#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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

10 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}

The difference of two squares factorization was used in two different steps.