Area of Quadrilaterals and Polygons (Advanced)
Area of Quadrilaterals (General Formula using Diagonal)
For a general quadrilateral (a four-sided polygon that does not necessarily have parallel sides or other special properties), we can find its area if we are given the length of one of its diagonals and the lengths of the perpendicular segments (altitudes) drawn from the other two vertices to that diagonal.
Method and Formula
Consider any quadrilateral ABCD. Let's choose one of its diagonals, say AC. Let the length of this diagonal be denoted by '$d$'.
From the other two vertices, B and D, draw perpendicular segments to the diagonal AC. Let the foot of the perpendicular from B to AC be $P$, and the foot of the perpendicular from D to AC be $Q$. The lengths of these perpendiculars are the heights of the triangles formed by the diagonal.
Let the length of the perpendicular from B to AC be $h_1$ (BP) and the length of the perpendicular from D to AC be $h_2$ (DQ). Note that $h_1$ and $h_2$ are the altitudes of the two triangles formed by the diagonal, with the diagonal as the common base.
The diagonal AC divides the quadrilateral ABCD into two triangles: $\triangle \text{ABC}$ and $\triangle \text{ADC}$. The area of the quadrilateral is the sum of the areas of these two triangles.
Area of $\triangle \text{ABC}$:
We can consider AC as the base of $\triangle \text{ABC}$. The length of the base is $d$. The corresponding height is the perpendicular distance from vertex B to the base AC, which is $h_1$.
Area($\triangle \text{ABC}$) $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \text{AC} \times h_1$
Area($\triangle \text{ABC}$) $= \frac{1}{2} d h_1$
... (1)
Area of $\triangle \text{ADC}$:
Similarly, we can consider AC as the base of $\triangle \text{ADC}$. The length of the base is $d$. The corresponding height is the perpendicular distance from vertex D to the base AC, which is $h_2$.
Area($\triangle \text{ADC}$) $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \text{AC} \times h_2$
Area($\triangle \text{ADC}$) $= \frac{1}{2} d h_2$
... (2)
Total Area of the Quadrilateral:
The total area of the quadrilateral ABCD is the sum of the areas of the two triangles it is composed of:
Area(ABCD) = Area($\triangle \text{ABC}$) + Area($\triangle \text{ADC}$)
... (3)
Substitute the expressions for the areas of the triangles (from equations 1 and 2) into equation (3):
Area(ABCD) $= \frac{1}{2} d h_1 + \frac{1}{2} d h_2$
[Substituting from (1) and (2)]
Factor out the common term $\frac{1}{2} d$ from both terms on the right side:
$\textbf{Area of Quadrilateral} = \mathbf{\frac{1}{2} d (h_1 + h_2)}$
... (4)
Formula:
The formula for the area of a general quadrilateral, given a diagonal and the perpendiculars to it from the other two vertices, is:
$\mathbf{A_{\text{quadrilateral}} = \frac{1}{2} \times (Length \$\$ of \$\$ Diagonal) \times (Sum \$\$ of \$\$ lengths \$\$ of \$\$ Perpendiculars \$\$ from \$\$ other \$\$ vertices \$\$ to \$\$ the \$\$ diagonal)}$
Or simply, $\mathbf{A = \frac{1}{2} d (h_1 + h_2)}$, where:
- $d$ is the length of the chosen diagonal.
- $h_1$ and $h_2$ are the lengths of the perpendiculars from the other two vertices to that diagonal.
Example
Example 1. Find the area of a quadrilateral PQRS where the diagonal PR = $20$ cm, and the perpendiculars from Q and S to PR are $6$ cm and $8$ cm respectively.
Answer:
Given:
Quadrilateral PQRS.
Length of the diagonal PR, $d = 20$ cm.
Length of the perpendicular from Q to PR, $h_1 = 6$ cm.
Length of the perpendicular from S to PR, $h_2 = 8$ cm.
To Find:
Area of quadrilateral PQRS.
