Question

Derive Heron's Formula.
Don't copy answers from other source. Don't be so greedy for points. ​

Answer 1

Given:-

A triangle ABC

Required answer:-

Derive the heron's formula

Derivation:-

We know that

$\sf \footnotesize{ar( \triangle \: ABC) = \frac{1}{2} \times base \times height}$

Now, let x + y = a

Let h be the height of the triangle

then, by applying PGT in two triangles,

$\sf \footnotesize{ {h}^{2} + {x}^{2} = {c}^{2} \: and \: {h}^{2} + {y}^{2} = {b}^{2} }$

Since, x + y = a, then, y = a - x

$\sf \footnotesize{ \implies{ {y}^{2} = {(a - x)}^{2} }}$

$\sf \footnotesize{ \implies{ {y}^{2} = {a}^{2} - 2ax + {y}^{2} }}$

$\sf \footnotesize{ Adding \: {h}^{2} on \: both \: sides, \: we \: get}$

$\sf \footnotesize{ {h}^{2} + {y}^{2} = {h}^{2} + {a}^{2} - 2ax + {y}^{2} }$

$\sf \footnotesize{ \implies{ {b}^{2} = {a}^{2} - 2ax + {c}^{2} }} \: \: \: \: \sf \footnotesize{ (\because \: {h}^{2} + {x}^{2} = {c}^{2} \: and \: {h}^{2} + {y}^{2} = {b}^{2}) }$

$\sf \footnotesize{ \implies{2ax = {a}^{2} + {c}^{2} - {b}^{2} }}$

$\sf \footnotesize{ \implies{x = \frac{ {a}^{2} + {c}^{2} - {b}^{2} }{2a} }}$

We have,

$\sf \footnotesize{ {h}^{2} = {c}^{2} - {x}^{2} }$

$\sf \footnotesize{ \implies{ {h}^{2} = (c + x)(c - x)}}$

$\sf \footnotesize{ \implies{ {h}^{2} = (c + \frac{ {a}^{2} + {c}^{2} - {b}^{2} }{2a})(c - \frac{ {a}^{2} + {c}^{2} - {b}^{2} }{2a} )}}$

$\sf \footnotesize{ \implies{ {h}^{2} = ( \frac{2ac + {a}^{2} + {c}^{2} - {b}^{2} }{2a} )( \frac{2ac - {a}^{2} - {c}^{2} + {b}^{2} }{2a} })}$

$\sf \footnotesize{ \implies{ {h}^{2} = ( \frac{ {(a + c)}^{2} - {b}^{2} }{2a})( \frac{ {b}^{2} - ( { {a}^{2} + {c}^{2} - 2ac)} }{2a} ) }}$

$\sf \footnotesize{ \implies{ {h}^{2} = ( \frac{ {(a + c)}^{2} - {b}^{2} }{2a})( \frac{ {b}^{2} - ( {a - c)}^{2} }{2a})}}$

$\sf \footnotesize{ \implies{ {h}^{2} = \frac{(a + c - b)(a + c + b)(b + a - c)(b - a + c)}{4 {a}^{2} }}}$

$\sf \footnotesize{ \implies{ {h}^{2} = \frac{2(s - b) \times 2s \times 2(s - c) \times 2(s - a)}{ 4{a}^{2} } }} \: \: \: \: \: \: \sf \footnotesize{( \because\: s = \frac{a + b+ c}{2} )}$

$\sf \footnotesize{ \implies{ {h}^{2} = \frac{4s \times (s - a)(s - b)(s - c)}{ {a}^{2} } }}$

$\sf \footnotesize{ \implies{ {h} = \frac{2 \times \sqrt{s \times (s - a)(s - b)(s - c)} }{a} }}$

$\sf \footnotesize{Since, \: area = \frac{1}{2} \times a \times h}$

$\sf \footnotesize{ \implies{ area = \frac{1}{2} \times a \times \frac{2 \times \sqrt{s \times (s - a)(s - b)(s - c)} }{a} }}$

$\sf \footnotesize{ \implies{area = \sqrt{s(s - a)(s - b)(s - c)}}}$

Answer 2

Step-by-step explanation:

Heron’s Formula can be used to find the area of a triangle given the lengths of the three sides. A triangle with side lengths

a,

b, and

c, its area

A can be calculated using the Heron’s formula

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

where

s = \displaystyle \frac{a+b+c}{2}

is the semiperimeter (half the perimeter) of the triangle.

