Quadrilaterals - Perimeters and Areas

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We start with the simplest case. Consider a rectangle of dimensions \(l \times b:\)

Area of rectangle

The perimeter of ABCD is

\[AB + BC + CD + DA = 2(l + b)\; \rm{units}\]

The area is

\[A = (l \times b)\; \rm{sq.\;units}\]

AC and BD  are the diagonal;  their lengths can be evaluated using the Pythagoras theorem:

\[AC = BD = \sqrt {{l^2} + {b^2}} \]

A square is a special case of a rectangle in which length is equal to breadth

Perimeter of rectangle

For this square ABCD,

  • \(\text{Perimeter}\,= 4a\,{\rm{  units}}\)
  • \({\rm{Area = }}{{\rm{a}}^2}\;{\rm{  sq.}}\;{\rm{units}}\)
  • \(AC = BD = \sqrt 2 a\;{\rm{  units}}\)

Next, consider a parallelogram:

Area of Parallelogram

We  have shown earlier that the area of a parallelogram equals base X height . For the figure above,

\[\text{Area}\left( {ABCD} \right) = b \times h\]

The perimeter P can be calculated as follows:

\[\begin{align}&P\; = AB + BC + CD + DA\\ &\quad = 2(AB + BC)\end{align}\]

Suppose that the parallelogram is a rhombus. Recall  that the diagonals of a rhombus are perpendicular, as shown in the following figure:

Area of rhombus

Suppose that the diagonal lengths are AC=dunits  and BD =dunits. We note that the diagonals divide the rhombus into four triangles of equal area, and so

\[\begin{align}&\rm{area}\,(ABCD) = 4\,area(\Delta AOD)\\&\qquad\qquad\qquad = 4\left( {\frac{1}{2} \times OD \times OA} \right)\\&\qquad\qquad\qquad = 4\left( {\frac{1}{2} \times \frac{{BD}}{2} \times \frac{{AC}}{2}} \right)\\&\qquad\qquad\qquad = \frac{1}{2} \times {d_1} \times {d_2}\,{\rm{ sq }}\,{\rm{.units}}\end{align}\]

Thus, the area of a rhombus  equals  half the product of its diagonals.

Another particular kind of quadrilateral is a trapezium, a quadrilateral in which two sides are parallel :

Area of trapezium

We know the base lengths (the lengths of the parallel sides) and the height of the trapezium (the distance between the parallel sides). Can we express the area of the trapezium in terms of these parameters?

Consider \(\Delta ABD\) and \(\Delta DBC\)  separately.

We have

\[\begin{align}&\rm{area}\left( {\Delta ABD} \right) = \frac{1}{2} \times AD \times height\\ & \qquad\qquad\qquad\;= \frac{1}{2} \times {b_1} \times h\\&\rm{area}\left( {\Delta DBC} \right) = \frac{1}{2} \times BC \times height\\ &\qquad\qquad\qquad\;= \frac{1}{2} \times {b_2} \times h\end{align}\]

Thus,

\[\begin{align}&\rm{area}\left( {ABCD} \right) = area\left( {\Delta ABD} \right) + area\left( {\Delta DBC} \right)\\&\qquad\qquad\qquad\; = \frac{1}{2} \times {b_1} \times h + \frac{1}{2} \times {b_2} \times h\\&\qquad\qquad\qquad\; = \frac{1}{2} \times \left( {{b_1} + {b_2}} \right) \times h\end{align}\]

Remember this: the area of a trapezium is

\[\frac{1}{2} \times \left( {{\rm{sum\;of\;bases}}} \right) \times \left( \rm{height} \right)\]

Now, consider an arbitrary quadrilateral:

Arbitrary quadrilateral

How can we find its area?

Let us consider a number of different scenarios:

  1. The length of the sides and one diagonal are known, Suppose that for the quadrilateral above, the lengths of the four sides and the diagonal AC  are known. In  this case, we can evaluate the areas of \(\Delta ABC\) and \(\Delta ADC\) separately and add them to obatain the area of the quadrilateral .
  2. The length of one diagonal and the lengths of the perpendiculars from opposite vertices to this diagonal are known.

