Algebra> prove...

plz state and proof herons formula in step by step explanation which i can understandi m in class 10

keshav kr , 13 Years ago

Grade 12

1 Answers

Chetan Mandayam Nayakar

Last Activity: 13 Years ago

In geometry, Heron's (or Hero's) formula, named after Heron of Alexandria, states that the area A of a triangle whose sides have lengths a, b, and c is

$A = \sqrt{s(s-a)(s-b)(s-c)}$

where s is the semiperimeter of the triangle:

$s=\frac{a+b+c}{2}.$

Proof

A modern proof, which uses algebra and is quite unlike the one provided by Heron (in his book Metrica), follows. Let a, b, c be the sides of the triangle and A, B, C the angles opposite those sides. We have

$\cos \widehat C = \frac{a^2+b^2-c^2}{2ab}$

by the law of cosines. From this proof get the algebraic statement:

$\sin \widehat C = \sqrt{1-\cos^2 \widehat C} = \frac{\sqrt{4a^2 b^2 -(a^2 +b^2 -c^2)^2 }}{2ab}.$

The altitude of the triangle on base a has length b·sin(C), and it follows

$\begin{align} A & = \frac{1}{2} (\mbox{base}) (\mbox{altitude}) \\ & = \frac{1}{2} ab\sin \widehat C \\ & = \frac{1}{4}\sqrt{4a^2 b^2 -(a^2 +b^2 -c^2)^2} \\ & = \frac{1}{4}\sqrt{(2a b -(a^2 +b^2 -c^2))(2a b +(a^2 +b^2 -c^2))} \\ & = \frac{1}{4}\sqrt{(c^2 -(a -b)^2)((a +b)^2 -c^2)} \\ & = \frac{1}{4}\sqrt{(c -(a -b))(c +(a -b))((a +b) -c)((a +b) +c)} \\ & = \sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}. \end{align}$