A painting at an art gallery is framed such that the area of the square frame is 1/4 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?
Suppose side of the painting(which is a square) is "a" and side of the outer square(paiting+frame) is "b"
Area of paiting = a^2 and we know that the area of the frame is (1/4) of that of the paiting so
Area of frame = (1/4) * a^2
Area of Frame + Paiting = a^2 + (1/4)*a^2 = (5/4)*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 = (5/4)*a^2
so, a^2 = (4/5)*b^2 = (4/5)*50 = 40
And area of paiting = a^2 = 40
So, answer will be D.
Hope it helps!
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