Triangle Area Calculator

The formula of Area of Triangular region 

In the previous part, we learned that area of the triangular region = 1/2 * base * height. In the following part, we have added some other relevant formulas

Right-angled triangle

Let in the right-angled triangle ABC, BC = a, and AB = b are the adjacent sides of the right angle. Here if we consider BC as the base and AB as the height

Area of  ABC = 1/2 × base × height = 1/2 ab

Two sides of a triangular region and the angle included between them are given

Let in ABC, the sides are 

BC = a, CA = b, AB =c.

AD is drawn perpendicular from A to BC. Let altitude (height) AD = h.

Considering the angle C we get, AD / CA = sinC

or, h/b = sinC or, h = bsinC

Area of ABC = 1/2 × BC × AD

Area of ABC = 1/2 × BC × AD

= 1/2 a × b sinC = 1/2 × absinC

Similarly, area of ABC = 1/2 bcsinA =1/2 casinB

Three sides of a triangle are given
Let in ABC, BC = a, CA = b and AB = c.

Perimeter of the triangle 2s = a + b + c.

We draw AD ⊥ BC.

Let, BD = x, then CD = a − x

In right-angled ABD and ACD

AD² = AB² − BD² and AD² = AC² − CD²

AB² − BD² =AC² − CD²

or, c² − x² = b² − (a − x)²

or, c² − x² = b² − a² + 2ax − x²

or, 2ax = c² + a² − b²

x = c² + a² − b²/ 2a

Again, 

AD²  = c² − x² 

AD²  = c² − (c² + a² − b²/ 2a)² 

AD²  = (c + c² + a² − b²/ 2a) (c - c² + a² − b²/ 2a)

AD²  = (2ac + c² + a² − b²/ 2a) (2ac - c² + a² − b²/ 2a)

AD²  = {(c + a)²  − b² } {b²  − (c − a)² }/4a² 

AD²  = (a + b + c) (a + b + c − 2b) (a+ b + c − 2a) (a+ b + c − 2c)/ 4a² 

AD²  =  2s(2s − 2b)(2s − 2a)(2s − 2c)/4a² 

AD²  =4s(s − a)(s − b)(s − c)/a²

AD = 2/a √s(s − a)(s − b)(s − c)

Area of ABC = 1/2 BC × AD

 =1/2 x a x 2/a  √{s(s − a)(s − b)(s − c)}

 =√{s(s − a)(s − b)(s − c)}

Equilateral triangle

Equilateral triangle: Let the length of each side of the equilateral triangular region ABC be a.

Draw AD ⊥ BC.

BD = CD = a/2

In right-angled ABD

BD² + AD² = AB²

or, AD² = AB² − BD² = a² − (a/2)²

or, AD² = a² − a²/4

or, AD² =4a² − a²/4

AD = √3a/2

Area of ABC = 1/2 BC × AD

=  1/2 × a × √3a/2=√3/4 a²

Isosceles triangle

Let ABC be an isosceles triangle in which

AB = AC = a and BC = b

Draw, AD is perpendicular BC. BD = CD = b/2

ABD is right angled.

AD² = AB² − BD²

AD²  = a² − (b/2)²

AD²  = a² − b²/4

AD²  = (4a² − b²)/4

AD = √(4a² − b²)/2

Area of isosceles  ABC = 1/2 BC × AD
 = 1/2 × b × √(4a² − b²)/2
 = b/4 √(4a² − b²)