Reference

Different Ways to Calculate the Area of a Triangle

Rule

Area of a Triangle

The area of a triangle is half the product of its base $b$ and its height $h .$

$A = \frac{1}{2} b h$

The triangle's base can be any of its sides. The height – or altitude – of the triangle is the segment that is perpendicular to the base and connects the base or its extension with its opposite vertex.

Proof

Proof for Right Triangles

First, consider the particular case of a right triangle. It is always possible to reflect a right triangle across its hypotenuse to form a rectangle.

Note that the area of the rectangle formed is twice the area of the original right triangle. Because of this, the formula for the area of the rectangle,

A_{r} = ℓ w,

can be used to find the area of the right triangle.

A_{r} = 2 A_{t} \Rightarrow ℓ w = 2 A_{t}

Furthermore, the height and base of the right triangle have the same measures as the width and length of the rectangle formed by reflecting the triangle. Based on this observation,

b

and

h

can be substituted for

ℓ

and

w,

respectively, to solve for the area of the original right triangle in terms of its base and height.

ℓ w = 2 A_{t}

SubstituteII

$ℓ = b$ , $w = h$

b h = 2 A_{t}

▼

Solve for

A_{t}

Div

Divide by $2$

\frac{b h}{2} = A_{t}

MoveNumRight

$\frac{a}{b} = \frac{1}{b} \cdot a$

\frac{1}{2} b h = A_{t}

RearrangeEqn

Rearrange equation

A_{t} = \frac{1}{2} b h

This shows that the area of a right triangle can be calculated by using the formula

A = \frac{1}{2} b h .

Proof for Non-Right Triangles

To generalize the previous result, it is useful to note that any non-right triangle can be split into two right triangles by drawing one of its heights.

Note that the area of the non-right triangle

A

is equal to the sum of the individual areas of the smaller right triangles

A_{1}

and

A_{2} .

Therefore, it is possible to calculate the area of the non-right triangle by using the previous result for the areas of the smaller right triangles.

A = \frac{b _{1} h}{2} + \frac{b _{2} h}{2}

▼

Simplify right-hand side

AddFrac

Add fractions

A = \frac{b _{1} h + b _{2} h}{2}

FactorOut

Factor out $h$

A = \frac{( b _{1} + b _{2} ) h}{2}

MoveNumRight

$\frac{a}{b} = \frac{1}{b} \cdot a$

A = \frac{1}{2} (b_{1} + b_{2}) h

Substitute

$b_{1} + b_{2} = b$

A = \frac{1}{2} b h

It has been found that the area of the non-right triangle is half the product of its base

b

and its height

h .

This is the same result as the area for a right triangle. Therefore, the area of any triangle is half the product of its base

b

and its height

h .

$A = \frac{1}{2} b h$

Rule

Area of a Triangle Using Sine

The area of a triangle is half the product of the lengths of any two sides and the sine of their included angle. There are three possible formulas for every triangle.

The proof of this theorem will be developed using the triangle shown, where

∠ B

is obtuse. However, the same proof is valid for all triangles.

Proof

To find the first formula, start by drawing the altitude from $B$ and let $h$ be its length. Since the altitude is perpendicular to the base, it generates two right triangles.

Because

△ B C D

is a right triangle, the height of the triangle can be related to the sine of

∠ C

using the sine ratio.

sin C = \frac{h}{a} \Leftrightarrow h = a sin C

Next, substitute the expression found for

h

into the general formula for the area of a triangle.

Area = \frac{1}{2} b h ⇓ Area = \frac{1}{2} a b sin C

The first formula was obtained. To obtain the second formula, notice that

△ A B D

is also a right triangle. Therefore, the sine ratio can be applied again, this time to connect

h

and

∠ A .

sin A = \frac{h}{c} \Leftrightarrow h = c sin A

By substituting this expression into the general formula for the area of a triangle, the second formula can be obtained.

