Area of Quadrilateral Using Heron’s Formula: Overview, Properties

Area of Quadrilateral Using Heron’s Formula: Heron of Alexandria gave Heron’s formula to calculate the area of a triangle whose sides are known. 

Heron’s formula calculates the area of different types of triangles like an equilateral triangle, isosceles triangle, scalene triangle etc. This formula is also used to find the area of a quadrilateral by dividing it into two triangles using any diagonal of the quadrilateral.

Introduction to Heron’s Formula

Heron of Alexandria was a Greek Engineer and Mathematician. He was the first person who used the lengths of the sides of a triangle to find its area in \(60 \,AD\). The formula is known as Heron’s formula.

He extended the basic formula to calculate the areas of quadrilaterals. He proved several trigonometric laws such as the Laws of cosines or Laws of cotangents using the same formula.

Thus, Heron’s formula calculates the area of different types of triangles like an equilateral triangle, isosceles triangle, scalene triangle etc. Heron’s formula is also used to find the quadrilateral area by dividing it into two triangles using any diagonal of the quadrilateral.

Heron’s Formula

Heron’s formula is used to calculate the area of a triangle whose sides are known. Let there be a triangle with the lengths of the sides \(a, b\) and \(c\) as shown below:

He extended the basic formula to calculate the areas of quadrilaterals. He proved several trigonometric laws such as the Laws of cosines or Laws of cotangents using the same formula.

Thus, Heron’s formula calculates the area of different types of triangles like an equilateral triangle, isosceles triangle, scalene triangle etc. Heron’s formula is also used to find the quadrilateral area by dividing it into two triangles using any diagonal of the quadrilateral.

Area of Triangle Using Heron’s Formula

Heron’s formula is used to calculate the area of a triangle whose sides are known. Let there be a triangle with the lengths of the sides \(a, b\) and \(c\) as shown below:

The area of the triangle is given by: \(\sqrt{s(s-a)(s-b)(s-c)}\).

The letter “\(s\)” used in Heron’s formula is known as the semi perimeter of the triangle and equals half of the perimeter (length of all sides) of the triangle.

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

Example: Find the area of a triangle whose lengths of the sides is \(5\) units, \(6\) units and \(9\) units, respectively.

Let \(a=5, b=6, c=9\), and \(s=\frac{5+6+9}{2}=10\)

Then, the area of the triangle is

\(\sqrt{10(10-6)(10-5)(10-9)}=\sqrt{10 \times 4 \times 5 \times 1}=\sqrt{200}=10 \sqrt{2} \,\text { sq.units }\)

Area of Equilateral Triangle Using Heron’s Formula

Let us consider an equilateral triangle with all sides equal to “\(a\)” units.

Then, the semi perimeter of the equilateral triangle is equal to half of the sum of the lengths of the sides of the equilateral triangle.

\(s=\frac{a+a+a}{2}=\frac{3 a}{2}\)

The area of the equilateral triangle can be found by using Heron’s formula as given below:

\(\sqrt{\frac{3 a}{2}\left(\frac{3 a}{2}-a\right)\left(\frac{3 a}{2}-a\right)\left(\frac{3 a}{2}-a\right)}=\sqrt{\frac{3 a}{2} \times \frac{a}{2} \times \frac{a}{2} \times \frac{a}{2}}=\frac{\sqrt{3} a^{2}}{4}\)

Define a Quadrilateral

A quadrilateral is a polygon with four sides. Thus, a quadrilateral has four sides.

Some special types of quadrilaterals are square, rhombus, rectangle and parallelogram etc. All these special types of quadrilaterals have some properties regarding sides and angles. We have certain formulas to calculate the area of these special quadrilaterals.

Then, how will you find the area of a general quadrilateral? Let us study the method of finding the area of the quadrilateral in detail.

Area of Quadrilateral Using Heron’s Formula

We don’t have any particular formula to calculate the area of general quadrilaterals. One way of calculating the quadrilateral area is by splitting the quadrilateral into two triangles along its diagonal. Then, by using Heron’s formula, we will calculate the areas of two triangles so formed, and their sum gives the area of the quadrilateral.

