Question 497161
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If you mean that the overall area of both the rectangle and the surmounting semi-circle is supposed to be 4 ft², then this is mighty weird looking stained glass window.  Since you didn't provide a dimension for the diameter of the semicircle, I have to assume that it is the same as the width of the rectangle.  I'm also assuming that *[tex \Large h] is the missing dimension of the rectangle since it cannot possibly be the overall height of the window.  The area of a semicircle with a diameter of 3 feet and therefore a radius of 1.5 feet is *[tex \Large \frac{2.25\pi}{2}\ \approx\ 3.53\ \text{ft^2}].  That means only *[tex \Large 0.47\ \text{ft^2}] of the given *[tex \Large 4\ \text{ft^2}] area remains for the rectangle.  Divided by 3, that leaves a height of less than *[tex \Large 0.16] feet for the rectangle part of the window -- not even 2 inches.  Standing back, you probably wouldn't notice that there was a rectanglual part of the window.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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