Question 282064
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Since the frame is square and measures 5 inches on a side, the area of the frame and the picture it contains is simply *[tex \Large 5^2\ =\ 25] and the actual size of the picture does not have any bearing on the total area of the frame and picture (as long as *[tex \Large x\ <\ 5] holds that is).


So, the area of the frame AND the picture is 25, but the area of just the *[tex \Large x] by *[tex \Large x] picture is *[tex \Large x^2].  Hence, the area of just the frame is the area of the frame and picture minus the area of the picture, or *[tex \Large 25\ -\ x^2]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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