Question

if X + Y + Z equal to 12 and X square + Y square + Z square equal to 64 then find X Y + Y Z plus ZX

Answer 1

x+y+z= 12
${x}^{2} + {y}^{2} + {z}^{2} = 64$
${(x + y + z)}^{2} = {x}^{2} + {y}^{2} + {z}^{2} + \: \: \: \: \: \\ 2(xy + yz + zx)$
therefore
$xy + yz + zx$
=
$\frac{1}{2} ( {(x + y + z)}^{2} - ( {x}^{2} + {y}^{2} + {z}^{2} ))$
$= \frac{1}{2} \times (12^{2} - 64)$
$= \frac{1}{2} \times \: (144 - 64)$
$= \frac{1}{2} \times 80$
=40

Answer 2

Answer: Solution is 40

Step-by-step explanation:

Given  X + Y + Z = 12

X² + Y²+ Z² = 64

Squaring the both sides

(X + Y + Z )² = 12²

X² + Y² + Z² + 2XY + 2YZ + 2ZX = 144

Regroup the term

( X² + Y²+ Z² ) + 2 ( XY +YZ + ZX ) = 144

Putting the value of X² + Y²+ Z² = 64

64 + 2 (XY + YZ + ZX ) = 144

2 ( XY + YZ + ZX ) = 144 - 64

2 ( XY + YZ + ZX ) = 80

XY + YZ + ZX = 80 /2

XY + YZ + ZX = 40

So the answer is 40