Question

Hero formula ka use karke triangle ka area find kaise karenge

Answer 1

Answer:

Hero formula kyaa hain?

Answer 2

Answer:

• Formulation
• Heron's formula states that the area of a triangle whose sides have lengths a, b, and c is
• {\displaystyle A={\sqrt {s(s-a)(s-b)(s-c)}},} A = \sqrt{s(s-a)(s-b)(s-c)},
• where s is the semi-perimeter of the triangle; that is,
• {\displaystyle s={\frac {a+b+c}{2}}.} s=\frac{a+b+c}{2}.[2]
• Heron's formula can also be written as
• {\displaystyle A={\frac {1}{4}}{\sqrt {(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}} A=\frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}
• {\displaystyle A={\frac {1}{4}}{\sqrt {2(a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2})-(a^{4}+b^{4}+c^{4})}}} A=\frac{1}{4}\sqrt{2(a^2 b^2+a^2c^2+b^2c^2)-(a^4+b^4+c^4)}
• {\displaystyle A={\frac {1}{4}}{\sqrt {(a^{2}+b^{2}+c^{2})^{2}-2(a^{4}+b^{4}+c^{4})}}} A=\frac{1}{4}\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}
• {\displaystyle A={\frac {1}{4}}{\sqrt {4(a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2})-(a^{2}+b^{2}+c^{2})^{2}}}.} {\displaystyle A={\frac {1}{4}}{\sqrt {4(a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2})-(a^{2}+b^{2}+c^{2})^{2}}}.}
• Example
• Let △ABC be the triangle with sides a = 4, b = 13 and c = 15. The semiperimeter is s =  
• 1
• /
• 2
• (a + b + c) =  
• 1
• /
• 2
• (4 + 13 + 15) = 16, and the area is
• {\displaystyle {\begin{aligned}A&={\sqrt {s\left(s-a\right)\left(s-b\right)\left(s-c\right)}}={\sqrt {16\cdot (16-4)\cdot (16-13)\cdot (16-15)}}\\&={\sqrt {16\cdot 12\cdot 3\cdot 1}}={\sqrt {576}}=24.\end{aligned}}}  
• \begin{align}
• A &= \sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} = \sqrt{16 \cdot (16-4) \cdot (16-13) \cdot (16-15)}\\
• &= \sqrt{16 \cdot 12 \cdot 3 \cdot 1} = \sqrt{576} = 24.
• \end{align}
• In this example, the side lengths and area are all integers, making it a Heronian triangle. However, Heron's formula works equally well in cases where one or all of these numbers is not an integer.

Step-by-step explanation: