Ant three right angle triange area and prove with herons formula
Answers
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))