Question

It is given that S=1/2(a+b+c)

show that 4a2c2-(a2-b2+c2)2=16s(s-a)(s-b)(s-c)

Answer 1

Jawab:

Proved

Penjelasan dengan langkah-langkah:

4a²c² - (a² - b² + c²)²

= (2ac)² - (a² - b² + c²)²

= [(2ac) + (a² - b² + c²)] [(2ac) - (a² - b² + c²)]

= [(a² + 2ac + c²) - b²] [b² - (a² - 2ac + c²)]

= [(a + c)² - b²] [b² - (a - c)²]

= [(a + c) + b] [(a + c) - b] [b + (a - c)] [b - (a - c)]

= (a + b + c) (a + b + c - 2b) (a + b + c - 2c) (a + b + c - 2a)

= 2s (2s - 2b) (2s - 2c) (2s - 2a)

= 2s [2(s - a)] [2(s - b)] [2(s - c)]

= 16s(s - a)(s - b)(s - c)