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.
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.
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
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.
A triangle has an area of $2x\sqrt{10}$2x√10 cm^{2}
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+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}$2x√10, 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.
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.
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.
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.
Apply trigonometric relationships, including the sine and cosine rules, in two and three dimensions
Apply trigonometric relationships in solving problems