A painting at an art gallery is framed such that the area of the square frame is 1/2 the area of the painting itself. If the diagonal line connecting corners of the frame has a length of 6, what is the area of the painting inside the frame?  Choose the correct choice from the following answers
A) 12 B) 20 C) 30 D) 40 E) 50
Suppose side of the painting(which is a square) is a and side of the outer square(painting+frame) is b   12rea of painting = a^2 and we know that the area of the frame is (1/2) of that of the painting so   12rea of frame = (1/2) * a^2   12rea of Frame + Painting = a^2 + (1/2)*a^2 = (3/2)*a^2 which is equal to b^2   Line connecting the corners of the frame is the diagonal of the outer square which is equal to b*sqrt2   so, b * sqrt2 = 6   so, b = 3 * sqrt2   b^2 = 18   we know that b^2 = (3/2)*a^2   so, a^2 = (2/3)*b^2 = (2/3)*18= 
<gadget id="calculator">(2/3)*18</gadget>
<output>12</output>
12   12nd area of painting= a^2 = 12   So, answer will be 12.
<result>A</result>
A painting at an art gallery is framed such that the area of the square frame is 1/2 the area of the painting itself. If the diagonal line connecting corners of the frame has a length of 6, what is the area of the painting inside the frame?