Heron's Formula

In a scalene triangle, if the length of each side is given but its height is not known and it cannot be obtained easily, we take the help of Heron's formula or Hero's formula given by Heron to find the area of such a triangle.

Heron's formula : If $a,b,c$ denote the lengths of the sides of a triangle ABC. Then,

Area of $△\mathrm{ABC}=\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}$ where $s$ is the semi-perimeter of $△ABC$. $\mathrm{s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}$

• This formula is applicable to all types of triangles whether it is right-angled or equilateral or isosceles and it is also useful in finding the area of a triangle when it is not possible to find the area of the triangle easily.

1.0Important formulas:

• Area of equilateral triangle $=\frac{\sqrt{3}}{4}\times {\mathrm{Side}}^2=\frac{\sqrt{3}}{4}{\mathrm{a}}^2$
  
• Area of scalene triangle $=\frac{1}{2}\times$ Base $\times$ Height
  
• Area of isosceles triangle $=\frac{1}{2}\times \sqrt{\left(a^2-\frac{b^2}{4}\right)}\times b$
  

• Q. Find the area of a triangle whose sides are $13\mathrm{cm},14\mathrm{cm}$ and 15 cm respectively. Solution: Let $\mathrm{a},\mathrm{b},\mathrm{c}$ be the sides of the given triangle and s be its semi-perimeter such that $\mathrm{a}=13\mathrm{cm}$, $\mathrm{b}=14\mathrm{cm}$ and $\mathrm{c}=15\mathrm{cm}$ Now, $s=\frac{(\mathrm{a}+\mathrm{b}+\mathrm{c})}{2}=\frac{(13+14+15)}{2}=21$ $\therefore \mathrm{s}-\mathrm{a}=21-13=8,\mathrm{s}-\mathrm{b}=21-14=7$ and $\mathrm{s}-\mathrm{c}=21-15=6$ Hence, Area of given triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ $=\sqrt{21\times 8\times 7\times 6}$ $=\sqrt{7\times 3\times 2\times 2\times 2\times 7\times 2\times 3}$ $=7\times 4\times 3=84{\mathrm{cm}}^2$
• Q. The perimeter of a triangular field is 450 m and its sides are in the ratio $13:12:5$. Find the area of triangle. Explanation: It is given that the sides $a,b,c$ of the triangle are in the ratio $13:12:5$ i.e., $a:b:c=13:12:5$ $\Rightarrow a=13x,b=12x$ and $c=5x$ $\therefore$ Perimeter $=450$ $\Rightarrow 13\mathrm{x}+12\mathrm{x}+5\mathrm{x}=450$ $\Rightarrow 30\mathrm{x}=450$ $\Rightarrow \mathrm{x}=15\mathrm{m}$ So, the sides of the triangle are $\mathrm{a}=13\times 15=195\mathrm{m},\mathrm{b}=12\times 15=180\mathrm{m}$ and $\mathrm{c}=5\times 15=75\mathrm{m}$ It is given that perimeter $=450$ $\Rightarrow 2\mathrm{s}=450$ $\Rightarrow \mathrm{s}=225\mathrm{m}$ Hence, Area $=\sqrt{s(s-a)(s-b)(s-c)}$ $=\sqrt{225(225-195)(225-180)(225-75)}$ $\Rightarrow$ Area $=\sqrt{225\times 30\times 45\times 150}$ $=\sqrt{5^2\times 3^2\times 3\times 5\times 2\times 3^2\times 5\times 5^2\times 2\times 3}$ $\Rightarrow$ Area $=\sqrt{5^6\times 3^6\times 2^2}$ $=5^3\times 3^3\times 2=6750{\mathrm{m}}^2$
• Squares of 25 natural numbers are $1^2=1,2^2=4,3^2=9,4^2=16$ $25^2=625$
• Q. Find the area of a triangle having perimeter 32 cm , one side 11 cm and difference of other two sides is $5cm$. Solution: Let $\mathrm{a},\mathrm{b}$ and c be the three sides of $△\mathrm{ABC}$. $\mathrm{a}=11\mathrm{cm}$ $\mathrm{a}+\mathrm{b}+\mathrm{c}=32\mathrm{cm}$ $\Rightarrow 11+\mathrm{b}+\mathrm{c}=32\mathrm{cm}$ or $b+c=21\mathrm{cm}$ Also, we are given that $\mathrm{b}-\mathrm{c}=5\mathrm{cm}$ Adding (i) and (ii), $2\mathrm{b}=26\mathrm{cm}$ i.e., $b=13\mathrm{cm}$ and $\mathrm{c}=8\mathrm{cm}$ Now, $s=\frac{a+b+c}{2}$ $=\frac{11+13+8}{2}=\frac{32}{2}=16\mathrm{cm}$ $(\mathrm{s}-\mathrm{a})=(16-11)\mathrm{cm}=5\mathrm{cm}$ $(\mathrm{s}-\mathrm{b})=(16-13)\mathrm{cm}=3\mathrm{cm}$ $(\mathrm{s}-\mathrm{c})=(16-8)\mathrm{cm}=8\mathrm{cm}$ $\therefore$ Area of $△ABC=\sqrt{s(s-a)(s-b)(s-c)}$ $$\begin{gathered} =\sqrt{16\times 5\times 3\times 8}{\mathrm{cm}}^2 \\ =\sqrt{64\times 30}{\mathrm{cm}}^2 \\ =8\sqrt{30}{\mathrm{cm}}^2 \end{gathered}$$
• Q. The length of the sides of a triangle are $5\mathrm{cm},12\mathrm{cm}$ and 13 cm respectively. Find the length of perpendicular from the opposite vertex to the side whose length is $13\mathbf{cm}$. Solution: Here, $\mathrm{a}=5\mathrm{cm},\mathrm{b}=12\mathrm{cm}$ and $\mathrm{c}=13\mathrm{cm}$ $$\begin{gathered} \therefore s=\frac{1}{2}(a+b+c) \\ =\frac{1}{2}(5+12+13) \\ =\frac{30}{2}=15\mathrm{cm} \end{gathered}$$ Let $A$ be the area of the given triangle. Then, $A=\sqrt{s(s-a)(s-b)(s-c)}$ $=\sqrt{15(15-5)(15-12)(15-13)}$ $\Rightarrow A=\sqrt{15\times 10\times 3\times 2}=30{\mathrm{cm}}^2$ Let p be the length of the perpendicular from vertex $A$ to the side BC. Then, A $=\frac{1}{2}\times 13\times p$ From (i) and (ii), we get $\Rightarrow \frac{1}{2}\times 13\times \mathrm{p}=30$ $\Rightarrow \mathrm{p}=\frac{60}{13}\mathrm{cm}$
• Q. In the figure, there is a triangular children park with sides, $\mathrm{AB}=7\mathrm{m},\mathrm{BC}=8\mathrm{m}$ and $AC=5m,AD\perp BC$ and $AD$ meets $BC$ at $D$. Trees are planted at $A,B,C$ and $D$. Find the distance between the trees at $A$ and $D$.
  
