A -mile track for racing stock cars consists
of two semicircles connected by parallel straightaways (see
the figure). In order to provide sufficient room for pit crews,
emergency vehicles, and spectator parking, the track must enclose an area of 100,000 square feet. Find the length of the
straightaways and the diameter of the semicircles to the nearest foot. Recall: The area A and circumference C of a circle of diameter d are given by A= pie times diameter squared divided by 4 and c= pie times diameter
i have no clue where to even start I am lost...
I dont expect the anwser
if someone could just explain to me how to get it into a number problem and out of a word problem and set it up so its ready to be solved...
as much details as possible would be great so i can understand it better... thanks a bunch for anyone that takes the time to help me... out... algebra will be the death of me
2007-09-06 09:49:15 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
http://i7.photobucket.com/albums/y294/joetta/cart.jpg
there is a pic of the diagram
see what i am thinking is that the atraighaways would be the length of the overal area which is 100,0o0 sq feet...
is that right or wrong?
2007-09-06 09:54:49 · update #1
what if the track was 1/4 of a mile would i just use 1/4th of 5280 instead of the full number
2007-09-06 10:16:27 · update #2
it's a 1-mile track, right? let x be length of straightaways, y be width of inner rectangle and hence diameter of half circles. we get an equation for length of track and one for area:
2x + πy = 5280
xy + π(y/2)² = 100,000
solve 1st for x and plug it into 2nd:
x = (5280 - πy)/2
x = 2640 - (π/2)y
[2640 - (π/2)y]y + (π/4)y² = 100,000
(π/4)y² - (π/2)y² + 2640y = 100,000
-(π/4)y² + 2640y = 100,000
(π/4)y² - 2640y + 100,000 = 0
there's your problem ready for quadratic formula.
2007-09-06 10:07:40 · answer #1 · answered by Philo 7 · 0⤊ 0⤋
The track components can be rearranged into a circle and a rectangle. If the radius of the circle is r, the width of the rectangle is 2r, and we can call its lengh h. The enclosed area is then pi r^2 + 2 r h, and the length of the track will be the circumference of the circle plus twice the length of the rectangle or 2 pi r + 2 h. If you equate the area to 100,000 square feet, and the length to 5280 feet, you have two equations in two unknowns which can be solved by the usual methods.
2007-09-06 16:57:23 · answer #2 · answered by Anonymous · 0⤊ 0⤋
The answer is = 2
2007-09-06 16:54:28 · answer #3 · answered by 7crows 2 · 0⤊ 0⤋
Can you paste a link to see the figure?
2007-09-06 16:54:26 · answer #4 · answered by plolol 2 · 0⤊ 0⤋