Solution:
We use the general formula for the area of a quadrilateral using a diagonal and the perpendiculars to it from the other vertices:
$\text{Area} (A) = \frac{1}{2} d (h_1 + h_2)$
[Formula (4)]
Substitute the given values:
$A = \frac{1}{2} \times 20 \$ \text{cm} \times (6 \$ \text{cm} + 8 \$ \text{cm})$
[Substituting $d=20$, $h_1=6$, $h_2=8$]
First, calculate the sum inside the parenthesis:
"$A = \frac{1}{2} \times 20 \$ \text{cm} \times (14 \$ \text{cm})$"
[Calculating $6+8$]
Now perform the multiplication:
"$A = 10 \$ \text{cm} \times 14 \$ \text{cm}$"
[Calculating $\frac{1}{2} \times 20$]
"$\mathbf{A = 140 \$\$ cm^2}$"
[Calculating $10 \times 14$]
Therefore, the area of the quadrilateral PQRS is 140 square centimetres ($\text{cm}^2$).
Area of Specific Quadrilaterals (Rhombus, Kite)
For certain quadrilaterals with specific properties related to their diagonals, we can derive simpler area formulas. The rhombus and the kite are two such examples where the area can be easily calculated using the lengths of their diagonals.
Area of a Rhombus
A rhombus is a parallelogram in which all four sides are equal in length. Key properties of a rhombus relevant to its area calculation using diagonals are:
- All four sides are equal.
- Opposite angles are equal.
- The diagonals bisect each other (divide each other into two equal parts).
- The diagonals are perpendicular to each other (intersect at a $90^\circ$ angle).
Let the lengths of the two diagonals of the rhombus be $d_1$ and $d_2$. Let the diagonals be AC and BD, intersecting at point O.
Since the diagonals bisect each other, $AO = OC = \frac{d_1}{2}$ and $BO = OD = \frac{d_2}{2}$. Since they are perpendicular, $\angle \text{AOB} = \angle \text{BOC} = \angle \text{COD} = \angle \text{DOA} = 90^\circ$.
Derivation using General Quadrilateral Formula:
We can use the general formula for the area of a quadrilateral: $A = \frac{1}{2} d (h_1 + h_2)$.
Choose the diagonal AC as the diagonal '$d$', so $d = d_1$.
The perpendicular from B to AC is BO, so $h_1 = BO = \frac{d_2}{2}$.
The perpendicular from D to AC is DO, so $h_2 = DO = \frac{d_2}{2}$.
Substitute these into the general formula:
$\text{Area of Rhombus} = \frac{1}{2} d_1 \left(\frac{d_2}{2} + \frac{d_2}{2}\right)$
[Using general quad formula]
$= \frac{1}{2} d_1 \left(\frac{2d_2}{2}\right)$
[Adding heights]
$= \frac{1}{2} d_1 (d_2)$
[Simplifying $\frac{2d_2}{2}$]
$\mathbf{A_{\text{rhombus}} = \frac{1}{2} d_1 d_2}$
... (1)
Derivation by Sum of Areas of Triangles:
The diagonals divide the rhombus into four congruent right-angled triangles (e.g., $\triangle \text{AOB}$, $\triangle \text{BOC}$, $\triangle \text{COD}$, $\triangle \text{DOA}$).
Area($\triangle \text{AOB}$) $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AO \times BO = \frac{1}{2} \times \frac{d_1}{2} \times \frac{d_2}{2} = \frac{d_1 d_2}{8}$
Since all four triangles are congruent, their areas are equal.
Area(Rhombus) = Area($\triangle \text{AOB}$) + Area($\triangle \text{BOC}$) + Area($\triangle \text{COD}$) + Area($\triangle \text{DOA}$)
$= 4 \times \text{Area}(\triangle \text{AOB})$
$= 4 \times \left(\frac{d_1 d_2}{8}\right)$
$\mathbf{A_{\text{rhombus}} = \frac{d_1 d_2}{2} = \frac{1}{2} d_1 d_2}$
Formula:
The formula for the area of a rhombus is:
$\textbf{Area of Rhombus} = \mathbf{\frac{1}{2} \times (Product \$\$ of \$\$ lengths \$\$ of \$\$ its \$\$ Diagonals)}$
Or simply, $\mathbf{A = \frac{1}{2} d_1 d_2}$, where $d_1$ and $d_2$ are the lengths of the two diagonals.
