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# Heron's Formula Study Guide

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Updated on Oct 2, 2011

## Heron's Formula

Heron's formula enables us to calculate the area of a non-right triangle, just by knowing the lengths of the three sides. We prove the formula in this lesson and show how it can be used to find the areas of triangles and polygons.

The area of a triangle is bh, where b is the length of the base and h is the height, as shown in Figure 19.1.

This is because the triangle has exactly half the area of a rectangle with the same base and height (shown in Figure 19.2).

If we know the three sides of a triangle, we can find the height with some work. For example, suppose the three sides of a triangle are 4, 6, and 7 inches long. First, turn the triangle so the longest side is down, as shown in Figure 19.3.

The height (sometimes called the altitude) will meet the longest side and divide it into two lengths. These can be called x and 7 – x, as illustrated in Figure 19.4.

We now have two right triangles. If we apply the Pythagorean theorem to the one on the left, we get

x2 + h2 = 42

If we apply the Pythagorean theorem to the triangle on the right, we get

(7 – x)2 + h = 62
72 – 2( 7 )x + x2 + h2 = 62

We already know that x2 + h2 = 42, so we substitute this in:

72 – 2( 7 )x + (x2 + h2) = 62
72 – 2( 7 )x + 42 = 62

We can solve for x.

72 + 42 – 62 = 2( 7 )x

Using x2 + h2 = 42, we can find the height of the triangle:

With the base b = 7, the triangle's area is thus

= 11.977 square inches

In general, if the triangle's sides are a, b, and c, then by the same computation as in the previous example, the area will be

There is an easier formula, which is credited to Heron, who lived in Alexandria, Egypt, around 100 A.D. It uses the semiperimeter which is half of the perimeter.

Using this, the area of a triangle with sides a, b, and c is

In the example with sides 4, 6, and 7 inches, the semiperimeter is and the area is

This is the same answer we obtained previously. If you replace each 5 in Heron's formula with and then multiply everything out, you would see that this is the same as the first formula. The algebra, however, would fill several pages, and so we omit doing this here.

#### Example 1

What is the area of the triangle with sides of length 10 feet, 15 feet, and 17 feet?

area = √s(sa)(sb)(sc)
area = √21(21 – 10)(21 – 15)(21 – 7)
area = √21(1l)(6)(4) = √5,544 ≈ 74.458 square feet

#### Example 2

What is the area of an equilateral triangle with all sides 6 inches in length?

area = √s(sa)(sb)(sc)
area = √9(9 – 6)(9 – 6)(9 – 6)
area = √9(3)(3)(3) = 9√13 ≈ 15.588 square inches

## Areas of Polygons

When trying to find the area of a figure made of straight lines, it helps to break it down into triangles.

#### Example

What is the area of the quadrilateral in Figure 19.11?

We divide this into two triangles with a diagonal line, as shown in Figure 19.12. Because one triangle is right, we can use the Pythagorean theorem to find the length d of the diagonal.

d2 = 42 + 52
d = √16 + 25 = √4l ≈ 6.4 feet

The right triangle has height 4 feet and base 5 feet, so its area is (4) (5) 10 square feet.

The larger triangle has sides of length 6 feet, 7 feet, and around 6.4 feet, so its semiperimeter is . Thus, we can use Heron's formula to find the area.

area = √s(s – a)(s – b)(s – c)
area = √9.7(9.7 – 6)(9.7 – 7)(9.7 – 6.4)
area = √9.7(3.7)(2.7)(3.3) ≈ 17.88 square feet

The area of the quadrilateral is thus 10 + 17.88 = 27.88 square feet.

Practice problems for this study guide can be found at:

Heron's Formula Practice Questions

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