A) 1
B) 2
C) 3
D) 4
Correct Answer: C
Solution :
\[X=\frac{1}{2}a{{(2)}^{2}}\Rightarrow X=2a\] \[X+Y=\frac{1}{2}a{{(4)}^{2}}=\frac{1}{2}a(16)\] \[X+Y=8a\] \[Y=8a-X\Rightarrow \,=8a-2a=6a\] \[\frac{Y}{x}=\frac{6a}{2a}\Rightarrow Y=3x\,\,(or)\,\,n=3\]You need to login to perform this action.
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