Question

find the value of AB + BC + CA if a + b + c = 15 and a^2 + b^2+c^2=77​

Answer 1

a+b+c=15

On squaring both sides

$( {a + b + c)}^{2} = ( {15)}^{2} \\ {a }^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca = 225$

It is given that

a^2+b^2+c^2=77

Putting this in the above equation

$77 + 2ab + 2bc + 2ca = 225 \\ 2(ab + bc + ca) = 225 - 77 \\ ab + bc + ca = \frac{148}{2} \\ ab + bc + ca = 74$

Answer 2

Answer:

74

Step-by-step explanation:

Given:

a + b + c = 15

a² + b² + c² = 77

To find:

ab + bc + ca

Solution:

We know,

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

By taking 2 common the Identity can be rewritten as:

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

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Plugging the values we know in (ii), we have:

(15)² = 77 + 2(ab + bc + ca)

225 = 77 + 2 (ab + bc + ca)

225 - 77 = 2 (ab + bc + ca)

148 = 2 (ab + bc + ca)

74 = (ab + bc + ca)