagwa Lesson Explainer: Heron’s Formula | Nagwa

Lesson Explainer: Heron’s Formula Mathematics • Second Year of Secondary School

In this explainer, we will learn how to use Heron’s formula to find the area of a triangle.

Before examining Heron’s formula in detail, let’s look at two other formulas we can use to find the area, 𝐴, of a triangle. The formula we choose will depend on the information that we have about the triangle.

Recall that, to calculate the area of any triangle that has a known base length and perpendicular height, we can use the formula 𝐴=12π‘β„Ž, where 𝑏 is the base length and β„Ž is the perpendicular height. To find the area of the triangle below, we can substitute 12 cm into this formula for 𝑏 and 4 cm for β„Ž:

𝐴=12(4)(12)=24.cm

Also remember that 𝐴=12π‘Žπ‘πΆsin is another formula we can use to calculate a triangle’s area. In it, π‘Ž and 𝑏 are the lengths of two of the triangle’s sides, and 𝐢 is the measure of the included angle, or the angle formed by the sides with lengths π‘Ž and 𝑏.

We can use this formula to find the area of any triangle that has an angle with a known measure and that is formed by two sides with known lengths. To find the area of the following triangle, we can substitute 15 cm and 18 cm into the formula for π‘Ž and 𝑏 and 30∘ for 𝐢:

𝐴=12(15)(18)(30)=12(15)(18)ο€Ό12=67.5.sincm∘

Now, suppose that we know the lengths of all three sides of a triangle but the measures of none of its angles. With only this information, we can use neither the formula 𝐴=12π‘β„Ž nor the formula 𝐴=12π‘Žπ‘πΆsin to calculate the triangle’s area. However, we can use Heron’s formula.

Definition: Heron’s Formula

Heron’s formula states that the area, 𝐴, of a triangle with side lengths of π‘Ž, 𝑏, and 𝑐 is 𝐴=βˆšπ‘ (π‘ βˆ’π‘Ž)(π‘ βˆ’π‘)(π‘ βˆ’π‘), where 𝑠 is the semiperimeter of the triangle, or half its perimeter. The triangle’s semiperimeter is given by the formula

𝑠=π‘Ž+𝑏+𝑐2.

Let’s test Heron’s formula to make sure that we get the same area, 𝐴, for an equilateral triangle with a side length of 2 cm that we get when using the formula 𝐴=√34π‘Ž.

In this formula, the variable π‘Ž is the triangle’s side length. Substituting π‘Ž=2 into this formula gives us 𝐴=√34(2)=√34(4)=√3.cm

We get √3 cm2 for the triangle’s area. To make our comparison, we will now use Heron’s formula to find the area. To do this, first, we will need to calculate the semiperimeter: 𝑠=2+2+22=62=3.cm

Heron’s formula then gives us 𝐴=√3(3βˆ’2)(3βˆ’2)(3βˆ’2)=√3(1)(1)(1)=√3.cm

We see that the triangle’s area is √3 cm2, the same answer that we arrived at previously. This is what we would expect.

In the examples that follow, we will look at some other instances in which Heron’s formula is used to calculate not only the areas of triangles but also the areas of figures composed of triangles.

Example 1: Finding the Area of a Triangle Using Heron’s Formula

The area of the triangle whose side lengths are 3 cm, 6 cm, and 7 cm equals cm2.

Answer

In this problem, we are given only the three side lengths of a triangle. Since we are given neither the triangle’s height nor any of its angle measures, we must use Heron’s formula to find its area. Recall that Heron’s formula states that the area, 𝐴, of a triangle with side lengths of π‘Ž, 𝑏, and 𝑐 is 𝐴=βˆšπ‘ (π‘ βˆ’π‘Ž)(π‘ βˆ’π‘)(π‘ βˆ’π‘), where 𝑠 is the triangle’s semiperimeter, or half its perimeter. The semiperimeter can be found using the formula 𝑠=π‘Ž+𝑏+𝑐2.

Let’s begin by finding the triangle’s semiperimeter. We must substitute each of the triangle’s side lengths into the semiperimeter formula for either π‘Ž, 𝑏, or 𝑐. It does not matter which side length we substitute for which variable. Here, we will substitute π‘Ž=3, 𝑏=6, and 𝑐=7 and then simplify to get a semiperimeter of 𝑠=3+6+72=162=8.cm

Now, we can substitute values into Heron’s formula to find the triangle’s area.

Since π‘Ž=3, 𝑏=6, 𝑐=7, and 𝑠=8, Heron’s formula gives us 𝐴=√8(8βˆ’3)(8βˆ’6)(8βˆ’7)=√8(5)(2)(1)=√80=4√5.cm

Thus, we can say that the area of the triangle whose side lengths are 3 cm, 6 cm, and 7 cm is 4√5 cm2.

In the next example, we will also calculate the area of a triangle with three given side lengths using Heron’s formula.

