AREA OF A TRIANGLE

Home » Straight lines » AREA OF A TRIANGLE

Consider a triangle as shown below:

clip_image002

The most common formula to find the area of the triangle is\frac{1}{2}bh.

In any triangle, the area of the triangle is given by

A = \frac{1}{2}bh

Where, b is the length of the base of the triangle and h is the length of the altitude drawn to an extension from the base.

Area of a triangle when three sides of a triangle are given (SSS):

A Greek philosopher and mathematician, Heron formulates a formula to find the area of a triangle, when the lengths of the three sides are given. This formula is known as Heron’s formula, which is written as follows:

\Delta s = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)}

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

Where a, b, and c are the lengths of the three sides of a triangle, and s is the semi perimeter.

Question 1

Find the area of the triangle when the lengths of base and height are12 meter and 16 meter.

Solution 1

Known the lengths of base and height, use the formula

A = \frac{1}{2}bh

Substitute the value as follows:

\begin{array}{c}
A = \frac{1}{2} \times 12 \times 16\\
 = \frac{1}{2} \times 192\\
 = 96
\end{array}

Therefore, the area of a triangle is 96 square meter.

Question 2

What is the area of a triangle when three sides are 6 cm, 4 cm and 20 cm.

Solution 2

Since three sides of a triangle are known then, use the formula

\Delta s = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)}

Find the value of s as follows:

\begin{array}{c}
s = \frac{{a + b + c}}{2}\\
 = \frac{{6 + 4 + 20}}{2}\\
 = \frac{{30}}{2}\\
 = 15
\end{array}

Substitute the value as follows:

\begin{array}{c}
\Delta s = \sqrt {15\left( {15 - 6} \right)\left( {15 - 4} \right)\left( {15 - 20} \right)} \\
 = \sqrt {15\left( 9 \right)\left( {11} \right)\left( { - 5} \right)} \\
 = \sqrt { - 7425} \\
 = 86.17
\end{array}

Therefore, the area of the triangle is 86. 17square centimeter.

Area of triangle when vertices are given:

Consider a triangle ABC, whose vertices are A(x1, y1), B(x2, y2) and C(x3, y3).

clip_image032

Draw AP \bot OX

\begin{array}{l}
BQ \bot OX\\
CR \bot OX
\end{array}

We know that,

 {\rm{Area of trapezium  =  }}\frac{1}{2} \times \left( {{\rm{sum}}\,{\rm{of}}\,{\rm{parallel}}\,{\rm{sides}}} \right){\rm{ }} \times \left( {{\rm{distance}}\,{\rm{between}}\,{\rm{them}}} \right)

 image

Area\,of\,\Delta ABC = \left| {\frac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]} \right|

Posted on