Question

To develop Herons formula for area of a triangle​

Answer 1

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Answer 2

Let:

ABC be a triangle with BM as an altitude on side AC

BM = h

To develop:

Herons formula for the area of a triangle​

Derivation:

AC = AM +MC

⇒ MC = AC - AM      - (1)

Squaring both sides we get,

⇒ $MC^2 = AC^2 +AM^2 -2 AC X AM$      -(2)

Adding h² on both sides we get,

$MC^2 + h^2 = AC^2 +AM^2 -2 AC X AM + h^2$    -(3)

In triangle MBC ∠M=90°

Applying Pythagoras Theorem  

MC² + h² = BC²       -(4)

Similarly in ΔMAB,

AM² + h² = AB²    -(5)

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

⇒  BC² = AC² + AB² - 2AC X AM

⇒ AM = (AC² + AB² - BC²)/2AC      -(6)

Applying identity in (5) [a² - b² = (a+b) (a-b)]

⇒ h² = AB² - AM² = (AB + AM) (AB - AM)     -(7)

Substituting (6) in (7) we get,

 h² = [(AC + AB + BC)(AC + AB - BC)(BC + AC - AB)(BC - AC + AB)]/4AC²)  -(8)

Perimeter of triangle is P = AB + BC +CA

P = 2s. (Here s = semi-perimeter)

∴ 2s = AB + BC +AC      -(9)

Substituting (9) in (8)

h = 2√(s(s - BC)(s - AC)(s - AB))/AC     -(10)

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

⇒ A = (1/2) × b × 2√(s(s - BC)(s - AC)(s - AB))/AC     (From 10)

⇒ A = √(s(s - BC)(s - AC)(s - AB))

Hence proved that area of the triangle ABC = $\sqrt{s(s - BC)(s - AC)(s - AB)}$unit².