# The Area of any Irregular Quadrilateral

A plane figure bounded by four straight line segments is called an irregular quadrilateral. The area of any irregular quadrilateral can be calculated by dividing it into triangles.

__Example__:

Find the area of a quadrilateral $$ABCD$$ whose sides are $$9$$ m, $$40$$ m, $$28$$ m and $$15$$ m respectively and the angle between the first two sides is a right angle.

__Solution__:

From the figure we notice that $$ABD$$ is a right-angled triangle, in which $$AB = 40$$ m, $$AD = 9$$ m. Also

\[\begin{gathered} B{D^2} = A{B^2} + A{D^2} \\ B{D^2} = {40^2} + {9^2} \\ BD = \sqrt {{{40}^2} + {9^2}} = \sqrt {1681} = 41 \\ \end{gathered} \]

Now, the area of $$\Delta ABD = \frac{1}{2} \times 40 \times 9 = 180$$ m

In $$\Delta BCD$$, $$BD = a = 41$$ m, $$DC = b = 28$$ m, $$CB = c = 15$$ m

$$\therefore s = \frac{{a + b + c}}{2} = \frac{{41 + 28 + 15}}{2} = 42$$ m

Now,

\[\begin{gathered} {\text{Area}}\,{\text{of}}\,\Delta BCD = \sqrt {s(s – a)(s – b)(s – c)} \, \\ {\text{Area}}\,{\text{of}}\,\Delta BCD = \sqrt {42(42 – 41)(42 – 28)(42 – 15)} = \sqrt {42 \times 14 \times 27} = 126\,sq\,m \\ \end{gathered} \]

The area of the quadrilateral $$ABCD$$$$ = $$ area of $$\Delta ABD$$ $$ + $$area of $$\Delta BCD$$

The area of the quadrilateral $$ABCD$$ $$ = (180 + 126) = 306$$ square meters.

__Example__:

In a quadrilateral the diagonal is $$42$$ cm and the two perpendiculars on it from the other vertices are $$8$$ cm and $$9$$ cm respectively. Find the area of the quadrilateral.

__Solution__:

Given that from the figure $$AC = 42$$ m, $$BN = 9$$ m, $$DM = 8$$ m

The area of $$ABCD = $$ area of $$\Delta ABC + $$area of $$\Delta ACD$$

The area of $$ABCD$$$$ = \frac{1}{2} \times 9 \times 42 + \frac{1}{2} \times 8 \times 42 = 189 + 168 = 357$$ square meters.

D . N. SAMANTA

July 12@ 4:35 pmVery good & very easy solution

K. Gomathi

September 23@ 7:48 pmVery easy to understand. But if the angle not known and only four sides are given. How to calculate

Nazim Laskar

February 12@ 10:00 amSir, will you please tell me what is the area of a quadrilater i.e. side AB=20’4″ BC=66’6″ CD=29′ DA=50′. In this angle ABC is 90 degree.

Sanjay Shyoran

September 8@ 5:57 pmSir my plot size is from South to north one west side is 198 ft and other east is 205.6 ft and from east to west from South side is 82.6ft and north side is 27.6 ft.There are 5 brothers.How to divide each share.plz solve this issue.Thanks.

Batraju Naga venkata Chaitanya

October 6@ 6:08 pmFor example if a chooses the constructing structure

It’s measurements of sides

8m width

8.5m width.

9 m length

9.3 m length

Here cross line is not available for measuring

How to calculate

Mrs.KALAISELVI

October 29@ 4:08 pmSir my plot is quadrilateral shape – total size land is 2513.58 feet . Side A=52 feet , Side B=61 feet, Side C=56 Feet and Side D=35 feet ( Side A facing road , Side B left side, Side C parallel to road and Side D right hand side). There are 2 children’s, how to divide two halves. Please give correct measures into two halves to solve this issue, by way of two method (1) horizontal and 2) straight way) and give correct feet size.

Mithun Mitra

October 30@ 3:47 amI have purchased a land of irregular side. What will be the area of the sides are 65 feets, 105 feets, 82 feets and 92 feets.

RAJESH KUMAR

November 19@ 3:30 pmWE HAVE A FIELD (KHET) WHICH SHAPE LIKE IRREGULAR Quadrilateral (LONG- SIDE-1 is 90 AND SIDE-2 is 100 & (WIDTH- SIDE-1 is 10 AND SIDE-2 is 30) HOW TO DIVIDE IN 2 PART OF THIS IN BOTH SIDE (LONG-2 & WIDTH-2)

PLS SUGGEST AND GIVE FORMULA