Area = \frac{1}{2} b h ⇓ Area = \frac{1}{2} b c sin A

To deduce the third formula, the altitude from $C$ or $A$ should be drawn. In this case, the altitude from $C$ will be arbitrarily drawn and labeled $D$ with a length of $h .$ Because $∠ B$ is obtuse, the altitude will lie outside the triangle.

In this case, the length of the base is

c

and the height is

h .

Since

△ B D C

is a right triangle, the sine ratio can be used to connect

∠ C B D

and

h .

sin C B D = \frac{h}{a}

Since

∠ C B D

and

∠ B

form a linear pair, they are supplementary angles. Recall that the sine of an angle is equal to the sine of its supplementary angle. With this information, and using the Substitution Property of Equality, a formula connecting

∠ B

and

h

can be written.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ s i n B = s i n C B D s i n C B D = \frac{h}{a} \Rightarrow s i n B = \frac{h}{a}

Multiplying both sides of the last equation by

a,

it is obtained that

h = a sin B .

Finally, substitute this expression for

h

into the formula for the area of

△ A B C .

Area = \frac{1}{2} c h ⇓ Area = \frac{1}{2} a c sin B

Rule

Area of a Triangle Using a Determinant

If the coordinates of the vertices of a triangle are known, its area can be calculated using a determinant.

A matrix can be written using the coordinate pairs of the vertices. To calculate a determinant, a square matrix is needed. Therefore, a third column — all of whose values are

1

— will be added.

M = ⎣ ⎢ ⎡ x_{1} x_{2} x_{3} y_{1} y_{2} y_{3} 111 ⎦ ⎥ ⎤

The area of the triangle is equal to half the absolute value of the determinant of

M .

$Area = \frac{1}{2} ∣ det M ∣$

Proof

The formula will be proven by drawing a rectangle so that the vertices of the triangle lie on the sides of the rectangle.

This proof will be developed based on the given diagram, but it is valid for any triangle. The sides of $△ A B C$ divide the rectangle into four different triangles. The area of $△ A B C$ will be calculated by following three steps.

Calculate the Area of Rectangle

A D E F

First, the side lengths of the rectangle will be calculated.

The area of the rectangle

A

is the product of its length and width.

A = ℓ w ⇓ A = (x_{2} - x_{1}) (y_{3} - y_{1})

Calculate the Areas of

△ A C F,

△ A B D,

and

△ B C E

Since

D,

E,

and

F

are vertices of the rectangle,

△ A C F,

△ A B D,

and

△ B C E

are right triangles. The area of a right triangle

A

is half the product of its legs.

A = \frac{1}{2} ℓ_{1} ℓ_{2}

The lengths of the legs of each triangle can be calculated using the coordinates.

The area of each triangle will be evaluated using a table.

Triangle	Length of the Legs	Area
$△ A B D$	$x_{2} - x_{1}$ and $y_{2} - y_{1}$	$A_{1} = \frac{1}{2} (x_{2} - x_{1}) (y_{2} - y_{1})$
$△ B C E$	$x_{2} - x_{3}$ and $y_{3} - y_{2}$	$A_{2} = \frac{1}{2} (x_{2} - x_{3}) (y_{3} - y_{2})$
$△ A C F$	$x_{3} - x_{1}$ and $y_{3} - y_{1}$	$A_{3} = \frac{1}{2} (x_{3} - x_{1}) (y_{3} - y_{1})$

Subtract the Areas and Simplify

The area of

△ A B C

is equal to the difference between the area of the rectangle and the sum of the areas of the right triangles.