Area of a Kite
A kite is a quadrilateral that has two distinct pairs of equal-length adjacent sides. The properties of a kite relevant to its area calculation using diagonals are:
- Two pairs of adjacent sides are equal.
- One diagonal is the perpendicular bisector of the other diagonal.
- The diagonals are perpendicular to each other (intersect at a $90^\circ$ angle).
Let the lengths of the two diagonals be $d_1$ and $d_2$. Let the diagonals be AC ($d_1$) and BD ($d_2$), intersecting perpendicularly at point O.
In a kite, the diagonal joining the vertices where equal sides meet (AC in the diagram) is the axis of symmetry, and it perpendicularly bisects the other diagonal (BD). So, $BO = OD = \frac{d_2}{2}$. The diagonals are perpendicular, $\angle \text{AOB} = 90^\circ$.
Derivation using General Quadrilateral Formula:
We can use the general formula for the area of a quadrilateral: $A = \frac{1}{2} d (h_1 + h_2)$.
Choose the diagonal AC as the diagonal '$d$', so $d = d_1$.
The perpendicular from B to AC is BO, so $h_1 = BO$.
The perpendicular from D to AC is DO, so $h_2 = DO$.
Substitute these into the general formula:
$\text{Area of Kite} = \frac{1}{2} d_1 (BO + DO)$
[Using general quad formula]
Since BO + DO = BD = $d_2$ (the length of the other diagonal):
$\mathbf{A_{\text{kite}} = \frac{1}{2} d_1 d_2}$
... (2)
Derivation by Sum of Areas of Triangles:
The diagonal AC divides the kite into two triangles: $\triangle \text{ABC}$ and $\triangle \text{ADC}$. Both triangles share the base AC ($d_1$). The height of $\triangle \text{ABC}$ from B to AC is BO, and the height of $\triangle \text{ADC}$ from D to AC is DO.
Area($\triangle \text{ABC}$) $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AC \times BO = \frac{1}{2} d_1 \times BO$
Area($\triangle \text{ADC}$) $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AC \times DO = \frac{1}{2} d_1 \times DO$
Area(Kite) = Area($\triangle \text{ABC}$) + Area($\triangle \text{ADC}$)
$= \frac{1}{2} d_1 \times BO + \frac{1}{2} d_1 \times DO$
Factor out the common term $\frac{1}{2} d_1$:
$= \frac{1}{2} d_1 (BO + DO)$
Since BO + DO = BD = $d_2$:
$\mathbf{A_{\text{kite}} = \frac{1}{2} d_1 d_2}$
The derivation is essentially the same as the one for a rhombus because both rely on the property that the diagonals are perpendicular. The formula for the area of a rhombus and a kite is the same.
Formula:
The formula for the area of a kite is:
$\textbf{Area of Kite} = \mathbf{\frac{1}{2} \times (Product \$\$ of \$\$ lengths \$\$ of \$\$ its \$\$ Diagonals)}$
Or simply, $\mathbf{A = \frac{1}{2} d_1 d_2}$, where $d_1$ and $d_2$ are the lengths of the two diagonals.
Examples
Example 1. Find the area of a rhombus whose diagonals measure $16$ cm and $12$ cm.
Answer:
Given:
Shape is a rhombus.
Length of one diagonal, $d_1 = 16$ cm.
Length of the other diagonal, $d_2 = 12$ cm.
To Find:
Area of the rhombus.