    Consider the following figure:

    Area of arbitrary quadrilateral

    We have :

    \[\begin{align}&\rm{area}\left( {ABCD} \right) \;= area\left( {\Delta ABD} \right) + area\left( {\Delta CBD} \right)\\ &\qquad \qquad \qquad\;= \frac{1}{2} \times l \times {h_1} + \frac{1}{2} \times l \times {h_2}\\ &\qquad \qquad \qquad\;= \frac{1}{2} \times l \times \left( {{h_1} + {h_2}} \right)\end{align}\]

  3. The diagonals are perpendicular and their lengths are known.

    Consider the following figure:

    Area of rhombus

    We have:

    \[\begin{align}&\rm{area}\left( {\Delta ABC} \right) = \frac{1}{2} \times AC \times BO\\&\rm{area}\left( {\Delta ADC} \right) = \frac{1}{2} \times AC \times DO\\&{\rm{Adding\;the\;two,\;we\;have}}\\&{\rm{area}}\left( {ABCD} \right)\quad= \frac{1}{2} \times AC \times \left( {BO + DO} \right)\\&\qquad\qquad\qquad\quad= \frac{1}{2} \times AC \times BD\\& \qquad\qquad\qquad\quad= \frac{1}{2} \times {d_1} \times {d_2}\end{align}\]

Thus, if the diagonals of a quadrilateral  are perpendicular, its area is given by half of the product of its diagonal lengths.

Example 1: Consider the following figure:

Area of arbitrary quadrilateral

Solution: Using the Pythagoras theorem, the length of BD will be \(\sqrt {{8^2} + {6^2}} \) or 10 cm. Now, we find the areas of \(\Delta ABD\) and \(\Delta CBD\) separately .

  • \(\rm{area}\,({\Delta ABD}) \)

                      \(\begin{align}&S = \frac{{AB + BD + DA}}{2} = 13cm\\&\rm{Area}\,= \sqrt {13 \times \left( {13 - 9} \right) \times \left( {13 - 10} \right) \times \left( {13 - 7} \right)} \\&\qquad = \sqrt {13 \times 4 \times 3 \times 6} \\&\qquad \approx 30.6\, \rm{cm^2}\end{align}\)

  • \(\rm{area}\,( {\Delta BCD}) \)

                           \(\begin{align}&\rm{Area}\;= \frac{1}{2} \times BC \times CD\\&\qquad  = \frac{1}{2} \times 8 \times 6\\&\qquad  = 24\,\rm{cm^2} \end{align}\)

The total area of ABCD is thus approximately 54.6 cm2.

Example 2: The adjacent sides of a parallelogram have lengths 12 cm and 8 cm.If the distance between the shorter sides is 6 cm, what is the distance between the longer sides ?

Solution: Consider the following figure:

Area of Parallelogram example

We can explain the area of ABCD as either

  •     \(\left( {BC \times x} \right)\; \rm{cm^2}\)

  •     \(or{\rm{ }}\left( {AB \times 6} \right)\;\rm{cm^2}\)

Thus,

\[\begin{align}&BC \times x = AB \times 6\\&\Rightarrow 12x = 48\\&\Rightarrow X = 4\end{align}\]

The distance between the longer sides is 4 cm.

Example 3:  A  park has the following dimensions:

Area of arbitrary pentagon

Find its  total area.

Solution: We have:

\[\begin{align}  & \text{area}\left( \Delta \,ABX \right)\text{=}\frac{\text{1}}{\text{2}}\times \text{14}\times \text{20=140}\,{{\text{m}}^{\text{2}}} \\  & \text{area}\left( \text{Trap}.\ BCZX \right)\text{=}\frac{\text{1}}{\text{2}}\times \left( \text{20+28} \right)\times \text{30} \\  & \qquad \qquad \qquad \qquad =720\,{{\text{m}}^{\text{2}}} \\  & \text{area}\left( \Delta \,CDZ \right)\text{=}\frac{\text{1}}{\text{2}}\times \text{20}\times \text{28=280}\,{{\text{m}}^{\text{2}}} \\  & \text{area}\left( \Delta \,ADF \right)\text{=}\frac{\text{1}}{\text{2}}\times AD\times FY \\  & \qquad \qquad \qquad \ =\frac{1}{2}\times 64\times 24=768\,{{\text{m}}^{\text{2}}} \\ \end{align}\]

By adding all these areas, we obtain the total area of the park as 1908 m2.

