Asked by TayB
A Norman window has the shape of a rectangle surmounted by a semicircle. If the perimeter of the window is 26 ft, express the area A of the window as a function of the width x of the window.
Answers
Answered by
Reiny
Now I would have started by defining the radius as r and the height of the rectangular window as h ( avoiding a lot of fractions)
Perimeter of half a circle = (1/2)(2πr)
= πr
the three sides of the rectangle = 2r + 2h
so
2r + 2h + πr = 26
h = (26 - 2r - πr)/2 = 13 - r - πr/2
area = (1/2)πr^2 + 2rh
= (1/2)πr^2 + 2r(13-r-πr/2)
= (1/2)πr^2 + 26r - 2r^2 - πr^2
= 26r - 2r^2 - (πr^2)/2
<b>now</b> if we let the base of the window be x
then the radius is x/2
let the height of the rectangle be y
(1/2)(2π(x/2)) + x + 2y = 26
(πx)/2 + x + 2y = 26
y = (26 - x - πx/2)/2
area = (1/2)π(x/2)^2 + xy
= πx^2/8 + x(26 - x - πx/2)/2
= πx^2/8 + 13x - x^2/2 - πx^2/4
= - πx^2/8 + 13x - x^2/4
check each setup for algebraic errors, I did not write it out first.
Perimeter of half a circle = (1/2)(2πr)
= πr
the three sides of the rectangle = 2r + 2h
so
2r + 2h + πr = 26
h = (26 - 2r - πr)/2 = 13 - r - πr/2
area = (1/2)πr^2 + 2rh
= (1/2)πr^2 + 2r(13-r-πr/2)
= (1/2)πr^2 + 26r - 2r^2 - πr^2
= 26r - 2r^2 - (πr^2)/2
<b>now</b> if we let the base of the window be x
then the radius is x/2
let the height of the rectangle be y
(1/2)(2π(x/2)) + x + 2y = 26
(πx)/2 + x + 2y = 26
y = (26 - x - πx/2)/2
area = (1/2)π(x/2)^2 + xy
= πx^2/8 + x(26 - x - πx/2)/2
= πx^2/8 + 13x - x^2/2 - πx^2/4
= - πx^2/8 + 13x - x^2/4
check each setup for algebraic errors, I did not write it out first.
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