Solution:
The formula for the area of a rhombus is $A = \frac{1}{2} d_1 d_2$. Using formula (1) derived above:
$\text{Area} = \frac{1}{2} d_1 d_2$
Substitute the given diagonal lengths:
$\text{Area} = \frac{1}{2} \times 16 \$ \text{cm} \times 12 \$ \text{cm}$
[Substituting $d_1=16$ cm and $d_2=12$ cm]
Perform the multiplication:
"$A = \frac{1}{2} \times (16 \times 12) \$ \text{cm}^2$"
"$A = \frac{1}{2} \times 192 \$ \text{cm}^2$"
"$\mathbf{A = 96 \$\$ cm^2}$"
Alternatively, simplify $\frac{1}{2}$ with one of the diagonal lengths first:
"$A = (\frac{1}{2} \times 16 \$ \text{cm}) \times 12 \$ \text{cm} = 8 \$ \text{cm} \times 12 \$ \text{cm} = 96 \$ \text{cm}^2$"
Or:
"$A = 16 \$ \text{cm} \times (\frac{1}{2} \times 12 \$ \text{cm}) = 16 \$ \text{cm} \times 6 \$ \text{cm} = 96 \$ \text{cm}^2$"
All methods give the same result. The area of the rhombus is 96 square centimetres ($\text{cm}^2$).
Example 2. Calculate the area of a kite whose diagonals are $18$ m and $10$ m long.
Answer:
Given:
Shape is a kite.
Length of one diagonal, $d_1 = 18$ m.
Length of the other diagonal, $d_2 = 10$ m.
To Find:
Area of the kite.
Solution:
The formula for the area of a kite is $A = \frac{1}{2} d_1 d_2$. Using formula (2) derived above:
$\text{Area} = \frac{1}{2} d_1 d_2$
Substitute the given diagonal lengths:
$\text{Area} = \frac{1}{2} \times 18 \$ \text{m} \times 10 \$ \text{m}$
[Substituting $d_1=18$ m and $d_2=10$ m]
Perform the multiplication:
"$A = \frac{1}{2} \times 180 \$ \text{m}^2$"
"$\mathbf{A = 90 \$\$ m^2}$"
Alternatively:
"$A = (\frac{1}{2} \times 18 \$ \text{m}) \times 10 \$ \text{m} = 9 \$ \text{m} \times 10 \$ \text{m} = 90 \$ \text{m}^2$"
The area of the kite is 90 square metres ($\text{m}^2$).
Application of Heron’s Formula for Finding Areas of Quadrilaterals (by dividing into triangles)
While there is a general formula for the area of a quadrilateral using a diagonal and the perpendiculars to it (as discussed in the previous section), sometimes this information (the perpendicular heights) is not directly available. However, if the lengths of all four sides of a quadrilateral and the length of at least one of its diagonals are known, we can still find the area of the quadrilateral by dividing it into two triangles and applying Heron's formula to each triangle.
Method
Consider a quadrilateral ABCD. Let the lengths of its sides be AB = $a$, BC = $b$, CD = $c$, and DA = $d$. Suppose the length of one of its diagonals, say AC, is known and its length is $e$. This diagonal divides the quadrilateral into two triangles: $\triangle \text{ABC}$ and $\triangle \text{ADC}$.
To find the area of the quadrilateral, we calculate the area of each of these two triangles separately using Heron's formula and then add them together.
Steps:
-
Divide the Quadrilateral:
Identify the two triangles formed by the given diagonal. For quadrilateral ABCD and diagonal AC, the triangles are $\triangle \text{ABC}$ and $\triangle \text{ADC}$.
-
Apply Heron's Formula to the First Triangle ($\triangle \text{ABC}$):
The side lengths of $\triangle \text{ABC}$ are the two adjacent sides of the quadrilateral connected by the diagonal, plus the diagonal itself. In this case, the sides are AB = $a$, BC = $b$, and AC = $e$.
- Calculate the semi-perimeter of $\triangle \text{ABC}$:
$\mathbf{s_1 = \frac{a+b+e}{2}}$
... (1)
- Calculate the area of $\triangle \text{ABC}$ using Heron's Formula:
$\mathbf{Area(\triangle ABC) = \sqrt{s_1(s_1-a)(s_1-b)(s_1-e)}}$
... (2)
- Calculate the semi-perimeter of $\triangle \text{ABC}$:
-
Apply Heron's Formula to the Second Triangle ($\triangle \text{ADC}$):
The side lengths of $\triangle \text{ADC}$ are the other two sides of the quadrilateral connected by the diagonal, plus the diagonal itself. In this case, the sides are CD = $c$, DA = $d$, and AC = $e$.
