Triangles - Perimeters and Areas

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The area of  a  triangle can be determined in a number of ways, based on what information  is available to us.

Consider the following  figure, which shows a triangle ABC with altitude AD equal to h units in length.

Area of triangle formula

The area of this triangle can be written as

\[\begin{align}&area\left( {\Delta ABC} \right) = \frac{1}{2} \times base \times height\\& \qquad\qquad\quad\quad\;= \frac{1}{2} \times BC \times h\end{align}\]

If we take a different side to be the base (say, AC),then we would  have to take the corresponding altitude (the altitude through B).

Suppose that the lengths of all the sides of a triangle are known, as shown in the following figure:

Example: Area of triangle

Note the naming convention. The length of the side opposite to \(\angle A\)  is represented using the letter a, and so on.

The perimeter of this triangle is a + b + c. We define the semi-perimeter s as

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

In terms of s and abc, the area of the triangle can be expressed using Heron’s formula:

\[area(\Delta ABC) = \sqrt {s(s - a)(s - b)(s - c)} \]

Thus, in general, a  triangle’s area can be evaluated using one of the following two techniques:

  •     \(\frac{1}{2} \times base \times height\)

  •     \(Heron's\;{\rm{  formula}}\)

The  following are some specific scenarios involving  area calculation of triangles:

  1. A right-angled triangle:

    If two sides are known in a right -angled triangle, the third can be calculated using the Pythagoras Theorem. Consider  the following figure:

    Right-angled triangle

    We can write the areas of this triangle as

    \[area\left( {\Delta ABC} \right) = \frac{1}{2} \times AB \times BC\]

  2.  An Isosceles triangle:

    In the following figure, \(\Delta ABC\)  is isosceles with AB = AC = a units and BC = b units:

    An Isosceles triangle

    We can write the area of this triangle as follows:

    \[\begin{align}&area\left( {\Delta ABC} \right) = \frac{1}{2} \times BC \times AD\\ &\qquad\qquad\qquad\,\;= \frac{1}{2} \times b \times \sqrt {{a^2} - {{\left( {\frac{b}{2}} \right)}^2}} \\&\qquad\qquad\qquad\,\;= \frac{1}{4}b\sqrt {4{a^2} - {b^2}} \end{align}\]

  3. An equilateral triangle:

    The  following figure shows an equilateral triangle with each side equal to a units:

    An equilateral triangle

    We have

    \[\begin{align}&AD = \sqrt {{a^2} - {{\left( {\frac{a}{2}} \right)}^2}}  = \sqrt {\frac{{3{a^2}}}{4}} \\& \qquad= \frac{{\sqrt 3 a}}{2}\end{align}\]

    Thus,

    \[\begin{align}&area\left( {\Delta ABC} \right) = \frac{1}{2} \times BC \times AD\\& \qquad\qquad\qquad\;= \frac{1}{2} \times a \times \frac{{\sqrt 3 a}}{2}= \frac{{\sqrt 3 }}{4}{a^2}\end{align}\]

    You  are urged to remember  this expression. For the area of an equilateral triangle.

Example 1: Find the area and perimeter of an equilateral triangle whose height is 10 cm.

Solution: Let the length of each of the triangle’s sides be a units. Consider the following figure:

Example: Equilateral triangle

We have in \(\Delta ABD\):

\[\begin{align}&{a^2} = {h^2} + {\left( {\frac{a}{2}} \right)^2}\\& \Rightarrow \frac{{3{a^2}}}{4} = {h^2}\\& \Rightarrow a = \frac{{2h}}{{\sqrt 3 }}\end{align}\]

The area A  of the triangle is

\[\begin{align}&A = \frac{{\sqrt 3 }}{4}{a^2} = \frac{{\sqrt 3 }}{4} \times \frac{{4{h^2}}}{3}\\&\frac{{{h^2}}}{{\sqrt 3 }} = \frac{{100}}{{\sqrt 3 }}cm \approx ~57.7 \,\rm{cm^2}\end{align}\]

The perimeter of the triangle is

\[\begin{align}&P = 3a = 3 \times \frac{{2h}}{{\sqrt 3 }}\\&\quad = 2\sqrt 3 h = 20\sqrt 3 cm \approx 34.6\, \rm{cm}\end{align}\]

Example 2: The sides of a triangular park are in the ratio 12:17:25, and its perimeter is 1080 m. What is its area?

Solution: We assume the triangle’s sides to be 12 km, 17 km and 25 km for some constant k . We have:

\[\begin{align}&12k + 17k + 25k = 1080\\& \Rightarrow 54k = 1080 \Rightarrow k = 20\end{align}\]

The sides of the triangle are  thus:

240 m, 340 m, 500 m

Now, the semi-perimeter of the triangle is

\[s = \frac{{1080}}{2} = 540m\]

Using the Heron’s Formula, the area A of the triangle is

\[\begin{align}&A = \sqrt {540\left( {540 - 240} \right)\left( {540 - 340} \right)\left( {540 - 500} \right)} \\ &\quad= \sqrt {540 \times 300 \times 200 \times 40} \\& \quad= 36,000\, \rm{m^2}\end{align}\]

Example 3: Find the area of the shaded region in the  following figure:

Heron’s Formula

Solution:  The altitude for \(\Delta ABC\)  is

\[\begin{align}&h \;= \sqrt {{{15}^2} - {5^2}} \left( {how?} \right)\\&\quad = 10\sqrt 2\, \rm{cm}\end{align}\]

Thus,

\[\begin{align}&area\left( {\Delta ABC} \right) = \frac{1}{2} \times BC \times h\\& \qquad\qquad\qquad\;= \frac{1}{2} \times 10 \times 10\sqrt 2 \\&\qquad\qquad\qquad\;= 50\sqrt 2 \,\rm{cm^2}\end{align}\]

Also,

\[\begin{align}&area\left( {\Delta BDC} \right) = \frac{1}{2} \times CD \times BD\\&\qquad\qquad\qquad\; = \frac{1}{2} \times 6 \times 8= 24\, \rm{cm^2}\end{align}\]

Thus, the area A of the shaded region is

  \[\begin{align}& A \;= \rm{area}(\Delta ABC) - area(\Delta BDC)\\&\quad = (50\sqrt 2 - 24)\,\rm{cm^2}\\&\quad \approx 46.7\, \rm{cm^2}\end{align}\]

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