Question

how to prove heron formula

Answer 1

We know that a triangle with sides 3,4 and 5 is a right triangle. Two such triangles would make a rectangle with sides 3 and 4, so its area is {\displaystyle {\frac {3\cdot 4}{2}}=6} {\displaystyle {\frac {3\cdot 4}{2}}=6} .

A triangle with sides 5,6,7 is going to have its largest angle smaller than a right angle, and its area will be less than {\displaystyle {\frac {5\cdot 6}{2}}=15} {\displaystyle {\frac {5\cdot 6}{2}}=15} . Let's see how much by, by calculating its area using Heron's formula.

{\displaystyle s={\frac {5+6+7}{2}}=9} {\displaystyle s={\frac {5+6+7}{2}}=9}

{\displaystyle A={\sqrt {s(s-a)(s-b)(s-c)}}={\sqrt {9(9-5)(9-6)(9-7)}}={\sqrt {9(4)(3)(2)}}={\sqrt {216}}=6{\sqrt {6}}\approx 14.7} {\displaystyle A={\sqrt {s(s-a)(s-b)(s-c)}}={\sqrt {9(9-5)(9-6)(9-7)}}={\sqrt {9(4)(3)(2)}}={\sqrt {216}}=6{\sqrt {6}}\approx 14.7}

Answer 2

Question-

Proof of heron's formula.

Answer-

There are two methods by which we can derive Heron’s formula.

• First, by using trigonometric identities and cosine rule.
• Secondly, solving algebraic expressions using the Pythagoras theorem.

The cosine rule is for higher grades and it isn't necessary right now, so I won't be explaining it. Just keep in mind that there is also another way to prove the heron's formula.

Let us take a triangle having lengths of sides, a, b, and c. Let the semi-perimeter of the triangle ABC is "s", the perimeter of the triangle ABC is "P" and the area of triangle ABC is "A". Let us assume the side length b is divided into two parts p and q as a perpendicular(h) falls from the vertex B on the side AC at point M.

As we know, the area of a triangle = (1/2) b × h where b is the base and h is the height of the triangle. Let us begin to calculate the value of h.

Thus, b = p + q

⇒ q = b - p ....(1)

On squaring both sides we get,

⇒ q² = b² + p² - 2bp ....(2)

Adding h² on both sides we get,

q² + h² = b² + p² - 2bp + h² ....(3)

Applying Pythagoras Theorem in the triangle ABM we get,

h² + q² = a² ....(4)

Applying Pythagoras Theorem in the triangle ACM we get,

p² + h² = c² ....(5)

Substituting the value of (4) and (5) in (3) we get,

q² + h² = b² + p² - 2bp + h²

⇒ a² = b² + c² - 2bp

⇒ p = (b² + c² - a²)/2b ....(6)

From (5)

p² + h² = c²

⇒ h² = c² - p² = (c + p) (c - p) ....(7) (As a² - b² = (a+b)(a-b))

Substituting (6) in (7) we get,

h² = (c + p) (c - p)

⇒ h² = (c + (b² + c² - a²)/2b) (c - (b² + c² - a²)/2b)

⇒ h² = ((2bc + b² + c² - a²)/2b) ((2bc - b² - c² + a²)/2b)

⇒ h² = ((b + c)² - a2)/2b) ((a² - (b - c)²)/2b)

⇒ h² = ((b + c + a)(b + c - a)(a + b - c)(a - b + c))/4b²) ....(8) (As a² - b² = (a+b)(a-b))

As perimeter of triangle is P = a + b + c and P = 2s. (Here s = semi-perimeter and s = P/2)

∴ 2s = a + b + c ....(9)

Substituting (9) in (8) we get,

h² = ((b + c + a)(b + c - a)(a + b - c)(a - b + c))/4b²)

⇒ h² = (2s × (2s - 2a) × (2s - 2b) × (2s - 2c))/4b²)

⇒ h² = (2s × 2(s - a) × 2(s - b) × 2(s - c))/4b²)

⇒ h² = 16s(s - a)(s - b)(s - c)/4bv

⇒ h = √(4s(s - a)(s - b)(s - c)/4b²)

⇒h = 2√(s(s - a)(s - b)(s - c))/b ....(10)

Area of triangle ABC, A = (1/2) × base × height

⇒ A = (1/2) × b × h

⇒ A = (1/2) × b × 2√(s(s - a)(s - b)(s - c))/b (From (10))

⇒ A = √(s(s - a)(s - b)(s - c))

∴ Area of the triangle ABC =

√(s(s - a)(s - b)(s - c)) unit²