It is called "Heron's Formula" after Hero of Alexandria (see below) Just use this two step process: Then, simplify the algebraic expression to find the value of the $x$.$\implies$ $\require{cancel} a^2-b^2+c^2$ $\,=\,$ $\cancel{x^2}-\cancel{x^2}+2cx$$\,\,\, \therefore \,\,\,\,\,\,$ $x \,=\, \dfrac{a^2-b^2+c^2}{2c}$In PQR, the base of the triangle is PR and it is equal to c but the height of the triangle is unknown. It is called as Heron’s formula or Hero’s formula.Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers. The perimeter of the triangle is $a+b+c$ but it is denoted by $2s$ in heron’s formula.Draw a perpendicular line to side $\overline{PR}$ from point $Q$ and it intersects the side $\overline{PR}$ at point $S$. ; Other proofs also exist, but they are more complex or they use the laws which are not so popular (such as e.g. a trigonometric proof using the law of cotangents).
History at your fingertips This article was most recently revised and updated by Take, the length of the sides $\overline{QS}$ and $\overline{PS}$ are $h$ and $x$ respectively, then the length of the side $\overline{SR}$ is equal to $c-x$.The $\Delta SPQ$ and $\Delta SRQ$ are two right triangles, which have $\overline{QS}$ as a common opposite side. Heron's formula proof. Let and . question_answer Answers(1) edit Answer . You can use: Algebra and the Pythagorean theorem;; Trigonometry and the law of cosines. Learn the geometrical proof of heron's formula with step by step procedure to derive the hero's formula in mathematical formula in geometry. Heron’s formula, formula credited to Heron of Alexandria (c. 62 ce) for finding the area of a triangle in terms of the lengths of its sides. So, let us evaluate the height of the triangle from the following equation, which was derived in previous step from the $\Delta SPQ$.Now, substitute the value of $x$ in this equation to find the height of the triangle.$\implies$ $h^2$ $\,=\,$ $a^2-\Bigg(\dfrac{a^2-b^2+c^2}{2c}\Bigg)^2$$\implies$ $h^2$ $\,=\,$ $\Bigg(a+\dfrac{a^2-b^2+c^2}{2c}\Bigg)$ $\Bigg(a-\dfrac{a^2-b^2+c^2}{2c}\Bigg)$$\implies$ $h^2$ $\,=\,$ $\Bigg(\dfrac{a\times 2c + (a^2-b^2+c^2)}{2c}\Bigg)$ $\Bigg(\dfrac{a\times 2c -(a^2-b^2+c^2)}{2c}\Bigg)$$\implies$ $h^2$ $\,=\,$ $\Bigg(\dfrac{2ac+a^2-b^2+c^2}{2c}\Bigg)$ $\Bigg(\dfrac{2ac-a^2+b^2-c^2}{2c}\Bigg)$$\implies$ $h^2$ $\,=\,$ $\Bigg(\dfrac{a^2+c^2+2ac-b^2}{2c}\Bigg)$ $\Bigg(\dfrac{-a^2-c^2+2ac+b^2}{2c}\Bigg)$$\implies$ $h^2$ $\,=\,$ $\Bigg(\dfrac{a^2+c^2+2ac-b^2}{2c}\Bigg)$ $\Bigg(\dfrac{-(a^2+c^2-2ac)+b^2}{2c}\Bigg)$$\implies$ $h^2$ $\,=\,$ $\Bigg(\dfrac{(a+c)^2-b^2}{2c}\Bigg)$ $\Bigg(\dfrac{-(a-c)^2+b^2}{2c}\Bigg)$$\implies$ $h^2$ $\,=\,$ $\Bigg(\dfrac{(a+c)^2-b^2}{2c}\Bigg)$ $\Bigg(\dfrac{b^2-(a-c)^2}{2c}\Bigg)$$\implies$ $h^2$ $\,=\,$ $\Bigg(\dfrac{(a+c+b)(a+c-b)}{2c}\Bigg)$ $\Bigg(\dfrac{(b+(a-c))(b-(a-c))}{2c}\Bigg)$$\implies$ $h^2$ $\,=\,$ $\Bigg(\dfrac{(a+b+c)(a+c-b)}{2c}\Bigg)$ $\Bigg(\dfrac{(b+a-c)(b-a+c)}{2c}\Bigg)$$\implies$ $h^2$ $\,=\,$ $\Bigg(\dfrac{(a+b+c)(a+c-b)}{2c}\Bigg)$ $\Bigg(\dfrac{(a+b-c)(-a+b+c)}{2c}\Bigg)$Now, express the each factor in terms of perimeter.Now, simplify this algebraic expression by substituting the equivalent value of every factor.$\implies$ $h^2$ $\,=\,$ $\Bigg(\dfrac{(2s)(2(s-b))}{2c}\Bigg)$ $\Bigg(\dfrac{(2(s-c))(2(s-a))}{2c}\Bigg)$$\implies$ $h^2$ $\,=\,$ $\Bigg(\dfrac{2s \times 2(s-b)}{2c}\Bigg)$ $\Bigg(\dfrac{2(s-c) \times 2(s-a)}{2c}\Bigg)$$\implies$ $h^2$ $\,=\,$ $\Bigg(\dfrac{2 \times s \times 2(s-b)}{2 \times c}\Bigg)$ $\Bigg(\dfrac{2 \times (s-c) \times 2(s-a)}{2 \times c}\Bigg)$$\implies$ $h^2$ $\,=\,$ $\require{cancel} \Bigg(\dfrac{\cancel{2} \times s \times 2(s-b)}{\cancel{2} \times c}\Bigg)$ $\Bigg(\dfrac{\cancel{2} \times (s-c) \times 2(s-a)}{\cancel{2} \times c}\Bigg)$$\implies$ $h^2$ $\,=\,$ $\Bigg(\dfrac{2s(s-b)}{c}\Bigg)\Bigg(\dfrac{2(s-a)(s-c)}{c}\Bigg)$$\implies$ $h^2$ $\,=\,$ $\dfrac{2s(s-b) \times 2(s-a)(s-c)}{c \times c}$$\implies$ $h^2$ $\,=\,$ $\dfrac{4s(s-a)(s-b)(s-c)}{c^2}$$\implies$ $h$ $\,=\,$ $\pm \sqrt{\dfrac{4s(s-a)(s-b)(s-c)}{c^2}}$According to Physics, the height is a positive factor.
Heron's Formula is used to calculate the area of a triangle with the three sides of the triangle.
Login Check out Britannica's new site for parents! Proof of this formula can be found in Hero of Alexandria’s book “Metrica”.
The lengths of sides of triangle $\overline{PQ}$, $\overline{QR}$ and $\overline{PR}$ are $a$, $b$ and $c$ respectively. You have to first find the semi-perimeter of the triangle with three sides and then area can be calculated based on the semi-perimeter of the triangle. For calculating the area of a triangle, it is required to know both base and height of triangle. Some also believe that this formula has Vedic roots and the credit should be given to the ancient Hindus. Heron's Formula. Area of a Triangle from Sides. Our editors will review what you’ve submitted and determine whether to revise the article. Many mathematicians believe that Archimedes already knew the formula before Heron. The formula was derived by Hero … Categories please give the simple proof for the heron's formula at the earliest.