A) 1 : 2 : 1
B) 1 : 2 : 3
C) 2 : 1 : 2
D) 1 : 1 : 1
Correct Answer: D
Solution :
\[{{S}_{1}}={{S}_{3}}=\int_{0}^{4}{\frac{{{x}^{2}}}{4}}dx\] \[=\frac{1}{4}\left[ \frac{{{x}^{3}}}{3} \right]_{0}^{4}=\frac{1}{12}\times 64=\frac{16}{3}sq\,units\] \[\therefore \] \[{{S}_{2}}+{{S}_{3}}=\int_{0}^{4}{\sqrt{4x}}dx\] \[=2\times 2\left[ \frac{{{x}^{3/2}}}{3} \right]_{0}^{4}=\frac{4}{3}\times 8=\frac{32}{3}sq\,units\] \[\Rightarrow \] \[{{S}_{2}}=\frac{16}{3}sq\,units\] \[\therefore \] \[{{S}_{1}}:{{S}_{2}}:{{S}_{3}}=\frac{16}{3}:\frac{16}{3}:\frac{16}{3}\] \[=1:1:1\]You need to login to perform this action.
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