- Calculate the semi-perimeter of $\triangle \text{ADC}$:
$\mathbf{s_2 = \frac{c+d+e}{2}}$
... (3)
- Calculate the area of $\triangle \text{ADC}$ using Heron's Formula:
$\mathbf{Area(\triangle ADC) = \sqrt{s_2(s_2-c)(s_2-d)(s_2-e)}}$
... (4)
- Calculate the semi-perimeter of $\triangle \text{ADC}$:
-
Sum the Areas:
The total area of the quadrilateral ABCD is the sum of the areas of the two triangles.
$\mathbf{Area(ABCD) = Area(\triangle ABC) + Area(\triangle ADC)}$
... (5)
This method is applicable whenever the lengths of all four sides and one diagonal of the quadrilateral are provided.
Example
Example 1. Find the area of a quadrilateral field ABCD in which AB = $9$ m, BC = $40$ m, CD = $28$ m, DA = $15$ m, and the diagonal AC = $41$ m.
Answer:
Given:
Quadrilateral field ABCD with side lengths:
AB ($a$) $= 9$ m
BC ($b$) $= 40$ m
CD ($c$) $= 28$ m
DA ($d$) $= 15$ m
Length of the diagonal AC ($e$) $= 41$ m.
To Find:
Area of quadrilateral ABCD.
Solution:
The diagonal AC divides the quadrilateral ABCD into two triangles: $\triangle \text{ABC}$ and $\triangle \text{ADC}$. We will find the area of each triangle using Heron's formula and then add them.
Step 1: Find Area of $\triangle \text{ABC}$ using Heron's Formula
The side lengths of $\triangle \text{ABC}$ are $a=9$ m, $b=40$ m, and $e=41$ m.
Calculate the semi-perimeter $s_1$:
"$s_1 = \frac{a+b+e}{2} = \frac{9+40+41}{2} \$ \text{m}$"
"$s_1 = \frac{90}{2} \$ \text{m} = 45 \$ \text{m}$"
Calculate the differences:
"$(s_1-a) = 45 - 9 = 36 \$ \text{m}$"
"$(s_1-b) = 45 - 40 = 5 \$ \text{m}$"
"$(s_1-e) = 45 - 41 = 4 \$ \text{m}$"
Apply Heron's Formula for Area($\triangle \text{ABC}$):
$\text{Area}(\triangle \text{ABC}) = \sqrt{s_1(s_1-a)(s_1-b)(s_1-e)}$
"$\text{Area}(\triangle \text{ABC}) = \sqrt{(45 \$ \text{m})(36 \$ \text{m})(5 \$ \text{m})(4 \$ \text{m})}$"
"$\text{Area}(\triangle \text{ABC}) = \sqrt{45 \times 36 \times 5 \times 4 \$ \text{m}^4}$"
Simplify the expression under the square root by prime factorization or grouping perfect squares:
"$45 = 9 \times 5 = 3^2 \times 5$"
"$36 = 6^2$"
"$4 = 2^2$"
"$\text{Area}(\triangle \text{ABC}) = \sqrt{(3^2 \times 5) \times 6^2 \times 5 \times 2^2 \$ \text{m}^4}$"
"$\text{Area}(\triangle \text{ABC}) = \sqrt{2^2 \times 3^2 \times 5^2 \times 6^2 \$ \text{m}^4}$"
Take the square root:
"$\text{Area}(\triangle \text{ABC}) = 2 \times 3 \times 5 \times 6 \$ \text{m}^2 = 180 \$ \text{m}^2$"
Alternative check for $\triangle \text{ABC}$: Notice that $9^2 + 40^2 = 81 + 1600 = 1681$. And $41^2 = 1681$. Since $9^2 + 40^2 = 41^2$, by the converse of the Pythagorean theorem, $\triangle \text{ABC}$ is a right-angled triangle with the right angle at B (opposite to the hypotenuse AC). The sides containing the right angle are AB ($9$ m) and BC ($40$ m). The area can also be calculated as:
"$\text{Area}(\triangle \text{ABC}) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 9 \$ \text{m} \times 40 \$ \text{m}$"
"$\text{Area}(\triangle \text{ABC}) = \frac{1}{2} \times 360 \$ \text{m}^2 = 180 \$ \text{m}^2$"
Both methods give the same area for $\triangle \text{ABC}$.
