By now we are reasonably familiar with the formula most often given for the area of a triangle $A=\frac{1}{2}bh$A=12bh, which uses the base ($b$b) and height ($h$h) dimensions (sometimes the height is referred to as the altitude of the triangle).

You may also have studied the law of sines, $A=\frac{1}{2}ab\sin C$A=12absinC. This calculates the area ($A$A), using the lengths of 2 sides ($a$a and $b$b) and the included angle $C$C.

Here is another method for finding the area of a triangle, it's called Heron's Formula.

Did you know?

Heron's Formula gets its name from Heron (Hero) of Alexandria. He was a Greek mathematician and engineer. An avid inventor, he also invented the first recorded windmill and steam engine.

Heron's Formula requires you to know the lengths of all 3 sides, (no perpendicular heights or angles necessary). An algebraic proof can be seen here.

Heron's formula

A triangle with sides of length a,b,c has area A and can be calculated using the formula,

$A=\sqrt{s(s-a)(s-b)(s-c)}$A=√s(s−a)(s−b)(s−c), where s is known as the semiperimeter of the triangle and is calculated using $s=\frac{a+b+c}{2}$s=a+b+c2

Worked examples

Question 1

Find the area of triangle $ABC$ABC, if $AB=2,BC=3$AB=2,BC=3 and $CA=4$CA=4.

Think: We need to identify $s$s, the semiperimeter, which is $s=\frac{a+b+c}{2}=\frac{2+3+4}{2}=4.5$s=a+b+c2=2+3+42=4.5

Do: Now we can use Heron's Formula

$A$`A`	$=$=	$\sqrt{s(s-a)(s-b)(s-c)}$√`s`(`s`−`a`)(`s`−`b`)(`s`−`c`)
$A$`A`	$=$=	$\sqrt{4.5(4.5-2)(4.5-3)(4.5-4)}$√4.5(4.5−2)(4.5−3)(4.5−4)
$A$`A`	$=$=	$\sqrt{4.5(2.5)(1.5)(0.5)}$√4.5(2.5)(1.5)(0.5)
$A$`A`	$=$=	$2.9$2.9 (to 1dp)

So the area of the triangle is $2.9$2.9 square units.

This applet will allow you to see the above calculation, but also manipulate the triangle to be any shape you want and see the Heron's formula calculation.

Created with Geogebra

Question 2

A triangle has an area of $2x\sqrt{10}$2x√10 cm²

and three sides that measure $x,x+2$x,x+2 and $2x-2$2x−2 (cm)

Find the value of $x$x, and hence calculate the perimeter of the triangle.

Think: We will need an expression for both $s$s (the semiperimeter) and $A$A.

$s$`s`	$=$=	$\frac{a+b+c}{2}$`a`+`b`+`c`2
$s$`s`	$=$=	$\frac{x+x+2+2x-2}{2}$`x`+`x`+2+2`x`−22
$s$`s`	$=$=	$\frac{4x}{2}$4`x`2
$s$`s`	$=$=	$2x$2`x`

$A$`A`	$=$=	$\sqrt{s(s-a)(s-b)(s-c)}$√`s`(`s`−`a`)(`s`−`b`)(`s`−`c`)
$A$`A`	$=$=	$\sqrt{2x(2x-a)(2x-b)(2x-c)}$√2`x`(2`x`−`a`)(2`x`−`b`)(2`x`−`c`)
$A$`A`	$=$=	$\sqrt{2x(2x-x)(2x-(x+2))(2x-(2x-2)}$√2`x`(2`x`−`x`)(2`x`−(`x`+2))(2`x`−(2`x`−2)
$A$`A`	$=$=	$\sqrt{2x(x)(x-2)(2)}$√2`x`(`x`)(`x`−2)(2)
$A$`A`	$=$=	$\sqrt{4x^2(x-2)}$√4`x`2(`x`−2)

Do: We are told the area is $2x\sqrt{10}$2x√10, so then

$2x\sqrt{10}$2`x`√10	$=$=	$\sqrt{4x^2(x-2)}$√4`x`2(`x`−2)
$40x^2$40`x`2	$=$=	$4x^2(x-2)$4`x`2(`x`−2)	(2)
$10$10	$=$=	$x-2$`x`−2
$x$`x`	$=$=	$12$12

See how on step 2, I divided by $4x^2$4x2, well I can only do this if $x$x is not zero. Can you see why?

If $x=0$x=0, then I have divided by $0$0 which we know we cannot do. So whilst we found that $x=12$x=12, we also need to say that $x$x could be $0$0. (substitute $x=0$x=0 into the original equation and you will see that it is indeed a solution).

Solve: So $x$x could be $0$0 or $12$12. But a triangle with side length of $0$0, would mean it isn't a triangle at all! So we can discount that as a valid solution.

So the value for $x$x is $12$12, which means the triangle has side lengths of $12,14$12,14 and $22$22 cm, and a total perimeter of $48$48 cm.

Practice questions

QUESTION 3

What is the area of the triangle with sides of length $5$5 cm, $6$6 cm and $5$5 cm?

First find the semi-perimeter $s$s.
Hence find the area using Heron's formula.

QUESTION 4

Find the area of a parallelogram with adjacent sides of length $9$9 cm and $10$10 cm, and a diagonal of length $17$17 cm.

QUESTION 5

A triangle has sides in the ratio $13:14:15$13:14:15 and a perimeter of $126$126 cm.

Let the sides, in centimeters, be $13x$13x, $14x$14x and $15x$15x.

Solve for $x$x.
Hence find the area of the triangle.

15.07 Heron's formula