A_{△ A B C} = A - (A_{1} + A_{2} + A_{3})

SubstituteValues

Substitute values

A_{△ A B C} = (x_{2} - x_{1}) (y_{3} - y_{1}) - (\frac{1}{2} (x_{2} - x_{1}) (y_{2} - y_{1}) + \frac{1}{2} (x_{2} - x_{3}) (y_{3} - y_{2}) + \frac{1}{2} (x_{3} - x_{1}) (y_{3} - y_{1}))

▼

Simplify right-hand side

Distr

Distribute $- 1$

A_{△ A B C} = (x_{2} - x_{1}) (y_{3} - y_{1}) - \frac{1}{2} (x_{2} - x_{1}) (y_{2} - y_{1}) - \frac{1}{2} (x_{2} - x_{3}) (y_{3} - y_{2}) - \frac{1}{2} (x_{3} - x_{1}) (y_{3} - y_{1})

NumberToFrac

$a = \frac{2 \cdot a}{2}$

A_{△ A B C} = \frac{2 ( x _{2} - x _{1} ) ( y _{3} - y _{1} )}{2} - \frac{1}{2} (x_{2} - x_{1}) (y_{2} - y_{1}) - \frac{1}{2} (x_{2} - x_{3}) (y_{3} - y_{2}) - \frac{1}{2} (x_{3} - x_{1}) (y_{3} - y_{1})

MoveNumRight

$\frac{a}{b} = \frac{1}{b} \cdot a$

A_{△ A B C} = \frac{1}{2} \cdot 2 (x_{2} - x_{1}) (y_{3} - y_{1}) - \frac{1}{2} (x_{2} - x_{1}) (y_{2} - y_{1}) - \frac{1}{2} (x_{2} - x_{3}) (y_{3} - y_{2}) - \frac{1}{2} (x_{3} - x_{1}) (y_{3} - y_{1})

FactorOut

Factor out $\frac{1}{2}$

A_{△ A B C} = \frac{1}{2} (2 (x_{2} - x_{1}) (y_{3} - y_{1}) - (x_{2} - x_{1}) (y_{2} - y_{1}) - (x_{2} - x_{3}) (y_{3} - y_{2}) - (x_{3} - x_{1}) (y_{3} - y_{1}))

Distr

Distribute $2$

A_{△ A B C} = \frac{1}{2} ((2 x_{2} - 2 x_{1}) (y_{3} - y_{1}) - (x_{2} - x_{1}) (y_{2} - y_{1}) - (x_{2} - x_{3}) (y_{3} - y_{2}) - (x_{3} - x_{1}) (y_{3} - y_{1}))

MultPar

Multiply parentheses

A_{△ A B C} = \frac{1}{2} (2 x_{2} y_{3} - 2 x_{2} y_{1} - 2 x_{1} y_{3} + 2 x_{1} y_{1} - (x_{2} y_{2} - x_{2} y_{1} - x_{1} y_{2} + x_{1} y_{1}) - (x_{2} y_{3} - x_{2} y_{2} - x_{3} y_{3} + x_{3} y_{2}) - (x_{3} y_{3} - x_{3} y_{1} - x_{1} y_{3} + x_{1} y_{1}))

Distr

Distribute $- 1$

A_{△ A B C} = \frac{1}{2} (2 x_{2} y_{3} - 2 x_{2} y_{1} - 2 x_{1} y_{3} + 2 x_{1} y_{1} - x_{2} y_{2} + x_{1} y_{2} + x_{2} y_{1} - x_{1} y_{1} - x_{2} y_{3} + x_{2} y_{2} + x_{3} y_{3} - x_{3} y_{2} - x_{3} y_{3} + x_{3} y_{1} + x_{1} y_{3} - x_{1} y_{1})

AddSubTerms

Add and subtract terms

A_{△ A B C} = \frac{1}{2} (x_{2} y_{3} - x_{2} y_{1} - x_{1} y_{3} + x_{1} y_{2} - x_{3} y_{2} + x_{3} y_{1})

Now, consider matrix

M

formed by the coordinates of the vertices of

△ A B C .

M = ⎣ ⎢ ⎡ x_{1} x_{2} x_{3} y_{1} y_{2} y_{3} 111 ⎦ ⎥ ⎤

The determinant of the

3 \times 3

matrix will be calculated.

