question_answer

The figure shows a triangle AOB and the parabola\[y={{x}^{2}}\]. The ratio of the area of the triangle AOB to the area of the region AOB of the parabola\[y={{x}^{2}}\]is equal to

A)  \[\frac{3}{5}\]                                  

B)  \[\frac{3}{4}\]

C)  \[\frac{7}{8}\]                                  

D)  \[\frac{5}{6}\]

E)  \[\frac{2}{3}\]

Correct Answer: B

Solution :

Area of curve OAB \[=2\int_{0}^{{{a}^{2}}}{x}dy\] \[=2\int_{0}^{{{a}^{2}}}{\sqrt{y}}dy=2\left[ \frac{{{y}^{3/2}}}{3/2} \right]_{0}^{{{a}^{2}}}\] \[=\frac{4}{3}[{{a}^{3}}]\] Now, Area of \[\Delta OAB=\frac{1}{2}\times AB\times OC\]                                 \[=\frac{1}{2}\times 2a\times {{a}^{2}}\]                                 \[={{a}^{3}}\] \[\therefore \]\[\frac{Area\,of\,\Delta AOB}{Area\,of\,curve\,AOB}=\frac{{{a}^{3}}}{\frac{4}{3}{{a}^{3}}}\]                                 \[=\frac{3}{4}\]