question_answer

ABCD is a square of side a, ABO is an equilateral triangle and OL is perpendicular to CD. Then, area of the trapezium AOLD is

A)  \[\frac{{{a}^{2}}}{2}+\frac{\sqrt{3}}{8}{{a}^{2}}\]

B)         \[\frac{{{a}^{3}}}{2}+\frac{\sqrt{3}}{4}{{a}^{3}}\]

C)  \[{{a}^{3}}+\sqrt{3}{{a}^{3}}\]      

D)  \[\frac{{{a}^{3}}}{2}+\frac{\sqrt{3}}{2}{{a}^{3}}\]

Correct Answer: A

Solution :

As, OAB is equilateral triangle.

\[\therefore \]      \[\angle OAM=60{}^\circ \]and AB = OA = OB = a

\[\therefore \]      (Altitude) \[OM=\frac{\sqrt{3}}{2}\]side \[=\frac{\sqrt{3}}{2}a\]

\[\therefore \]      \[OL=OM+ML=\frac{\sqrt{3}}{2}a+a\]

\[\therefore \] Area of trapezium \[\frac{1}{2}(AD+OL)AM\]

\[=\frac{1}{2}\left( a+\frac{\sqrt{3}}{2}a+a \right)\frac{a}{2}\left( \because AM=\frac{1}{2}AB \right)\]

\[=\frac{\sqrt{3}}{2}{{a}^{2}}+\frac{{{a}^{2}}}{2}\]