Step 2: Find Area of $\triangle \text{ADC}$ using Heron's Formula
The side lengths of $\triangle \text{ADC}$ are $c=28$ m, $d=15$ m, and $e=41$ m.
Calculate the semi-perimeter $s_2$:
"$s_2 = \frac{c+d+e}{2} = \frac{28+15+41}{2} \$ \text{m}$"
"$s_2 = \frac{84}{2} \$ \text{m} = 42 \$ \text{m}$"
Calculate the differences:
"$(s_2-c) = 42 - 28 = 14 \$ \text{m}$"
"$(s_2-d) = 42 - 15 = 27 \$ \text{m}$"
"$(s_2-e) = 42 - 41 = 1 \$ \text{m}$"
Apply Heron's Formula for Area($\triangle \text{ADC}$):
$\text{Area}(\triangle \text{ADC}) = \sqrt{s_2(s_2-c)(s_2-d)(s_2-e)}$
"$\text{Area}(\triangle \text{ADC}) = \sqrt{(42 \$ \text{m})(14 \$ \text{m})(27 \$ \text{m})(1 \$ \text{m})}$"
"$\text{Area}(\triangle \text{ADC}) = \sqrt{42 \times 14 \times 27 \times 1 \$ \text{m}^4}$"
Simplify the expression under the square root by prime factorization:
"$42 = 2 \times 3 \times 7$"
"$14 = 2 \times 7$"
"$27 = 3 \times 3 \times 3 = 3^3$"
"$\text{Area}(\triangle \text{ADC}) = \sqrt{(2 \times 3 \times 7) \times (2 \times 7) \times (3^3) \times 1 \$ \text{m}^4}$"
Group the factors in pairs:
"$\text{Area}(\triangle \text{ADC}) = \sqrt{(2 \times 2) \times (3 \times 3^3) \times (7 \times 7) \$ \text{m}^4}$"
"$\text{Area}(\triangle \text{ADC}) = \sqrt{2^2 \times 3^4 \times 7^2 \$ \text{m}^4}$"
Take the square root:
"$\text{Area}(\triangle \text{ADC}) = 2^{2/2} \times 3^{4/2} \times 7^{2/2} \$ \text{m}^{4/2}$"
"$\text{Area}(\triangle \text{ADC}) = 2^1 \times 3^2 \times 7^1 \$ \text{m}^2$"
"$\text{Area}(\triangle \text{ADC}) = 2 \times 9 \times 7 \$ \text{m}^2 = 18 \times 7 \$ \text{m}^2$"
"$\text{Area}(\triangle \text{ADC}) = 126 \$ \text{m}^2$"
Step 3: Find Total Area of Quadrilateral ABCD
The total area of the quadrilateral is the sum of the areas of the two triangles:
$\text{Area(ABCD)} = \text{Area}(\triangle \text{ABC}) + \text{Area}(\triangle \text{ADC})$
[Formula (5)]
"$\text{Area(ABCD)} = 180 \$ \text{m}^2 + 126 \$ \text{m}^2$"
[Substituting areas]
Perform the addition:
$\begin{array}{cc} & 1 & 8 & 0 \\ + & 1 & 2 & 6 \\ \hline & 3 & 0 & 6 \\ \hline \end{array}$"$\mathbf{Area(ABCD) = 306 \$\$ m^2}$"
Therefore, the area of the quadrilateral field ABCD is 306 square metres ($\text{m}^2$).
Area of General Polygons (using triangulation or other methods)
To find the area of a polygon with more than four sides (e.g., pentagon - 5 sides, hexagon - 6 sides, octagon - 8 sides, etc.), where a direct formula might not be available or applicable, we can typically break down the complex polygon into simpler shapes whose areas we already know how to calculate. The most common simpler shape used for this purpose is the triangle, in a process called triangulation.
