NZ Level 7 (NZC) Level 2 (NCEA)
Area of a triangle - herons formula
Lesson

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 sine rule, $A=\frac{1}{2}ab\sin C$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(sa)(sb)(sc), where s is known as the semiperimeter of the triangle and is calculated using $s=\frac{a+b+c}{2}$s=a+b+c2

#### 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.

##### Question 2

A triangle has an area of $2x\sqrt{10}$2x10 cm2

and three sides that measure $x,x+2$x,x+2 and $2x-2$2x2 (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+c2​ $s$s $=$= $\frac{x+x+2+2x-2}{2}$x+x+2+2x−22​ $s$s $=$= $\frac{4x}{2}$4x2​ $s$s $=$= $2x$2x
 $A$A $=$= $\sqrt{s(s-a)(s-b)(s-c)}$√s(s−a)(s−b)(s−c) $A$A $=$= $\sqrt{2x(2x-a)(2x-b)(2x-c)}$√2x(2x−a)(2x−b)(2x−c) $A$A $=$= $\sqrt{2x(2x-x)(2x-(x+2))(2x-(2x-2)}$√2x(2x−x)(2x−(x+2))(2x−(2x−2) $A$A $=$= $\sqrt{2x(x)(x-2)(2)}$√2x(x)(x−2)(2) $A$A $=$= $\sqrt{4x^2(x-2)}$√4x2(x−2)

Do: We are told the area is $2x\sqrt{10}$2x10, so then

 $2x\sqrt{10}$2x√10 $=$= $\sqrt{4x^2(x-2)}$√4x2(x−2) $40x^2$40x2 $=$= $4x^2(x-2)$4x2(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.

#### Worked Examples

##### QUESTION 1

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

1. First find the semi-perimeter $s$s.

2. Hence find the area using Heron's formula.

##### QUESTION 2

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 3

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

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

Solve for $x$x.

2. Hence find the area of the triangle.

### Outcomes

#### M7-4

Apply trigonometric relationships, including the sine and cosine rules, in two and three dimensions

#### 91259

Apply trigonometric relationships in solving problems