Question

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Answer 1

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Here, we having two equations

a + b + c = 12 ...(eq.1)

a^2 + b^2 + c^ =64 …(eq.2)

by eq.1 we can calculate the value of a, b & c as

a = 12 - b - c

b = 12 - a - c

c = 12 - a - b

then we put the value of a in second equation.

we get,

(12 - b - c)^2 + b^2 + c^2 = 64

after solving this,

we get, b^2 + bc + c^2 - 12b - 12c + 40 =0

by using above equation we calculate the term bc.

therefore, bc = 12b + 12c - b^2 -c^2 - 40

similarly by putting the values of b and c in eq.2 we get,

ac = 12a + 12c - a^2 - c^2 - 40

ab = 12a + 12b - a^2 - b^2 - 40

and now,

ab + bc + ac = (12a + 12b - a^2 - b^2 - 40) + (12b + 12c - b^2 -c^2 - 40) + (12a + 12c - a^2 - c^2 - 40)

ab + bc + ac = 24a + 24b + 24c - 2a^2 - 2b^2 – 2c^2 – 120

ab + bc + ac = 24( a + b + c ) - 2(a^2 + b^2 + c^) - 120

by eq.1 and eq.2,

ab + bc + ac = 24(12) - 2(64) - 120

ab + bc + ac = 40

Hence, ab + bc + ac = 40

Answer 2

Answer:

40

Step-by-step explanation:

(a+b+c) ka whole square = a square + b square+ c square+2ab+2bc+2ca , 12 ka square = 64+ 2(ab + bc +ca), 144 _64 = 2(ab+bc+ca) ,80 =2(ab+bc+ca),(ab+bc+ca) = 40