heron's formula proof

Therefore, you do not have to rely on the formula for area that uses base and height.Diagram 1 below illustrates the general formula where S represents the semi-perimeter of the triangle. It can be applied to any shape of triangle, as long as we know its three side lengths. For calculating the area of a triangle, it is required to know both base and height of triangle. There are many ways to prove the Heron's area formula, but you need to know some geometry basics. I think the following argument also constitutes a proof: consider the square of the area of a triangle of sides a, b, c. ... must be a factor, etc. thumb_up Like (6) visibility Views (33.8K) edit Answer . Heron's formula is a formula that can be used to find the area of a triangle, when given its three side lengths. Heron’s formula has been known to mathematicians for nearly 2000 years. It has been suggested that Archimedes knew the formula, and since Metrica is a collection of the mathematical knowledge available in the ancient world, it is possible that it predates the reference given in the work.. A formula equivalent to Heron's namely: Now, write the relation between sides of each triangle by the Now, replace the $h^2$ by its equivalent value as per right triangle $SPQ$. Some also believe that this formula has Vedic roots and the credit should be given to the ancient Hindus. You can calculate the area of a triangle if you know the lengths of all three sides, using a formula that has been known for nearly 2000 years. $\Delta PQR$ is a triangle.

how_to_reg Follow . 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. Features  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.

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