Triangulation Method
This is a general and powerful method applicable to any simple polygon (a polygon that does not intersect itself). The idea is to divide the polygon into a set of non-overlapping triangles.
Steps:
-
Choose a Vertex:
Select any one vertex of the polygon.
-
Draw Diagonals:
Draw all possible diagonals from the chosen vertex to all other non-adjacent vertices. A diagonal connects two non-adjacent vertices of a polygon.
-
Formation of Triangles:
An $n$-sided polygon will be divided into $(n-2)$ non-overlapping triangles by drawing diagonals from one vertex.
For example, in a pentagon (n=5), drawing diagonals from one vertex (say, A) to the non-adjacent vertices C and D divides the pentagon into $5-2=3$ triangles: $\triangle \text{ABC}$, $\triangle \text{ACD}$, and $\triangle \text{ADE}$.
-
Calculate the Area of Each Triangle:
Find the area of each of the $(n-2)$ triangles formed. The method for calculating each triangle's area depends on the information available (side lengths for Heron's formula, or base and height for the $\frac{1}{2}bh$ formula). You might need to use given side lengths and diagonal lengths, or heights perpendicular to the diagonals.
-
Sum the Areas:
The total area of the polygon is the sum of the areas of all the triangles into which it was divided.
$\text{Area(Polygon)} = \text{Area(Triangle 1)} + \text{Area(Triangle 2)} + ... + \text{Area(Triangle n-2)}$
For the pentagon ABCDE:
Area(ABCDE) = Area($\triangle \text{ABC}$) + Area($\triangle \text{ACD}$) + Area($\triangle \text{ADE}$)
This method requires sufficient information to calculate the area of each resulting triangle. This could mean knowing all side lengths and diagonal lengths used, or knowing base and corresponding height for each triangle.
Decomposition into Simpler Shapes
Sometimes, depending on the shape of the polygon and the given information, it may be more convenient to divide the polygon into a combination of shapes like triangles, rectangles, or trapeziums.
This method is often used when dealing with polygons drawn on a grid or when coordinates are involved, or when the polygon has several parallel or perpendicular sides that make it easy to draw dividing lines that form rectangles or trapeziums.
You identify simpler, non-overlapping shapes within the polygon by drawing auxiliary lines. Calculate the area of each simpler shape using its appropriate formula (e.g., $l \times w$ for rectangles, $\frac{1}{2}bh$ for triangles, $\frac{1}{2}(a+b)h$ for trapeziums). The total area of the polygon is the sum of the areas of these simpler shapes.
Coordinate Geometry Method (Shoelace Formula)
If the coordinates of the vertices of the polygon are known in a coordinate system, there is a direct formula called the Shoelace Formula (or Surveyor's Formula) to find its area.
Let the vertices of the polygon be $P_1(x_1, y_1), P_2(x_2, y_2), ..., P_n(x_n, y_n)$, listed in either counterclockwise or clockwise order.
Formula (Shoelace Formula):
$\mathbf{A = \frac{1}{2} |(x_1y_2 + x_2y_3 + ... + x_{n-1}y_n + x_ny_1) - (y_1x_2 + y_2x_3 + ... + y_{n-1}x_n + y_nx_1)|}$
This formula involves summing the products of coordinates in a specific cross-multiplication pattern (resembling tying a shoelace) and taking half of the absolute value of the difference between the two sums.
This method is very efficient when coordinates are given but requires a different set of information compared to side lengths or base/heights.
Example (Using Decomposition Method)
Example 1. Find the area of the pentagonal field ABCDE shown below, where perpendiculars BF, CH, and EG are drawn to the diagonal AD. Given the lengths: AD = $12$ cm, BF = $4$ cm, CH = $6$ cm, EG = $5$ cm, and the segments on AD are AF = $3$ cm, FH = $4$ cm, HG = $2$ cm, GD = $3$ cm.
Answer:
Given:
Pentagonal field ABCDE.
Diagonal AD = $12$ cm.