  Solution: In figure, $\mathrm{a}=8\mathrm{m},\mathrm{b}=5\mathrm{m}$ and $\mathrm{c}=7\mathrm{m}$ $\mathrm{s}=\frac{8+5+7}{2}\mathrm{m}=\frac{20}{2}=10\mathrm{m}$ The area of $△ABC=\sqrt{s(s-a)(s-b)(s-c)}$ $=\sqrt{10\times (10-8)\times (10-5)\times (10-7)}{\mathrm{m}}^2$ $=\sqrt{10\times 2\times 5\times 3}{\mathrm{m}}^2=10\sqrt{3}{\mathrm{m}}^2$ Now, $AD$ is perpendicular to $BC$. $\Rightarrow \frac{1}{2}\times \mathrm{BC}\times \mathrm{AD}=10\sqrt{3}$ $\Rightarrow \frac{1}{2}\times 8\times \mathrm{AD}=10\sqrt{3}$ $\Rightarrow \mathrm{AD}=\frac{10\sqrt{3}}{4}\mathrm{m}=\frac{5\sqrt{3}}{2}\mathrm{m}$ Hence, the distance between the trees at A and D is $\frac{5\sqrt{3}}{2}\mathrm{m}$.
• Sometimes it is not required to use Heron's formula as the length of sides forms "Pythagorean Triplet". Examples of Pythagorean Triplet are: 3 units, 4 units, 5 units; 6 units, 8 units, 10 units; 12 units, 5 units, 13 units etc.
• You might have observed that while solving the questions related to Heron's formula, we get the answers as irrational numbers (having radical sign). In most of the questions the values of these irrational terms are given but if the value is not given than we can follow the method of estimation to find the approx. answer. E.g. Let us try to find the value of $\sqrt{10}$ by estimation. By inspection we can see $3^2<10<4^2$ $3<\sqrt{10}<4$ By hit and trial, we can find that value of $\sqrt{10}$ will be between 3.1 and 3.2 as $(3.1{)}^2=9.61$ and $(3.2{)}^2=10.24$ But 10 is more closer to $(3.2{)}^2=10.24$ so we can take estimated value of $\sqrt{10}$ as 3.2.
• To find estimated roots quickly and accurately, it is advisable to learn the squares of first 25 natural numbers.

