Nov 1, 2024

Different ways of finding the Area of a Triangle

few formulae

#math 1 min read

There are multiple ways of finding the area of a triangle, from unconventional to completely straight forward ways. These are a few of them:

the 1/2 x base x height

\begin{array}{l} \frac{1}{2} \times b a s e \times h e i g h t \end{array}

That's what we all know.

Heron's Formula

\begin{array}{l} \sqrt{s (s - a) (s - b) (s - c)} \end{array}

where $s$ is the semi-perimeter ( $\frac{1}{2} \times p e r i m e t e r$ ) and $a, b, c$ are the length of the sides.

See Proof (Part 1, 2)

Coordinate Formula

\begin{array}{l} \frac{1}{2} \cdot [x_{1} \cdot (y_{2} - y_{3}) + x_{2} \cdot (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] \end{array}

where $(x_{i}, y_{i})$ are the coordinates of the three vertices of the triangle in a coordinate plane.

See Proof

1/2 x Determinant of a 2x2 matrix

\begin{array}{l} \frac{1}{2} \times det A \\ \frac{1}{2} \times | \begin{array}{c} a & b \\ c & d \end{array} | \end{array}

Well, $det A$ gives you the area of a parallelogram. Which means, twice the area of a triangle.

See Proof

1/2 x magnitude of vector cross-product

\begin{array}{l} \frac{1}{2} \times | \vec{a} \times \vec{b} | \end{array}

Since $| \vec{a} \times \vec{b} |$ is essentially the area of the parallelogram ( $∵ | \vec{a} \times \vec{b} | = a \times b \times \sin θ \to b a s e \times h e i g h t$ ), the half of it will equal to the area of the triangle subtended by the vectors.

And all you have to do now is to use the right method at the right time, which is the hardest step of it all!

Happy Math!

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Different ways of finding the Area of a Triangle

few formulae

the 1/2 x base x height #

Heron's Formula #

Coordinate Formula #

1/2 x Determinant of a 2x2 matrix #

1/2 x magnitude of vector cross-product #