The area is 2*h*(45-2*h)/(2+pi) where h is the height of the rectangular part of the door.
To find the area of the Norman window, you need to divide the perimeter by 2 to get the width of the rectangle, which is 22.5 feet. The radius of the semicircle is half of the rectangle's width, so it's 11.25 feet. The area of the rectangle is 11.25 * 22.5 = 253.13 square feet, and the area of the semicircle is 0.5 * pi * 11.25^2 = 198.93 square feet. So, the total area of the window is 253.13 + 198.93 = 452.06 square feet.
75
To solve this problem, let's break it down step by step. Let ( r ) be the radius of the semicircle, which is also the width of the rectangle. The perimeter of this figure (Norman window) is given as 45 feet: The perimeter, ( P ), is the sum of the parts: the semicircle's circumference and the perimeter of the rectangle. The semicircle's circumference: ( \pi r ) The perimeter of the rectangle: ( 2r + 2r = 4r ) So, the total perimeter equation is: [ \pi r + 4r = 45 ] This simplifies to: [ \pi r + 4r = 45 ] [ (\pi + 4) r = 45 ] [ r = \frac{45}{\pi + 4} ] Now, we need to find the total area of the figure. The area of the semicircle is: [ \frac{\pi r^2}{2} ] The area of the rectangle is: [ r \times 2r = 2r^2 ] The total area, ( A ), is the sum of these two parts: [ A = \frac{\pi r^2}{2} + 2r^2 ] Substitute the value of ( r ) derived earlier: [ A = \frac{\pi (\frac{45}{\pi + 4})^2}{2} + 2(\frac{45}{\pi + 4})^2 ] Calculating this would give the area of the largest possible Norman window with a perimeter of 45 feet.
These windows could get extremely large, but let's play with Andersen Windows' catalog. If you build a Norman window using Andersen's largest semicircle window, which is six feet wide, and their tallest twin double-hung window, which is 6'4" high, you get an area of 52.11 square feet.
A Norman window is a semicircle on top of a rectangle. You have given one dimension of 15.5 ft. There is no indication as to which dimension this refers. If it is the width of the window, then the semi-circular bit at the top has a radius of 7.75 ft and there is no indication of the height of the rectangular bit; thus a rectangle that can contain the window is at least 15.5 ft wide and at least 7.75 ft high. If 15.5 ft is the height of the window, then the window can have any width greater than 0 ft and less than 31 ft - the greater the width the shorter the rectangular bit of the window. which means the rectangle will have varying dimensions between a width greater than 0 ft and a height greater than 15.5 ft, to a width of 31 ft and a height greater than 15.5 ft; in this case a rectangle greater than 31 ft by greater than 15.5 ft will contain any possible Norman window that is 15.5 ft high. (At 31ft by 15.5 ft the rectangle will cover the widest Norman window possible.)
Let x be the width and y be the length of the rectangle. x/2 is the radius of the semicircle Perimeter of the Norman window is x+2y+(π x)/2 Let P be the perimeter --- 288 in this problem. P = x+2y+(π x)/2--------(1) Solving for y from equation (1) 2y = P-x-πx/2 y = P/2-x/2-πx/4--------(2) Area = xy + π x^2 / 8 A = x(P/2-x/2-π x/4) + π x^2/8 A= Px/2-x^2 /2 -πx^2/4 +πx^2/8 dA/dx = P/2 -2x/2-2πx /4 +2πx / 8 =0 (4p-8x-2πx)/8=0 4p-2x(π+4)=0 4p=2x(π+4) x= 2P / (4+π) The radius is x/2 = P/(4+PI) Substitute P with 288 radius = 288 / (4+PI) will maximize the area of the window. d^2A/dx^2 =-1-π/2+π/4 < 0, indicates that the area is maximized. You'll have to simplify x and y if you want them in numeric format.
An Anglo-Norman is a Norman who settled in England after the Norman Conquest, or a descendant of one.
An Anglo-Norman is a Norman who settled in England after the Norman Conquest, or a descendant of one.
Norman Lear's birth name is Norman Milton Lear.
Norman Field's birth name is Field, Norman Randolph.
Yes. He was a Norman. He was the first Norman King in England.
Irving Norman has written: 'Irving Norman'
Norman Taurog went by Uncle Norman.
The address of the Norman Public Library is: 225 N Webster, Norman, 73069 7133