A) 10
B) 8
C) 15
D) 12
Correct Answer: C
Solution :
(c): Let the two numbers are x and y. \[x+y=8\] and \[{{x}^{2}}+{{y}^{2}}=34\] Now, we have \[{{\left( x+y \right)}^{2}}=\left( {{x}^{2}}+{{y}^{2}} \right)+2xy\] \[\Rightarrow {{(x+y)}^{2}}-({{x}^{2}}+{{y}^{2}})=2xy\] \[\Rightarrow xy=\frac{1}{2}\left[ {{\left( x+y \right)}^{2}}-\left( {{x}^{2}}+{{y}^{2}} \right) \right]\] \[=\frac{1}{2}\left[ {{\left( 8 \right)}^{2}}-\left( 34 \right) \right]\] \[=\frac{1}{2}\left[ 64-34 \right]=\frac{1}{2}\times 30=15\]
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