Applications of Heron's formula in finding area of a quadrilateral

Heron's formula can be applied to find the area of a quadrilateral by dividing the quadrilateral into two triangular parts. If we join any of the two diagonals of the quadrilateral, then we get two triangles. Area of each triangle is calculated and the sum of two areas is the area of the quadrilateral.

• Q. Prove that the area of the quadrilateral $ABCD$ is $3(4+3\sqrt{3}){\mathrm{m}}^2$, If $AB=5\mathrm{m},\mathrm{BC}=5$ $\mathrm{m},\mathrm{CD}=6\mathrm{m},\mathrm{AD}=6m$, and diagonal $\mathrm{AC}=6m$. Solution:
  
  Diagonal AC divides the quadrilateral $ABCD$ into two triangles $△ACD$ and $△ABC$. For $△ACD$, sides are $6\mathrm{m},6\mathrm{m}$ and 6 m . semi-perimeter, $s=\frac{6m+6m+6m}{2}=9\mathrm{m}$ $\therefore$ Area of $△ACD=\sqrt{s(s-a)(s-b)(s-c)}$ $=\sqrt{9\times (9-6)(9-6)\times (9-6)}{\mathrm{m}}^2$ $=\sqrt{9\times 3\times 3\times 3}=9\sqrt{3}{\mathrm{m}}^2$ For $△\mathrm{ABC}$, sides are $5\mathrm{m},5\mathrm{m}$ and 6 m . semi-perimeter, $s=\frac{5m+5m+6m}{2}=8\mathrm{m}$ Area of $△ABC=\sqrt{s(s-a)(s-b)(s-c)}$ $=\sqrt{8(8-5)(8-5)(8-6)}$ $=\sqrt{8\times 3\times 3\times 2}$ $=\sqrt{16\times 9}{\mathrm{m}}^2=12{\mathrm{m}}^2$ Thus, the area of the quadrilateral $\mathrm{ABCD}=(12+9\sqrt{3}){\mathrm{m}}^2=3(4+3\sqrt{3}){\mathrm{m}}^2$ Hence proved.
• Q. Sanya has a piece of land which is in the shape of a rhombus. She wants her one daughter and one son to work on the land and produce different crops to suffice the needs of their family. She divided the land in two equal parts. If the perimeter of the land is $400\mathbf{m}$ and one of the diagonals is $160\mathbf{m}$, how much area each of them will get for their crops? Explanation: Let ABCD be the field which is divided by the diagonal $\mathrm{BD}=160\mathrm{m}$ into two equal parts. Since ABCD is a rhombus of perimeter 400 m . Therefore, $\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}=\frac{400}{4}\mathrm{m}=100\mathrm{m}$
  
  Now, consider $△BCD$ Let $s$ be the semi-perimeter of $△BCD$. Then, $s=\frac{BC+CD+BD}{2}$ $=\frac{100+100+160}{2}\mathrm{m}$ $=180\mathrm{m}$ So, area of $△BCD=\sqrt{s(s-a)(s-b)(s-c)}$ $=\sqrt{180(180-100)(180-100)(180-160)}$ $=\sqrt{180\times 80\times 80\times 20}=4800{\mathrm{m}}^2$ Since, diagonal of rhombus divides it into two equal areas. Hence each of two children will get an area of $4800{\mathrm{m}}^2$.

2.0Memory map