A painting at an art gallery is framed such that the area of the square frame is 2/3 the area of the painting itself. If the diagonal line connecting corners of the frame has a length of 10, what is the area of the painting inside the frame?	Choose the correct answer.
A) 10	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   Area of painting = a^2 and we know that the area of the frame is (2/3) of that of the painting so   Area of frame = (2/3) * a^2   Area of Frame + Painting = a^2 + (2/3)*a^2 = (5/3)*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 = 10   so, b = 5 * sqrt2   b^2 = 50   we know that b^2 = (5/3)*a^2   so, a^2 = (3/5)*b^2 = (3/5)*50= 
<gadget id="calculator">(3/5)*50</gadget>
<output>30</output>
30   And area of painting= a^2 = 30   So, answer will be 30.
<result>C</result>
A painting at an art gallery is framed such that the area of the square frame is 2/3 the area of the painting itself. If the diagonal line connecting corners of the frame has a length of 10, what is the area of the painting inside the frame?