By the Commutative Property of Addition and Multiplication, the right-hand side of the above equation can be rewritten.

det M = x_{1} y_{2} + y_{1} x_{3} + x_{2} y_{3} - x_{3} y_{2} - y_{3} x_{1} - x_{2} y_{1}

CommutativePropAdd

Commutative Property of Addition

det M = x_{2} y_{3} - x_{2} y_{1} - y_{3} x_{1} + x_{1} y_{2} - x_{3} y_{2} + y_{1} x_{3}

CommutativePropMult

Commutative Property of Multiplication

det M = x_{2} y_{3} - x_{2} y_{1} - x_{1} y_{3} + x_{1} y_{2} - x_{3} y_{2} + x_{3} y_{1}

The value of the determinant is twice the the area of the triangle previously found.

Area of $△ A B C$	$\frac{1}{2} (x_{2} y_{3} - x_{2} y_{1} - x_{1} y_{3} + x_{1} y_{2} - x_{3} y_{2} + x_{3} y_{1})$
Determinant	$x_{2} y_{3} - x_{2} y_{1} - x_{1} y_{3} + x_{1} y_{2} - x_{3} y_{2} + x_{3} y_{1}$

This means that the area of

△ A B C

is half the determinant of

M .

Additionally, since the area is always positive, the determinant is also positive. Therefore, by the definition of absolute value,

det M = ∣ det M ∣ .

{Area = \frac{1}{2} d e t M d e t M = ∣ det M ∣

Finally, using the Transitive Property of Equality, the formula for the area of

△ A B C

can be obtained.

$Area = \frac{1}{2} ∣ det M ∣$

Rule

Heron's Formula

Let $a,$ $b,$ and $c$ be the side lengths of a triangle and $s$ its semi-perimeter — half the perimeter.

The area of the triangle $A$ can be calculated by using the following formula.

$A = s (s - a) (s - b) (s - c)$

Proof

Consider a triangle with side lengths $a,$ $b,$ $c,$ and height $h .$

The height divides the base into two segments. If

d

represents the length of one of these segments, the second segment has length

b - d .

Also, the height divides the original triangle into two right triangles. Therefore, the Pythagorean Theorem can be used to write a system of equations for

h

and

d .

{h^{2} + d^{2} = c^{2} h^{2} + (b - d)^{2} = a^{2} (I) (II)

The value of

d

can be found by subtracting Equation (II) from Equation (I) and simplifying.

h^{2} + d^{2} - (h^{2} + (b - d)^{2}) = c^{2} - a^{2}

▼

Solve for

d

Distr

Distribute $- 1$

h^{2} + d^{2} - h^{2} - (b - d)^{2} = c^{2} - a^{2}

SubTerm

Subtract term

d^{2} - (b - d)^{2} = c^{2} - a^{2}

ExpandNegPerfectSquare

$(a - b)^{2} = a^{2} - 2 a b + b^{2}$

d^{2} - (b^{2} - 2 b d + d^{2}) = c^{2} - a^{2}

Distr

Distribute $- 1$

d^{2} - b^{2} + 2 b d - d^{2} = c^{2} - a^{2}

SubTerm

Subtract term

- b^{2} + 2 b d = c^{2} - a^{2}

AddEqn

$LHS + b^{2} = RHS + b^{2}$

2 b d = c^{2} - a^{2} + b^{2}

DivEqn

$LHS / 2 b = RHS / 2 b$

d = \frac{c ^{2} - a ^{2} + b ^{2}}{2 b}

CommutativePropAdd

Commutative Property of Addition

d = \frac{- a ^{2} + b ^{2} + c ^{2}}{2 b}

Now

h^{2}

can be calculated by substituting the expression for

d

into Equation (I).

