Question

please help !!!! solving this question ​

Answer 1

Answer:

Area of triangle ABC = √s(s-a)(s-b)(s-c), where s = Perimeter/2 = (a + b + c)/2

Example: Find the area of a triangle whose lengths are 5 units, 6 units, and 9 units respectively.

Solution: As we know, a = 5 units, b = 6 units and c = 9 units

Thus, Semi-perimeter, s = (a + b + c)/2 = (5 + 6 + 9)/2 = 10 units

Area of triangle = √(s(s-a)(s-b)(s-c)) = √(10(10-5)(10-6)(10-9))

⇒ Area of triangle = √(10 × 5 × 4 × 1) = √200 = 14.142 unit2

∴ The area of the triangle is 14.142 unit2

Derivation of Heron's Formula for Area of Triangle

We will use some Pythagoras theorem, area of a triangle formula, and algebraic identities to derive 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. Consider the triangle below:

Derivation of Heron's Formula

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, as per the image, b = p + q

⇒ q = b - p ....(1)