SOLUTION: A window is in the shape of a rectangle surmounted by a semicircle. The height of the rectangle is 0.4m more than the width. The total area of the window is 10.4m2 find the width a

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: A window is in the shape of a rectangle surmounted by a semicircle. The height of the rectangle is 0.4m more than the width. The total area of the window is 10.4m2 find the width a      Log On


   



Question 1104847: A window is in the shape of a rectangle surmounted by a semicircle. The height of the rectangle is 0.4m more than the width. The total area of the window is 10.4m2 find the width and height of the window.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let the width of the rectangle = +w+ m
The height of the rectangle is +w+%2B+.4+ m
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The area of the semicircle = +%281%2F2%29%2Api%2A%28w%2F2%29%5E2+ m2
The total area is:
+w%2A%28+w+%2B+.4+%29+%2B+%281%2F2%29%2A%281%2F4%29%2Api%2Aw%5E2+=+10.4+ m2
+w%5E2+%2B+.4w+%2B+%28+pi%2F8+%29%2Aw%5E2+=+10.4+ m2
+%28%28+pi+%2B+8+%29+%2F+8+%29%2Aw%5E2+%2B+.4w%5E2+-+10.4+=+0+
+1.3927w%5E2+%2B+.4w+-+10.4+=+0+
Rounding off:
+1.4w%5E2+%2B+.4w+-+10.4+=+0+
Use quadratic formula
+w+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
+a+=+1.4+
+b+=+.4+
+c+=+-10.4+
+w+=+%28-.4+%2B-+sqrt%28+.4%5E2-4%2A1.4%2A%28-10.4%29+%29%29%2F%282%2A1.4%29+
+w+=+%28-.4+%2B-+sqrt%28+.16+%2B+58.24+%29%29%2F+2.8+
+w+=+%28+-.4+%2B+sqrt%28+58.4+%29%29+%2F+2.8+
+w+=+%28+-.4+%2B+7.642+%29+%2F+2.8+
+w+=+7.242+%2F+2.8+
+w+=+2.586+
and
+w+%2B+.4+=+2.986+
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The height, including semicircle is
+2.986+%2B+w%2F2+
+2.986+%2B+2.586%2F2+
+2.986+%2B+1.293=+4.279+
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width = 2.586 m
height = 4.279 m
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check:
+1.4w%5E2+%2B+.4w++=+10.4+
+1.4%2A2.586%5E2+%2B+.4%2A2.586+=+10.4+
+1.4%2A6.687+%2B+1.0344+=+10.4+
+9.3618+%2B++1.0344+=+10.4+
+10.3962+=+10.4+
Looks close enough -error due to rounding off