h^{2} + d^{2} = c^{2}

Substitute

$d = \frac{- a ^{2} + b ^{2} + c ^{2}}{2 b}$

h^{2} + (\frac{- a ^{2} + b ^{2} + c ^{2}}{2 b})^{2} = c^{2}

SubEqn

$LHS - (\frac{- a ^{2} + b ^{2} + c ^{2}}{2 b})^{2} = RHS - (\frac{- a ^{2} + b ^{2} + c ^{2}}{2 b})^{2}$

h^{2} = c^{2} - (\frac{- a ^{2} + b ^{2} + c ^{2}}{2 b})^{2}

▼

Simplify right-hand side

PowQuot

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

h^{2} = c^{2} - \frac{( - a ^{2} + b ^{2} + c ^{2} ) ^{2}}{( 2 b ) ^{2}}

PowProd

$(a \cdot b)^{m} = a^{m} \cdot b^{m}$

h^{2} = c^{2} - \frac{( - a ^{2} + b ^{2} + c ^{2} ) ^{2}}{4 b ^{2}}

NumberToFrac

$a = \frac{4 b ^{2} \cdot a}{4 b ^{2}}$

h^{2} = \frac{4 b ^{2} c ^{2}}{4 b ^{2}} - \frac{( - a ^{2} + b ^{2} + c ^{2} ) ^{2}}{4 b ^{2}}

SubFrac

Subtract fractions

h^{2} = \frac{4 b ^{2} c ^{2} - ( - a ^{2} + b ^{2} + c ^{2} ) ^{2}}{4 b ^{2}}

WritePow

Write as a power

h^{2} = \frac{2 ^{2} b ^{2} c ^{2} - ( - a ^{2} + b ^{2} + c ^{2} ) ^{2}}{4 b ^{2}}

ProdPow

$a^{m} \cdot b^{m} = (a \cdot b)^{m}$

h^{2} = \frac{( 2 b c ) ^{2} - ( - a ^{2} + b ^{2} + c ^{2} ) ^{2}}{4 b ^{2}}

FacDiffSquares

$a^{2} - b^{2} = (a + b) (a - b)$

h^{2} = \frac{( 2 b c - a ^{2} + b ^{2} + c ^{2} ) ( 2 b c + a ^{2} - b ^{2} - c ^{2} )}{4 b ^{2}}

CommutativePropAdd

Commutative Property of Addition

h^{2} = \frac{( b ^{2} + 2 b c + c ^{2} - a ^{2} ) ( a ^{2} - b ^{2} + 2 b c - c ^{2} )}{4 b ^{2}}

FactorOut

Factor out $- 1$

h^{2} = \frac{( b ^{2} + 2 b c + c ^{2} - a ^{2} ) ( a ^{2} - ( b ^{2} - 2 b c + c ^{2} ) )}{4 b ^{2}}

$a^{2} \pm 2 a b + b^{2} = (a \pm b)^{2}$

h^{2} = \frac{( ( b + c ) ^{2} - a ^{2} ) ( a ^{2} - ( b - c ) ^{2} )}{4 b ^{2}}

FacDiffSquares

$a^{2} - b^{2} = (a + b) (a - b)$

h^{2} = \frac{( b + c + a ) ( b + c - a ) ( a + b - c ) ( a - b + c )}{4 b ^{2}}

CommutativePropAdd

Commutative Property of Addition

h^{2} = \frac{( a + b + c ) ( b + c - a ) ( a + b - c ) ( a + c - b )}{4 b ^{2}}

Next, the semi-perimeter

s

of the triangle will be incorporated into the right-hand side of the obtained equation. To do so, recall that the semi-perimeter is half the perimeter.

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

Therefore, the expression

a + b + c

can be written in terms of

s .

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

MultEqn

$LHS \cdot 2 = RHS \cdot 2$

2 s = a + b + c

RearrangeEqn

Rearrange equation

a + b + c = 2 s

The other factors of the numerator in the expression for

h^{2},

which are

b + c - a,

a + b - c,

and

a + c - b,

can be written in terms of

s

as well.

$a + b + c = 2 s$
Original Expression	Rewrite	Substitute
$b + c - a$	$a + b + c - 2 a$	$2 s - 2 a$
$a + b - c$	$a + b + c - 2 c$	$2 s - 2 c$
$a + c - b$	$a + b + c - 2 b$	$2 s - 2 b$