Question 918440: A Norman window has the shape of a rectangle surmounted by a semicircle. If the perimeter of the window is 36.900 ft. give the area A of the window in square feet when the width is 7.500 ft. Give the answer to two decimal places.
I'm stuck and I need this one to finish my homework! Even my dad can't help me and I got no results from google and my math book only uses rectangular examples!!! PLEASE HELP ME!!!!
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39823)   (Show Source):
You can put this solution on YOUR website! The problem description is somewhat ambiguous. Width of the window is 7.500 feet; but is THE semicircle attached to a width or is it attached to a length? Any figure or diagram included in your book or page might help tell you this nonverbally.
I can explain generally.
x and y are the dimensions of the rectangle portion of the window. You know perimeter of the window p and one of the diensions, either x or y;
but you do not know A area of the window.
p=36.9000
A is unknown
either x or y is known but not both.
Either x or y serves as diameter of the semicricle, but only one of those dimensions; NOT both, and not both at the same time.
If the semicircle has diameter of y, then  , and  .
Nothing more specific without the exact and complete problem description.
Answer by MathTherapy(10839)  (Show Source):
You can put this solution on YOUR website!
A Norman window has the shape of a rectangle surmounted by a semicircle. If the perimeter of the window is 36.900 ft. give
the area A of the window in square feet when the width is 7.500 ft. Give the answer to two decimal places.
I'm stuck and I need this one to finish my homework! Even my dad can't help me and I got no results from google and my math
book only uses rectangular examples!!! PLEASE HELP ME!!!!
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Respondent @josgaritmetic presumes that additional info is needed to determine the area in this problem. Quite the contrary!
 .
This problem was redone and corrected, since the width and perimeter were erroneously entered before as 7,500 and 36,900,
respectively, instead of 7.5' and 36.9', respectively
Width (W), or DC = 7.5’
Diameter of semi-circle = AB = DC = W = 7.5'
Perimeter (GIVEN) = 36.9’
Perimeter of window = BCDA + length of arc of semi-circle AB
= L + W + L + circumference of semicircle (length of arc AB)
= W + 2L + circumference of semicircle (length of arc AB)
36.9 = 7.5 + 2L +  ꙥD ---- Substituting 36.9 for perimeter, 7.5 for W, and ꙥD for circumference of
semi-circle, AB
36.9 = 7.5 + 2L + ꙥAB
36.9 = 7.5 + 2L + 7.5ꙥ
36.9 = 7.5 + 2L + 3.75ꙥ
36.9 - 7.5 - 3.75ꙥ = 2L
29.4 - 3.75ꙥ = 2L
8.81' = L
Area of rectangle, ABCD: LW = 8.81(7.5) = 66.075 sq ft
Diameter of semi-circle, AB = AB
Radius (r) =  D = AB = 7.5 = 3.75’
Area of semi-circle AB = 
=  ꙥ ---- Substituting 3.75 for r
= (14.0625ꙥ)
= (44.17864) = 22.08932 sq ft
Area of the window = Area of rectangle, ABCD + Area of semi-circle AB
= 66.075 sq ft + 22.08932 sq ft = 88.16432, or approximately 88.16 sq ft
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