Question

DERIVE THE HERON'S FORMULA
$\sqrt{s(s - a)(s - b)(s - c)}$
CORRECT SOLUTION WILL BE MARKED AS BRAINLIEST ANSWER​

Answer 1

$\large\text{\underline{Required}}$

Heron's formula.

$\large\text{\underline{Solution}}$

Let's consider a triangle $\triangle ABC$.

First, we are going to clarify the equation by using some letters.

• $a\text{, the opposite side of the angle }\angle A$
• $b\text{, the opposite side of the angle }\angle B$
• $c\text{, the opposite side of the angle }\angle C$

$\large\text{\underline{Step 1. Al-Kashi's law of cosines}}$

According to Al-Kashi's law of cosines, two sides and an angle is required to find the remaining side.

$\implies a^{2}=b^{2}+c^{2}-2bc\cos\angle A$

$\implies b^{2}=c^{2}+a^{2}-2ca\cos\angle B$

$\implies c^{2}=a^{2}+b^{2}-2ab\cos\angle C$

Let's choose one of the equations.

$\red{\bigstar}a^{2}=b^{2}+c^{2}-2bc\cos\angle A\text{ (Al-Kashi's law of cosines)}$

$\large\text{\underline{Step 2. Area of a triangle}}$

The basis of the following starts from this equation.

$\implies\triangle ABC=\dfrac{1}{2}bh$

As we can find the height by using the value of $\sin$ values, the following formula is derived.

$\implies\triangle ABC=\dfrac{1}{2}ab\sin\angle C$

$\implies\triangle ABC=\dfrac{1}{2}bc\sin\angle A$

$\implies\triangle ABC=\dfrac{1}{2}ca\sin\angle B$

We can find the area of the triangle when we are given two sides and the contained angle.

$\red{\bigstar}\triangle ABC=\dfrac{1}{2}bc\sin\angle A\text{ (Area of a triangle)}$

$\large\text{\underline{Step 3. Trigonometric identity}}$

This is the most complicated step in the answer, so please read carefully.

We here need to find the value of $\sin\angle A$, since we need to find the area.

The chosen equation gives the following.

$\implies \cos\angle A=\dfrac{b^{2}+c^{2}-a^{2}}{2bc}$

$\implies \cos^{2}\angle A=\left(\dfrac{b^{2}+c^{2}-a^{2}}{2bc}\right)^{2}$

This becomes,

$\implies1-\sin^{2}\angle A=\left(\dfrac{b^{2}+c^{2}-a^{2}}{2bc}\right)^{2}$

Manipulating the equations we get,

$\implies\sin^{2}\angle A=1-\left(\dfrac{b^{2}+c^{2}-a^{2}}{2bc}\right)^{2}$

Further becomes,

$\implies\sin^{2}\angle A=(1+\dfrac{b^{2}+c^{2}-a^{2}}{2bc})(1-\dfrac{b^{2}+c^{2}-a^{2}}{2bc})$

$\implies\sin^{2}\angle A=(\dfrac{b^{2}+2bc+c^{2}-a^{2}}{2bc})(\dfrac{a^{2}-b^{2}+2bc-c^{2}}{2bc})$

$\implies \sin^{2}\angle A=\dfrac{(b+c)^{2}-a^{2}}{2bc}\times\dfrac{a^{2}-(b-c)^{2}}{2bc}$

$\implies \sin^{2}\angle A=\dfrac{(b+c+a)(b+c-a)}{2bc}\times\dfrac{(a-b+c)(a+b-c)}{2bc}$

$\implies \sin\angle A=\dfrac{\sqrt{(b+c+a)(b+c-a)(a-b+c)(a+b-c)}}{2bc}$

Now, Step 3 is finished.

$\red{\bigstar}\sin\angle A=\dfrac{\sqrt{(b+c+a)(b+c-a)(a-b+c)(a+b-c)}}{2bc}$

$\large\text{\underline{Step 4. Integrating Step 2, Step 3}}$

According to the area of a triangle and $\sin\angle A$, we can find the area of the triangle.

$\implies\triangle ABC=\dfrac{1}{2}\times bc\sin\angle A$

$\implies \triangle ABC=\dfrac{1}{2}\times bc\times\dfrac{\sqrt{(b+c+a)(b+c-a)(a-b+c)(a+b-c)}}{2bc}$

$\implies \triangle ABC=\dfrac{1}{4}\times\sqrt{(b+c+a)(b+c-a)(a-b+c)(a+b-c)}$

Distributing fractions into the square root,

$\implies \triangle ABC=\sqrt{\dfrac{(b+c+a)(b+c-a)(a-b+c)(a+b-c)}{16}}$

$\implies \triangle ABC=\sqrt{\dfrac{b+c+a}{2}\times\dfrac{b+c-a}{2}\times\dfrac{a-b+c}{2}\times\dfrac{a+b-c}{2}}$

For the semi-perimeter, we have $s=\dfrac{a+b+c}{2}$.

$\implies \triangle ABC=\sqrt{s(s-a)(s-b)(s-c)}$

Now, here is our result.

$\red{\bigstar}\triangle ABC=\sqrt{s(s-a)(s-b)(s-c)}$

$\large\text{\underline{Conclusion}}$

The area of a triangle which sides are $a,b,c$ and the semi-perimeter $s=\dfrac{a+b+c}{2}$ is $\sqrt{s(s-a)(s-b)(s-c)}$. (Heron's formula)

Answer 2

Answer:

given :

• DERIVE THE HERON'S FORMULA
• $\sqrt{s(s - a)(s - b)(s - c)}$

to find :

• [tex] \sqrt{s(s - a)(s - b)(s - c)} [/tex

solution :

• 2(s-a)=- a+b+c

• 2(s-b) a-b+c => 2[s-c) = a +b-c

• Let p+q=cas indicated.

• Then, ha-p² (1)

• Also, h²=b²-q²

• From (i) and (i)

• a-p²b²q²

• =q² =-a²+p²b²

• Since, q=c-pq²(cp)²q²c²+p²-2pc

• Then, +p²-2pc-a²+p²+b²

• 2p = a+ b- c = ( a - b + c)

• h²= (a-p) (a+p)

• h²= (a-(a²-b+c)/2c) (a+(ab+c)/2c)

• h² = ((2ac-a²+ b² /2c)x((2ac+ a²b²+ c/2c)

• h²((b-(ac)²(a + c)²-b²/4c²

• h =((b-a+c) (b+a- c)(a+c+ b)(a +c-b)

• h²=(2[s-a) x 2[s-c) x 25 x2(s-b))/4c²

• h²=14s (s-a) x (s-c) x[s-b))/c²