Let us consider the quadrilateral \(ABCD\) as shown below:

The area of the given quadrilateral can be determined using Heron’s formula, such as dividing the given quadrilateral \(ABCD\) into two triangles (\(ABD\) and \(BDC\)) along the diagonal \(BD\).

Later, the quadrilateral area can be found by adding the areas of two triangles (\(ABD\) and \(BDC\)) so formed.

The area of the two triangles can be found using Heron’s formula. The area of the quadrilateral \(ABCD\) is given by

\(ar(A B C D)=ar(\Delta A B D)+ar(\Delta B C D)\)

The steps to be followed for calculating the area of a general quadrilateral is given below:

Step 1: Consider a quadrilateral \(ABCD\) as given below

Step 2: Join the points \(B\) and \(D\) of the given quadrilateral \(ABCD\).

Step 3: Now, quadrilateral \(ABCD\) is divided into two triangles \(ABD\) and \(BDC\), as given below

Step 4: Area of each triangle \(A B D\) and \(B C D\) as found by using Heron’s formula.

The area of the triangle \(A B D\) with semiperimeter \(s=\frac{A B+A D+B D}{2}\), is found by using

\(\sqrt{s(s-A B)(s-A D)(s-B D)}\)

The area of the triangle \(B D C\) with semiperimeter \(s=\frac{B C+B D+C D}{2}\), is found by using

\(\sqrt{s(s-B D)(s-B C)(s-C D)}\)

Step 5: Now, the area of the quadrilateral \(A B C D\) is obtained by adding the areas of the two triangles \(A B D\) and \(B D C\)

\(ar(A B C D)=\operatorname{ar}(\Delta A B D)+ar(\Delta B C D)\)

Solved Examples on Area of Quadrilateral Using Heron’s Formula

Q.1. The puppy has landed in the shape of a rhombus. She told her one son and one daughter to work on the land to produce the different crops. She divided the land into two equal parts. The perimeter of the land is \(400 \,\text {m}\), and the length of one of the diagonals is \(160 \,\text {m}\). How much area will each of them get to produce the crop? Also, find the total area of the part of the land.
Ans:
Given the part of the land of the Puppy is in the shape of a rhombus.
We know that the perimeter of the rhombus is the sum of all its sides. Given perimeter of the rhombus is \(400 \,\text {m}\).
\(4 \times \text {side} =400 \mathrm{~m}\)
So, the length of the side is given by \(\frac{400}{4}=100 \mathrm{~m}\).
So, the rhombus \(A B C D\), such that \(A B=B C=C D=A D=100 \mathrm{~m}\) and the diagonal \(B D=160 \mathrm{~m}\).

Given that Puppy divides the part of the land (rhombus) into two equal parts (triangles)
So, the area of the triangles \(A B D\) and \(B D C\), so formed are equal, and each of them is found by using Heron’s formula as shown below:
In triangle \(A D B\), the semi perimeter of the triangle is \(s=\frac{100+100+160}{2}=\frac{360}{2}=180 \mathrm{~m}\)
The area of the triangle \(A D B\) is found by using Heron’s formula as
\(\Rightarrow ar(\Delta A B D)=\sqrt{180(180-100)(180-100)(180-160)}\)
\(\Rightarrow ar(\triangle A B D)=\sqrt{180 \times 80 \times 80 \times 20}\)
\(\Rightarrow ar(\Delta A B D)=4800 \mathrm{~m}^{2}\)
Similarly, the area of the triangle \(B D C\) is \(4800 \mathrm{~m}^{2}\).
Each of the people will get the area of \(4800 \mathrm{~m}^{2}\) for the crops.
The area of the part of the land (rhombus) equals the sum of the areas of two parts (area of each triangle).
\(ar(A B C D)=ar(A D B)+ar(B D C)=4800+4800=9600 \mathrm{~m}^{2}\)
Hence, the total area of the part of the land of the Puppy is \(9600 \mathrm{~m}^{2}\).

