Question

Question 3 If (a-456)2 +(b-567)2 +(c-228)2 = 0, then find the value of (ab)/2c.​

Answer 1

SOLUTION

GIVEN

$\displaystyle \sf \: {(a - 456)}^{2} + {(b - 567)}^{2} + {(c - 228)}^{2} = 0$

TO DETERMINE

$\displaystyle \sf \: \frac{ab}{2c}$

EVALUATION

Here it is given that

$\displaystyle \sf \: {(a - 456)}^{2} + {(b - 567)}^{2} + {(c - 228)}^{2} = 0$

We know if sum of the squares of three Real Numbers are zero then they are separately zero

Which gives

a - 456 = 0 , b - 567 = 0 , c - 228 = 0

Consequently we get

a = 456 , b = 567 , c = 228

Thus we get

$\displaystyle \sf \: \frac{ab}{2c}$

$\displaystyle \sf \: = \frac{456 \times 567}{2 \times 228}$

$\displaystyle \sf \: = \frac{456 \times 567}{456}$

$\displaystyle \sf \: =567$

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