In this post, I will provide a detailed derivation of this formula.

Some Preliminaries

The area of a triangle is half the product of its base and its altitude. In the figure below,

\overline{AM} is the altitude of triangle

ABC. If the length of the altitude is not given, and an angle measure is given, we can use Trigonometry to calculate the altitude.

Heron's Formula

In the figure above, the altitude

\overline{AM} forms right triangle

AMC. We know that

\sin \theta = (length of opposite side)/(length of hypotenuse)

so,

\sin C = \displaystyle\frac{AM }{AC}.

Simplifying, we have

AM = AC \sin C which is equivalent to

AM = b \sin C

Since the base of triangle is

\overline{BM} = a and its altitude is

b \sin C, its area

A is given by the formula

A = ab \sin C. (1)

The Derivation

Now we apply the preceding formula, the Cosine Law and the Pythagorean identity

\sin \theta + \cos \theta = 1 to derive the Heron’s formula.

Using the Pythagorean identity, and manipulating algebraically

\sin^2 C + \cos^2 C = 1

\sin^2 C = 1 - \cos^2 C

\sin^2 C = (1 + \cos C)(1- \cos C) (2)

By the Cosine Law, in a triangle

ABC with side lengths

a,

b, and

c

c^2 = a^2 + b^2 -2ab \cos C.

Calculating for

\cos C, we have

\cos C = \displaystyle\frac{a^2 + b^2 - c^2}{2ab}.

Substituting the preceding equation to (2), we have

\sin^2 C = \left (1 + \displaystyle\frac{a^2 + b^2 - c^2}{2ab} \right ) \left ( 1 - \displaystyle \frac{a^2 + b^2 - c^2} {2ab} \right ).

= \left (\displaystyle\frac{2ab + a^2 + b^2 - c^2}{2ab} \right ) \left ( \displaystyle \frac{2ab - a^2 - b^2 + c^2} {2ab} \right ).

=\displaystyle \frac{ [(a + b)^2 - c^2][c^2 - (a-b)^2 ]}{4a^2b^2}.

\sin^2 C = \displaystyle \frac {( a + b + c)(a + b - c)( c + a - b) (c - a + b)}{4a^2b^2}.

Getting the square root of both sides, we have

\sin C = \displaystyle \frac{ \sqrt{(a + b +c)(a + b - c)(c + a - b) (c -a + b)}}{2ab} (3).

Using (1) and (3), we calculate the area of the triangle ,

A = ab \sin C

A = \frac{1}{2} ab \displaystyle \frac{ \sqrt{(a + b +c)(a + b - c)(c + a - b) (c -a + b)}}{2ab}

A = \displaystyle \frac{1}{4} \sqrt{(a + b + c)(a + b - c)(c + a - b)(c - a + b)} (4)

Now, if we let

s be the semiperimeter (half the perimeter) of triangle

ABC, then

a + b + c = 2s.

Now,

a + b + c - 2a = b + c - a = 2s -2a = 2(s - a).

Also,

a - b + c = 2(s - b) and

a + b - c = 2(s - c).

Substituting the expressions with s to (4), we have

A = \frac{1}{4} \sqrt{2s [2(s-a)][2(s-b)][2(s-c)]}

= \frac{1}{4} \sqrt{16s(s-a)(s-b)(s-a)}

which is equivalent to

A = \frac{4}{4} \sqrt{s(s-a)(s-b)(s-a)}.

Simplifying, we have

A = \sqrt{s(s-a)(s-b)(s-c)} , heron formula

hope it helps