Q.2. Students of a school staged a rally for a cleanliness campaign. They walked through the lanes in two groups. Then they cleaned the area enclosed within their lanes. One group walked through the lanes \(AB, BD\) and \(DA\), while the other through \(BC, CD\) and \(DB\). If \(AD = 9 \,\text {m}, AB= 40 \,\text {m}, CB= 15 \,\text {m}, DC = 28 \,\text {m}\) and \(\angle A=90^{\circ}\). Find the total area cleaned by the students. Which group cleaned more area and by how much?

Ans:
Given, in a triangle \(A B D, \angle A=90^{\circ}\)
By using Pythagoras theorem, \(H y p^{2}=\operatorname{side}^{2}+\operatorname{sid} e^{2}\).
\(\Rightarrow B D^{2}=A D^{2}+A B^{2}\)
\(\Rightarrow B D^{2}=9^{2}+40^{2}=81+1600=1681\)
\(\Rightarrow B D=\sqrt{1681}=41 \mathrm{~m}\)
Semi perimeter is \(s=\frac{9+40+41}{2}=\frac{90}{2}=45 \mathrm{~m}\)
The area of the triangle \(A D B\) is found by using Heron’s formula is
\(\Rightarrow ar(\Delta A D B)=\sqrt{45(45-40)(45-41)(45-9)}\)
\(\Rightarrow ar(\Delta A D B)=\sqrt{45 \times 5 \times 4 \times 36}=180 \mathrm{~m}^{2}\)
In a triangle, \(B C D\), semi perimeter is \(s=\frac{41+28+15}{2}=\frac{84}{2}=42 \mathrm{~m}\)
The area of the triangle \(B C D\) is found by using Heron’s formula is
\(ar(\Delta B C D)=\sqrt{42(42-28)(42-15)(42-41)}\)
\(\Rightarrow ar(\Delta B C D)=\sqrt{42 \times 1 \times 27 \times 14}=126 \mathrm{~m}^{2}\)
The total area cleaned by the students is the sum of the areas cleaned by two groups (triangles \(B C D, A D B)\)
\(ar(A B C D)=180+126=306 \mathrm{~m}^{2}\)
Hence, the total area cleaned by the students is \(306 \mathrm{~m}^{2}\).
Comparing the areas clean by the two groups, the first group cleaned more area as compared to the second group by \(180 \mathrm{~m}^{2}-126 \mathrm{~m}^{2}=54 \mathrm{~m}^{2}\).

Q.3. Find the area of the quadrilateral \(ABCD\) as shown below, such that triangle \(ABC\) is a right triangle with a right angle at \(B\).

Ans:
Given the angle, \(B\) is the right angle.
\(\angle B=90^{\circ}\)
In a right triangle \(A B C\), by using the Pythagorean theorem,
Hypotenuse \(^{2}=\operatorname{side}^{2}+\operatorname{sid} e^{2}\)
\(\Rightarrow A C^{2}=A B^{2}+B C^{2}\)
\(\Rightarrow A C^{2}=3^{2}+4^{2}=9+16=25\)
\(\Rightarrow A C=\sqrt{25}=5\,{\rm{units}}\)
Semiperimeter of the triangle \(A B C\) is \(s=\frac{5+4+3}{2}=6\)
Area of the triangle \(A B C\) is found by using Heron’s formula as follows:
\(ar(\Delta A B C)=\sqrt{6(6-5)(6-4)(6-3)}\)
\(\Rightarrow ar(\Delta A B C)=\sqrt{6 \times 1 \times 2 \times 3}=6\,{\rm{sq}}{\rm{.units}}\)…….(i)
Semiperimeter of the triangle \(A D C\) is \(s=\frac{5+4+5}{2}=7\)
Area of the triangle \(A D C\) is found by using Heron’s formula as follows:
\(ar(\Delta A D C)=\sqrt{7(7-5)(7-4)(7-5)}\)
\(\Rightarrow ar(\Delta A D C)=\sqrt{7 \times 2 \times 2 \times 3}=2 \sqrt{21}\,{\rm{sq}}{\rm{.units}}\)……..(ii)
The area of the quadrilateral \(A B C D\) is the sum of the areas of the triangles \(A B C\) and \(A D C\).
\(ar(\triangle A B C D)=6+2 \sqrt{21}=2(3+\sqrt{21})\,{\rm{sq}}{\rm{.units}}.\)
Hence, the area of the quadrilateral \(A B C D\) is \(2(3+\sqrt{21})\,{\rm{sq}}{\rm{.units}}.\)

Q.4. Asit made the kite as shown below with the given measurements. Find the area of the kite made by Asit.

