A painting at an art gallery is framed such that the area of the square frame is 1/5 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:	A) 10	B) 15	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   Area of painting = a^2 and we know that the area of the frame is (1/5) of that of the painting so   Area of frame = (1/5) * a^2   Area of Frame + Painting = a^2 + (1/5)*a^2 = (6/5)*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 = (6/5)*a^2   so, a^2 = (5/6)*b^2 = (5/6)*18= 
<gadget id="calculator">(5/6)*18</gadget>
<output>15</output>
15   And area of painting= a^2 = 40   So, answer will be 15.
<result>B</result>
A painting at an art gallery is framed such that the area of the square frame is 1/5 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?