Question

(a-c):(b-c)=a^2:b^2, find (a+c):(b+c)= ?

Answer 1

Answer:

If a+b+c = 12 and a^2 + b^2 + c^2 = 64, find the value of ab+bc+AC?

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

A/Q, the given values are

a+b+c = 12

a^2 + b^2 + c^2 = 64

Now, the value we need to find is ab + bc + ca. This can be determined by using the expansion of (a + b + c)^2

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

2 (ab+bc+ca) = (a+b+c)^2 - (a^2+b^2+c^2)

(ab+bc+ca) = [(a+b+c)^2 - (a^2+b^2+c^2)]/2

= [(12^2) - (64)]/2

= (144 - 64)/2

= 80/2

ab + bc + ca = 40

Therefore,

ab + bc + ca = 40

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