Example 2: Finding the Area of a Triangle Using Heron’s Formula

𝐴𝐡𝐢 is a triangle, where 𝐡𝐢=28cm, 𝐴𝐢=20cm, and 𝐴𝐡=24cm. Find the area of 𝐴𝐡𝐢 giving the answer to the nearest square centimetre.

Answer

First, let’s draw a sketch of triangle 𝐴𝐡𝐢. We are told that 𝐡𝐢=28cm, 𝐴𝐢=20cm, and 𝐴𝐡=24cm.

Since we know the triangle’s three side lengths, we can use Heron’s formula to find its area. According to Heron’s formula, the area, 𝐴, of a triangle with side lengths of π‘Ž, 𝑏, and 𝑐 is 𝐴=βˆšπ‘ (π‘ βˆ’π‘Ž)(π‘ βˆ’π‘)(π‘ βˆ’π‘), where 𝑠 is the triangle’s semiperimeter. The semiperimeter can be calculated using the formula 𝑠=π‘Ž+𝑏+𝑐2.

Substituting π‘Ž=28, 𝑏=20, and 𝑐=24 into the semiperimeter formula gives us 𝑠=28+20+242=722=36.cm

Now, since π‘Ž=28, 𝑏=20, 𝑐=24, and 𝑠=36, Heron’s formula tells us that the triangle’s area is 𝐴=√36(36βˆ’28)(36βˆ’20)(36βˆ’24)=√36(8)(16)(12)=√55296=235.151015….cm

The area of triangle 𝐴𝐡𝐢 to the nearest square centimetre is 235 cm2.

Next, we will use Heron’s formula to calculate the area of a rhombus. Remember that the sides of a rhombus all have the same length.

Example 3: Finding the Area of a Rhombus Using Heron’s Formula

The perimeter of the given rhombus is 292 cm and the length of 𝐴𝐢 is 116 cm. Use Heron’s formula to calculate the area of the rhombus, giving the answer to three decimal places.

Answer

Recall that Heron’s formula is 𝐴=βˆšπ‘ (π‘ βˆ’π‘Ž)(π‘ βˆ’π‘)(π‘ βˆ’π‘), where 𝐴 is the area of a triangle that has sides with lengths of π‘Ž, 𝑏, and 𝑐 and a semiperimeter of 𝑠. The semiperimeter is given by the formula 𝑠=π‘Ž+𝑏+𝑐2.

In order to use Heron’s formula to calculate the area of the given rhombus, we must decompose the rhombus into two triangles: triangle 𝐴𝐡𝐢 and triangle 𝐴𝐷𝐢. Since a rhombus is a quadrilateral that has four sides of equal length, each of the sides of rhombus 𝐴𝐡𝐢𝐷 must have a length of 2924=73.cm

Therefore, we know that the lengths of the sides of triangle 𝐴𝐡𝐢 are 𝐴𝐡=73,𝐡𝐢=73,𝐴𝐢=116.cmcmandcm

With this information, we can use Heron’s formula to calculate the triangle’s area. We can begin by finding its semiperimeter: 𝑠=73+116+732=2622=131.cm

Next, we can substitute π‘Ž=73, 𝑏=116, 𝑐=73, and 𝑠=131 into Heron’s formula to get an area of 𝐴=√131(131βˆ’73)(131βˆ’116)(131βˆ’73)=√131(58)(15)(58)=√6610260.cm

We know that triangle 𝐴𝐷𝐢 must have the same area as triangle 𝐴𝐡𝐢, because the two triangles have the same side lengths. Therefore, since rhombus 𝐴𝐡𝐢𝐷 is composed of the two triangles, the rhombus’s area must be equal to 2√6610260=5142.085180….cm

The area of the rhombus to three decimal places is 5β€Žβ€‰β€Ž142.085 cm2.

In the example that follows, we will calculate the area of a figure composed of two triangles using Heron’s formula. One of the two triangles is a right triangle, and we will need to use the Pythagorean theorem to determine one of its leg lengths.

Example 4: Finding the Area of a Quadrilateral Using Heron’s Formula

Find the area of the figure below using Heron’s formula, giving the answer to three decimal places.

Answer

We know that Heron’s formula is 𝐴=βˆšπ‘ (π‘ βˆ’π‘Ž)(π‘ βˆ’π‘)(π‘ βˆ’π‘), where 𝐴 is the area of a triangle that has sides with lengths of π‘Ž, 𝑏, and 𝑐 and a semiperimeter of 𝑠. We also know that the semiperimeter can be calculated with the formula 𝑠=π‘Ž+𝑏+𝑐2.

In order to use Heron’s formula to calculate the area of the figure, we must decompose it into a pair of triangles.