Lengths of perpendiculars (offsets) from vertices to diagonal AD:
BF = $4$ cm (Perpendicular from B to AD)
CH = $6$ cm (Perpendicular from C to AD)
EG = $5$ cm (Perpendicular from E to AD)
Lengths of segments on the diagonal AD formed by the feet of the perpendiculars:
AF = $3$ cm
FH = $4$ cm
HG = $2$ cm
GD = $3$ cm
Let's verify the sum of the segments on the diagonal AD:
"$AF + FH + HG + GD = 3 + 4 + 2 + 3 = 12 \$ \text{cm}$"
[Sum of segments]
This matches the given length of AD, so the segments cover the entire diagonal.
To Find:
Area of pentagon ABCDE.
Solution:
The diagonal AD and the perpendiculars from the other vertices divide the pentagon into a set of simpler shapes: triangles and trapeziums. Observing the diagram, we can see the pentagon is divided into three triangles ($\triangle \text{ABF}$, $\triangle \text{HCD}$, $\triangle \text{ADE}$) and one trapezium (FBCH).
We need to calculate the area of each of these regions and sum them up to get the total area of the pentagon.
1. Area of $\triangle \text{ABF}$ (above AD)
Base = AF = $3$ cm.
Height = BF = $4$ cm.
Area($\triangle \text{ABF}$) $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AF \times BF$
"$= \frac{1}{2} \times 3 \$ \text{cm} \times 4 \$ \text{cm} = \frac{1}{2} \times 12 \$ \text{cm}^2 = 6 \$ \text{cm}^2$"
2. Area of Trapezium FBCH (above AD)
This is a trapezium with parallel sides BF and CH (both perpendicular to AD). The height is the distance between the parallel sides along AD, which is FH.
Parallel side $a$ = BF = $4$ cm.
Parallel side $b$ = CH = $6$ cm.
Height $h$ = FH = $4$ cm.
Area(Trapezium FBCH) $= \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height} = \frac{1}{2} \times (BF + CH) \times FH$
"$= \frac{1}{2} \times (4 \$ \text{cm} + 6 \$ \text{cm}) \times 4 \$ \text{cm}$"
"$= \frac{1}{2} \times (10 \$ \text{cm}) \times 4 \$ \text{cm} = \frac{1}{2} \times 40 \$ \text{cm}^2 = 20 \$ \text{cm}^2$"
3. Area of $\triangle \text{HCD}$ (above AD)
Base = HD. HD is the segment on AD from H to D. HD = HG + GD = $2 + 3 = 5$ cm.
Height = CH = $6$ cm (This is the perpendicular from C to AD, which is the base HD).
Area($\triangle \text{HCD}$) $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times HD \times CH$
"$= \frac{1}{2} \times 5 \$ \text{cm} \times 6 \$ \text{cm} = \frac{1}{2} \times 30 \$ \text{cm}^2 = 15 \$ \text{cm}^2$"
4. Area of $\triangle \text{ADE}$ (below AD)
Base = AD = $12$ cm.
Height = EG = $5$ cm (This is the perpendicular from E to AD).
Area($\triangle \text{ADE}$) $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AD \times EG$
"$= \frac{1}{2} \times 12 \$ \text{cm} \times 5 \$ \text{cm} = \frac{1}{2} \times 60 \$ \text{cm}^2 = 30 \$ \text{cm}^2$"
5. Total Area of the Pentagon ABCDE
The total area is the sum of the areas of all the non-overlapping regions:
Area(ABCDE) = Area($\triangle \text{ABF}$) + Area(Trapezium FBCH) + Area($\triangle \text{HCD}$) + Area($\triangle \text{ADE}$)
"$= 6 \$ \text{cm}^2 + 20 \$ \text{cm}^2 + 15 \$ \text{cm}^2 + 30 \$ \text{cm}^2$"
"$= (6 + 20 + 15 + 30) \$ \text{cm}^2$"
"$= (26 + 15 + 30) \$ \text{cm}^2$"
"$= (41 + 30) \$ \text{cm}^2$"
"$\mathbf{= 71 \$\$ cm^2}$"
Therefore, the area of the pentagonal field ABCDE is 71 square centimetres ($\text{cm}^2$).