A) \[{{S}^{2}}\]
B) \[{{d}^{2}}\]
C) \[{{S}^{2}}-{{d}^{2}}\]
D) \[{{S}^{2}}+{{d}^{2}}\]
Correct Answer: C
Solution :
Given that, \[l+b+h=S\] ...(i) and \[\sqrt{{{l}^{2}}+{{b}^{2}}+{{h}^{2}}}=d\] ...(ii) So, \[{{l}^{2}}+{{b}^{2}}+{{h}^{2}}={{d}^{2}}\] \[\therefore \] \[{{(l+b+h)}^{2}}={{S}^{2}}\] \[\Rightarrow \] \[{{l}^{2}}+{{b}^{2}}+{{h}^{2}}+2\,(lb+bh+hl)={{S}^{2}}\] \[\Rightarrow \] \[{{d}^{2}}+2\,(lb+bh+hl)={{S}^{2}}\] From Eq. (ii), \[2(lb+bh+hl)=({{S}^{2}}-{{d}^{2}})\] \[\therefore \] Surface area \[=({{S}^{2}}-{{d}^{2}})\]
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