Example 4:  What  is the area of the following quadrilateral?

Example of area of quadrilateral

Solution: We find the areas of both \(\Delta ABC\) and  \(\Delta ADC\) using the Heron’s formula.

  • \(\Delta ABC\)

\(\begin{align}&\quad\qquad s = \frac{{AB + BC + CA}}{2} = 13 \;\rm{cm}\\&\quad
\rm{Area} = \sqrt {13 \times \left( {13 - 9} \right) \times \left( {13 - 5} \right) \times \left( {13 - 12} \right)} \\&\qquad\quad= \sqrt {13 \times 4 \times 8 \times 1} \\&\qquad\quad{\rm{}} \approx 20.4\;\rm{cm^2}\end{align}\)

  • \(\Delta ADC\)
    \(\begin{align}&\qquad s = \frac{{AD + DC + CA}}{2} = 13\,\rm{cm}\\&\rm{Area} = \sqrt {13 \times \left( {13 - 6} \right) \times \left( {13 - 8} \right) \times \left( {13 - 12} \right)} \\&\qquad= \sqrt {13 \times 7 \times 5 \times 1} \\&\qquad \approx 21.3\; \rm{cm^2}\end{align}\)

The total area is thus approximately equal to 41.7cm2 .

Example 5: Consider  the following figure ABCD is a rhombus while DCEF is a trapezium.

Areas of rhombus and trapezium

What is the total area of the polygon ABCEFD?

Solution: We note that the area  of the rhombus ABCD is

\[\frac{1}{2} \times AC \times BD = \frac{1}{2} \times 10 \times 20 = 100c{m^2}\]

We can also write this same area as \(CD \times \left( {{\text{distance between AB  and CD}}} \right)\) .Thus,

\[\begin{align}&CD \times 8 = 100\\&\Rightarrow CD = 12.5\,\rm{cm}\end{align}\]

Now,

\[\begin{align}  & \ \text{area}\left( trap\ DCEF \right)\text{=}\frac{\text{1}}{\text{2}}\times \left( CD\text{+}EF \right)\times \text{12} \\  & \qquad \qquad \qquad \quad \quad =\frac{1}{2}\times \left( 12.5+7.5 \right)\times 12 \\  & \qquad \qquad \qquad \quad \quad =120\,\text{c}{{\text{m}}^{2}}\end{align}\]

Thus, the total area is (100+120) or 220cm2.

Example 6:  Find the area of the following trapezium:

Example of area of trapezium

Solution:Observe that  PS||QR. However, we do no know the distance between these parallels. How do we calculate the area in this case?

Through P, draw PX||SR meeting QR at X, and \(PY \bot QR\), as shown:

Solution of area of trapezium

Note that PSRX is a parallelogram, and thus, PX = SR=25 cm, .Also, XR = PS = 60 cm, and so,

\[QX = QR - XR = 77 - 60 = 17 \,\rm{cm}\]

Now, let us calculate the area of \(\Delta PQX\) using Heron’s formula . We have:

\[s = \frac{{PQ + QX + XP}}{2}\;=\;\frac{{26 + 17 + 25}}{2} = 34\, \rm{cm}\]

Thus,

\[\begin{align}&\rm{area}\,\left( {\Delta PQX} \right) = \sqrt {34 \times \left( {34 - 26} \right) \times \left( {34 - 17} \right) \times \left( {34 - 25} \right)} \\&\qquad\qquad\qquad = \sqrt {34 \times 8 \times 17 \times 9} \\&\qquad\qquad\qquad = 204 \;\rm{cm^2}\end{align}\]

However, we can also express this area as \(\frac{1}{2} \times QX \times PY\) .

Thus,

\[\begin{align}&\frac{1}{2} \times QX \times PY = 204\\&\Rightarrow \frac{1}{2} \times 17 \times PY = 204\\ &\Rightarrow PY = 24\; \rm{cm}\end{align}\]

Thus, the distance between the parallels of the trapezium is 24 cm,and hence, its area is

\[\begin{align}&\frac{1}{2} \times \left( {PS + QR} \right) \times PY\\& = \frac{1}{2} \times \left( {60 + 77} \right) \times 24\\ &= 1644\,\rm{cm^2}\end{align}\]

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