Heron's formula

The famil­iar for­mula for the area of a tri­an­gle, $\displaystyle \frac{1}{2} \times \text{base} \times \text{height}$, needs the height. Often we know only the three sides: a sur­veyor mea­sures the sides of a tri­an­gu­lar field, not its height. Heron's for­mula, named after Heron of Alexan­dria, gives the area from the three sides alone. In this les­son we use it for tri­an­gles and then for quadri­lat­er­als, by split­ting them into tri­an­gles.

The for­mula#

For a tri­an­gle with sides $a$, $b$ and $c$, first find the semi-perime­ter (half the perime­ter):

$$\displaystyle s = \frac{a + b + c}{2}$$

Then

$$\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}$$

Work in this order: find $s$, then the three dif­fer­ences $s - a$, $s - b$, $s - c$, then mul­ti­ply and take the square root. A quick check: the three dif­fer­ences always add up to $s$.

Exam­ple 1#

Find the area of a tri­an­gle with sides $13$ cm, $14$ cm and $15$ cm.

$$\displaystyle s = \frac{13 + 14 + 15}{2} = 21$$

$$s - a = 8, \quad s - b = 7, \quad s - c = 6$$

(Check: $8 + 7 + 6 = 21$.)

$$\text{Area} = \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = 84 \text{ cm}^2$$

To find the square root with­out a cal­cu­la­tor, break the num­bers into prime fac­tors and pair them: $21 \times 8 \times 7 \times 6 = (3 \times 7) \times 2^3 \times 7 \times (2 \times 3) = 2^4 \times 3^2 \times 7^2$, whose square root is $2^2 \times 3 \times 7 = 84$.

Exam­ple 2#

Find the area of an isosce­les tri­an­gle with equal sides $10$ cm and base $12$ cm.

$$\displaystyle s = \frac{10 + 10 + 12}{2} = 16, \qquad \text{Area} = \sqrt{16 \times 6 \times 6 \times 4} = \sqrt{2304} = 48 \text{ cm}^2$$

We can check this another way. The height to the base meets it at its mid­point, so the height is $\sqrt{10^2 - 6^2} = 8$ cm and the area is $\displaystyle \frac{1}{2} \times 12 \times 8 = 48$ cm². The two meth­ods agree.

Exam­ple 3#

Find the area of an equi­lat­eral tri­an­gle of side $10$ cm.

$$s = 15, \qquad \text{Area} = \sqrt{15 \times 5 \times 5 \times 5} = \sqrt{1875} = 25\sqrt{3} \approx 43.30 \text{ cm}^2$$

For any equi­lat­eral tri­an­gle of side $a$, Heron's for­mula sim­pli­fies to $\displaystyle \text{Area} = \frac{\sqrt{3}}{4}a^2$. For $a = 10$ that is $\displaystyle \frac{\sqrt{3}}{4} \times 100 = 25\sqrt{3}$.

Exam­ple 4#

The sides of a tri­an­gle are in the ratio $3 : 4 : 5$ and its perime­ter is $48$ cm. Find its area.

The sides are $3k$, $4k$, $5k$ with $12k = 48$, so $k = 4$: the sides are $12$, $16$ and $20$ cm.

$$s = 24, \qquad \text{Area} = \sqrt{24 \times 12 \times 8 \times 4} = \sqrt{9216} = 96 \text{ cm}^2$$

Since $12^2 + 16^2 = 20^2$ the tri­an­gle is right-angled, and indeed $\displaystyle \frac{1}{2} \times 12 \times 16 = 96$.

Quadri­lat­er­als#

A diag­o­nal splits a quadri­lat­eral into two tri­an­gles. If we know the sides of both tri­an­gles, Heron's for­mula gives each area.

Exam­ple 5#

A field $ABCD$ has $AB = 8$ m, $BC = 15$ m, $CD = 25$ m, $DA = 26$ m and $\angle ABC = 90°$. Find its area.

Draw the diag­o­nal $AC$. Tri­an­gle $ABC$ is right-angled at $B$, so

$$\displaystyle AC = \sqrt{8^2 + 15^2} = \sqrt{289} = 17 \text{ m}, \qquad \text{ar}(ABC) = \frac{1}{2} \times 8 \times 15 = 60 \text{ m}^2$$

Tri­an­gle $ACD$ has sides $17$, $25$ and $26$ m:

$$\displaystyle s = \frac{17 + 25 + 26}{2} = 34, \qquad \text{ar}(ACD) = \sqrt{34 \times 17 \times 9 \times 8} = \sqrt{41616} = 204 \text{ m}^2$$

The area of the field is $60 + 204 = 264$ m².

Prac­tice#

1. Find the area of a tri­an­gle with sides $5$ cm, $12$ cm and $13$ cm.
2. Find the area of a tri­an­gle with sides $7$ cm, $24$ cm and $25$ cm.
3. Find the area of a tri­an­gle with sides $9$ cm, $10$ cm and $17$ cm.
4. Find the area of an isosce­les tri­an­gle with equal sides $5$ cm and base $8$ cm.
5. Find the area of an equi­lat­eral tri­an­gle of side $6$ cm.
6. The sides of a tri­an­gle are in the ratio $5 : 12 : 13$ and its perime­ter is $60$ cm. Find its area.
7. A tri­an­gu­lar plot has sides $25$ m, $17$ m and $12$ m. Find its area, and the cost of lev­el­ling it at ₹20 per m².
8. A rhom­bus has side $10$ cm and one diag­o­nal $12$ cm. Find its area using Heron's for­mula.
9. In quadri­lat­eral $ABCD$, $AB = 5$ cm, $BC = 12$ cm, $CD = 14$ cm, $DA = 15$ cm and $\angle ABC = 90°$. Find its area.

Answers#