One of the triangles into which the figure can be decomposed has side lengths of 16 cm, 20 cm, and 23 cm. Let’s begin by finding the area of this triangle using Heron’s formula. Substituting the triangle’s three side lengths into the semiperimeter formula gives us 𝑠=16+20+232=592=29.5.cm

Next, substituting the triangle’s three side lengths and its semiperimeter into Heron’s formula gives us 𝐴=√29.5(29.5βˆ’16)(29.5βˆ’20)(29.5βˆ’23)=√29.5(13.5)(9.5)(6.5)=√24591.9375=156.818166….cm

Looking at the other triangle, we can see that it is a right triangle with a leg that is 16 cm long and a hypotenuse that is 20 cm long. However, we are not given the length of the other leg of the triangle.

In order to find the length, we can use the Pythagorean theorem, which states that if a right triangle has legs of lengths π‘Ž and 𝑏 and a hypotenuse of length 𝑐, then π‘Ž+𝑏=𝑐.

We can substitute π‘Ž=16 and 𝑐=20, and then solve for 𝑏 to determine the length as follows: 16+𝑏=20,256+𝑏=400,𝑏=144,𝑏=12.cm

Note that we only need to consider the positive root of 144, since a length cannot be negative. Thus, the lengths of the three sides of the second triangle are 12,16,20.cmcmandcm

Now that we know all three side lengths, let’s find the triangle’s area. First, we can substitute the side lengths into the semiperimeter formula to get 𝑠=12+16+202=482=24.cm

Next, we can substitute the side lengths and the semiperimeter into Heron’s formula to get 𝐴=√24(24βˆ’12)(24βˆ’16)(24βˆ’20)=√24(12)(8)(4)=√9216=96.cm

Note that we could have also used the formula 𝐴=12π‘β„Ž to find the triangle’s area, since it is a right triangle with a base of 12 cm and a height of 16 cm. This formula also gives us an area of 𝐴=12(12)(16)=96.cm

Combining the areas of both triangles gives us a total area of 156.818166…+96=252.818166…cm.

The area of the figure to three decimal places is 252.818 cm2.

In the final example, we use Heron’s formula to help us find the radius of a circle inscribed in a triangle.

Example 5: Using Heron’s Formula to Find the Radius of a Circle That is Inside a Triangle

The lengths of a triangle are 12 cm, 5 cm, and 11 cm. Find the radius of the interior circle touching the sides using the formula π‘Ÿ=△𝐴𝐡𝐢𝑃Area, where 𝑃 is half of the triangle’s perimeter.

Answer

According to the formula π‘Ÿ=△𝐴𝐡𝐢𝑃,Area in order to calculate the radius of the interior circle touching the sides of the triangle described in the problem, we need to know both the triangle’s area and half its perimeter.

Since we are given the side lengths of the triangle, we can use Heron’s formula to calculate its area. Recall that Heron’s formula states that the area, 𝐴, of a triangle with side lengths of π‘Ž, 𝑏, and 𝑐 and a semiperimeter of 𝑠 is 𝐴=βˆšπ‘ (π‘ βˆ’π‘Ž)(π‘ βˆ’π‘)(π‘ βˆ’π‘).

The semiperimeter can be found using the formula 𝑠=π‘Ž+𝑏+𝑐2.

Let’s begin by finding the triangle’s semiperimeter. Note that this will also be the value of 𝑃 in π‘Ÿ=△𝐴𝐡𝐢𝑃,Area because it is half the triangle’s perimeter. Using the triangle’s side lengths of 12 cm, 5 cm, and 11 cm, its semiperimeter is 𝑠=12+5+112=282=14.cm

Now, we can substitute the triangle’s side lengths into Heron’s formula for π‘Ž, 𝑏, and 𝑐, and its semiperimeter for 𝑠, giving 𝐴=√14(14βˆ’12)(14βˆ’5)(14βˆ’11)=√14(2)(9)(3)=√756=6√21.cm

Since we have now established that half the perimeter of the triangle described in the problem is 14 cm and that its area is 6√21 cm2, we can use the formula π‘Ÿ=△𝐴𝐡𝐢𝑃Area to calculate the radius of the interior circle touching the triangle’s sides. Substituting into the formula gives us π‘Ÿ=6√2114=37√21.cm

The radius of the circle is 37√21 cm.

Now let’s finish by recapping some key points.

Key Points

  • Heron’s formula states that the area, 𝐴, of a triangle with side lengths of π‘Ž, 𝑏, and 𝑐 is 𝐴=βˆšπ‘ (π‘ βˆ’π‘Ž)(π‘ βˆ’π‘)(π‘ βˆ’π‘), where 𝑠 is the semiperimeter of the triangle, or half its perimeter.
  • A triangle’s semiperimeter is given by the formula 𝑠=π‘Ž+𝑏+𝑐2, where π‘Ž, 𝑏, and 𝑐 are its side lengths.
  • The radius of the interior circle touching the sides of a triangle can be calculated using the formula π‘Ÿ=△𝐴𝐡𝐢𝑃,Area where 𝑃 is half of the triangle’s perimeter.

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