Area of a Triangle

Every school-age student knows that the area of a triangle is one half the base times its height, or

Area = ½bh

A rather clever proof for this is to show that two identical triangles can be cut and rotated into a single rectangle of the same base and height, and since two identical triangles would have an area of A = bh, one triangle would have an area of A = ½bh. 

However, what if the height or base of a triangle is not given?  Can the area still be found?  You may recall from geometrical congruency proofs that a triangle can be defined by a combination of its angles and sides, namely side-angle-side (SAS), side-side-side (SSS), angle-side-angle (ASA) or angle-angle-side (AAS).  Therefore, there must be an area formula for each of these congruencies.

SAS Triangle Area Formula

If two sides and an included angle of a triangle are given (SAS), the area is

Area = ½ a b sin C

where a and b are the two sides and ∠C is the measure of the included angle.  This formula is often taught in high school trigonometry, and is handy for answering some SAT and ACT questions.

The proof is fairly straightforward.  In the diagram above, sin C = ^h/_a, or h = a sin C.  Substituting this into A = ½bh, we get A = ½b(a sin C), or A = ½ a b sin C.

SSS Triangle Area Formula (Heron’s Formula)

If all three sides of a triangle are given (SSS), the area is

where a, b, and c are the three sides, and s = ½(a + b + c).  This formula is also known as Heron’s Formula and is somewhat obscure in high school math curricula, likely because of the complexity of the derivation.

Heron’s formula can be derived by manipulating Pythagorean’s Theorem from the two right angle triangles in the above diagram.  For the left-hand right triangle,

b12 + h2 = a2

(Pythagorean’s Theorem)

and for the right-hand right triangle,

b22 + h2 = c2	(Pythagorean’s Theorem)
(b – b1)2 + h2 = c2	(since b = b1 + b2)
b2 – 2b1b + b12 + h2 = c2	(expand)
b12 + h2 = c2 – b2 + 2b1b	(rearrange)

This means that b₁² + h² is equal to both a² and c² – b² + 2b₁b, so

a2 = c2 – b2 + 2b1b	(transitive)
2b1b = a2 + b2 – c2	(rearrange)
b1 = 1/2b(a2 + b2 – c2)	(divide)

Going back to the left-hand triangle formula, substituting b₁, and solving and simplifying h:

b12 + h2 = a2	(Pythagorean’s Theorem)
(1/2b(a2 + b2 – c2))2 + h2 = a2	(substitution)
h2 = a2 – (1/2b(a2 + b2 – c2))2	(subtraction)
h = √(a2 – (1/2b(a2 + b2 – c2))2)	(square root)
h = √(a2 – 1/4b^2(a2 + b2 – c2)2)	(square)
h = √(1/4b^24a2b2 – 1/4b^2(a2 + b2 – c2)2)	(common denominator)
h = √(1/4b^2(4a2b2 – (a2 + b2 – c2)2))	(factor out 1/4b^2)
h = 1/2b√(4a2b2 – (a2 + b2 – c2)2)	(square root 1/4b^2)
h = 1/2b√((2ab – (a2 + b2 – c2)) (2ab + (a2 + b2 – c2)))	(difference of squares)
h = 1/2b√((2ab – a2 – b2 + c2) (2ab + a2 + b2 – c2))	(distribute)
h = 1/2b√((c2 – (a2 – 2ab + b2)) ((a2 + 2ab + b2) – c2))	(rearrange, factor)
h = 1/2b√((c2 – (a – b)2)((a + b)2 – c2))	(factor)
h = 1/2b√((c – (a – b))(c + (a – b)) ((a + b) – c)((a + b) + c))	(difference of squares)
h = 1/2b√((c – a + b)(c + a – b) (a + b – c)(a + b + c))	(distribute negative)
h = 1/2b√((-a + b + c)(a – b + c) (a + b – c)(a + b + c))	(rearrange)
h = 1/2b√((a+b+c – 2a)(a+b+c – 2b) (a+b+c – 2c)(a+b+c))	(rearrange)
h = 1/2b√((16)½(a+b+c – 2a)½(a+b+c – 2b) ½(a+b+c – 2c)½(a+b+c))	(rearrange)
h = 1/2b√(16(½(a+b+c) – a)(½(a+b+c) – b) (½(a+b+c) – c)(½(a+b +c))	(distribute)
h = 1/2b√(16(s – a)(s – b)(s – c)s)	(substitute s = ½(a + b + c))
h = 1/2b4√(s(s – a)(s – b)(s – c))	(square root, rearrange)
h = 2/b√(s(s – a)(s – b)(s – c))	(simplify)

Lastly, substituting h into the area formula:

A = ½bh	(area formula)
A = ½b(2/b√(s(s – a)(s – b)(s – c)))	(substitute)
A = √(s(s – a)(s – b)(s – c))	(simplify)

which is Heron’s Formula.

ASA Triangle Area Formula

If two angles and an included side of a triangle are given (ASA), the area is

which is also equivalent to

The proof involves the SAS Triangle Area Formula (A = ½ a b sin C), the Law of Sines (^{sin A}/_a = ^{sin B}/_b = ^{sin C}/_c), and the trigonometric identity sin(180° – x ) = sin x.  From the Law of Sines, a = ^{b sin A}/_{sin B}.  Substituting this into the SAS Triangle Area Formula we get:

A = ½ a b sin C	(SAS area formula)
A = ½ (b sin A/sin B) b sin C	(substitute)
A = b^2 sin A sin C/2 sin B	(simplify)
A = b^2 sin A sin C/2 sin (180° – (A + C))	(angle sum of triangle)
A = b^2 sin A sin C/2 sin (A + C)	(trig identity)

The secondary formula can be derived by expanding sin (A + C) and dividing the numerator and denominator by cos A cos C:

A = b^2 sin A sin C/2 (sin A cos C + cos A sin C)	(sine addition formula)
A = b^2 tan A tan C /2 (tan A + tan C)	(divide numerator and denominator by cos A cos C)

AAS Triangle Area Formula

If two angles and a non-included side of a triangle are given (AAS), the area is

The proof is similar to the ASA area proof, and also involves the SAS Triangle Area Formula (A = ½ a b sin C), the Law of Sines (^{sin A}/_a = ^{sin B}/_b = ^{sin C}/_c), and the trigonometric identity sin(180° – x ) = sin x.  From the Law of Sines, a = ^{b sin A}/_{sin B}.  Substituting this into the SAS Triangle Area Formula we get:

A = ½ a b sin C	(SAS area formula)
A = ½ (b sin A/sin B) b sin C	(substitute)
A = b^2 sin A sin C/2 sin B	(simplify)
A = b^2 sin A sin (180° – (A + B))/2 sin B	(angle sum of triangle)
A = b^2 sin A sin (A + B)/2 sin B	(trig identity)

Conclusion

The area of a triangle can be expressed in many different ways other than the traditional formula of Area = ½bh.  If two sides and an included angle are given (SAS), the area can be expressed as Area = ½ a b sin C.  If all three sides are given (SSS), the area can be expressed as Area = √(s(s – a)(s – b)(s – c)), which is also known as Heron’s Formula.  If two angles and an included side of a triangle are given (ASA), the area can be expressed as either Area = ^{b^2 sin A sin C}/_{2 sin (A + C)} or Area = ^{b^2 tan A tan C} /_{2 (tan A + tan C)}.  Finally, if two angles and a non-included side of a triangle are given (AAS), the area can be expressed as Area = ^{b^2 sin A sin (A + B)}/_{2 sin B}.