Question

If a+b+c= 16 ,a²+b²+c²= 250 Find 1/3(ab+bc+ca) *
a.1
b.-1
c.0
d.none​

Answer 1

Answer:

a 2+b 2+c 2

=250 & ab+bc+ac=3, find a+b+c

→ the general formula

(a+b+c) 2=a ^2+b ^2+c ^2+2(ab+bc+ac)

=250+2(3)

=250+6

∴(a+b+c)

2

=256

∴a+b+c=

256

∴a+b+c=16

Answer 2

Answer:

option a) 1 is correct

Step-by-step explanation:

Given :

a + b + c = 16

a² + b² + c² = 250

Formula :

(a + b + c )² = a² + b² + c² +2 (ab + bc + ca)

16² = 250 + 2 (ab + bc + ca)

256 - 250 = 2 (ab + bc + ca)

2(ab + bc + ca) = 6

(ab + bc + ca) = 3

∴ 1/3 (ab + bc + ca) = 1