NZ Level 7 (NZC) Level 2 (NCEA)
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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(sa)(sb)(sc)
$A$A $=$= $\sqrt{4.5(4.5-2)(4.5-3)(4.5-4)}$4.5(4.52)(4.53)(4.54)
$A$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+2x22
$s$s $=$= $\frac{4x}{2}$4x2
$s$s $=$= $2x$2x
$A$A $=$= $\sqrt{s(s-a)(s-b)(s-c)}$s(sa)(sb)(sc)
$A$A $=$= $\sqrt{2x(2x-a)(2x-b)(2x-c)}$2x(2xa)(2xb)(2xc)
$A$A $=$= $\sqrt{2x(2x-x)(2x-(x+2))(2x-(2x-2)}$2x(2xx)(2x(x+2))(2x(2x2)
$A$A $=$= $\sqrt{2x(x)(x-2)(2)}$2x(x)(x2)(2)
$A$A $=$= $\sqrt{4x^2(x-2)}$4x2(x2)

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

$2x\sqrt{10}$2x10 $=$= $\sqrt{4x^2(x-2)}$4x2(x2)  
$40x^2$40x2 $=$= $4x^2(x-2)$4x2(x2) (2)
$10$10 $=$= $x-2$x2  
$x$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

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