Ans:
Given Asit made a kite \(P Q R S\).
The area of the kite \(P Q R S\) is equal to the sum of the two parts, as shown in the figure.
In triangle \(P Q S\), semi perimeter \((s)=\frac{3+3+5}{2}=\frac{11}{2}=5.5\,{\rm{units}}.\)
\(ar(\Delta P Q S)=\sqrt{5.5(5.5-4)(5.5-4)(5.5-5)}=\sqrt{5.5 \times 1.5 \times 1.5 \times 0.5}\)
\(=1.5 \sqrt{0.5 \times 0.5 \times 11}\)
\(=1.5 \times 0.5 \sqrt{11}=0.75 \sqrt{11}{\rm{sq}}{\rm{.units}}.\)
In triangle \(Q R S\), semi perimeter \((s)=\frac{6+6+5}{2}=\frac{17}{2}=8.5\,{\rm{units}}.\)
\(ar(\Delta Q R S)=\sqrt{8.5(8.5-6)(8.5-6)(8.5-5)}=\sqrt{8.5 \times 2.5 \times 2.5 \times 3.5}\)
\(=2.5 \sqrt{17 \times 0.5 \times 7 \times 0.5}=2.5 \times 0.5 \times \sqrt{119}=1.25 \sqrt{119}\,{\rm{sq}}{\rm{.units}}.\)
The area of the kite \(P Q R S=ar(\Delta P Q S)+ar(\Delta Q R S)=(0.75 \sqrt{11}+1.25 \sqrt{119})\,{\rm{sq}}{\rm{.units}}.\)
Hence, the area of the kite made by Asit is \((0.75 \sqrt{11}+1.25 \sqrt{119})\,{\rm{sq}}{\rm{.units}}.\)

Q.5. Find the quadrilateral area formed by joining the two congruent triangles of area equals \(25\) sq. units.
Ans:
Given a quadrilateral is made by joining two congruent triangles.
The area of the quadrilateral equals the sum of the areas of two congruent triangles.
Area of the quadrilateral \(=25+25=50\,{\rm{sq}}{\rm{.units}}.\)
Hence, the area of the quadrilateral is \(50\,{\rm{sq}}{\rm{.units}}.\)

Summary

Heron of Alexandria was a Greek Engineer and Mathematician. He discovered The formula which is known as Heron’s formula. Heron’s formula can be utilised to calculate the area of the triangles, quadrilaterals. The Heron’s formula is articulated as \(s=\frac{a+b+c}{2}\). Here the letter “s” indicates the semi perimeter of the triangle.

Furthermore, a quadrilateral is a polygon with four sides. In this article, we discussed about how to calculate area of quadrilateral using heron’s formula.

Learn About Area of a Triangle

FAQs on Area of Quadrilateral Using Heron’s Formula

Q.1. Does Heron’s formula work for quadrilaterals?
Ans: Yes. Heron’s formula is used to find the area of quadrilaterals by dividing them into two triangles.

Q.2. What is Heron’s formula?
Ans: The formula used for the area of the triangle with sides \(a, b, c\) is \(\sqrt{s(s-a)(s-b)(s-c)}\), where \(s\) is the semi perimeter given by \(s=\frac{a+b+c}{2}\).

Q.3. When is Heron’s formula used?
Ans: Heron’s formula is used to find the area of the triangles, quadrilaterals.

Q.4. Can we use Heron’s formula in an equilateral triangle?
Ans: Yes, we can use Heron’s formula in the equilateral triangle to find its area.

Q.5. What does “s” stands in Heron’s formula?
Ans: The letter “s” stands for the semi perimeter of the triangle given by \(s=\frac{a+b+c}{2}\).

Q.6. Who found Heron’s formula?
Ans: Heron of Alexandria discovered Heron’s Formula. He was a Greek Engineer and Mathematician. 

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