If a cell tower can service an area of 75 square miles how far away can you be and still connect?

[@tomtke7351]

A = pi × r^2 75 = pi × r^2 r^2 = 75/pi = 23.8732 r = 4.886miles

[@chrisdissanayake6979]

Area= π r^2 = 75 sq. miles π = 22/7 22/7 r^2 = 75 sq. miles r^2 = (75 * 7)/22 r^2 = 525/22 r = √525/22 = √ 23.86 = 4.88 miles The radius is 4.88 miles. You must be within the circular area, with a radius of 4.88 miles, around the tower to pick up a signal.

[@trinitron40237]

Who said that the tower is omnidirectional? In that range of frequency the antenna has a totally different characteristic, far from the circular shape.

[@doughuffman5790]

Not to get too technical here, perhaps the assumption of smooth coverage is unwarranted.

[@WoozyPolarBear]

Not to get technical here but since the question only had 2 significant digits, the answer is actually more correctly 4.9 mi, not 4.88.

[@christianbaughn199]

And to 2 decimal places would be 4.89 miles, not 4.88.

[@francisdelpuech6415]

To check you do the problem backwards S=π•r^2=3.1416•4.9^2=75.43, When 4.88 gives 74.82 which is closer to the given data. The closest being 4.9 which gives 75.12

[@christianbaughn199]

@francisdelpuech6415 Re your last sentence, you mean 4.89. Like you said, pi x 4.9² = 75.43

[@StephenJWalter]

Yeah, this guy consistently truncates rather than rounding up or down. I would’ve just as consistently failed maths tests in high school if I did that.

[@francisdelpuech6415]

@@christianbaughn199 🙌

[@russelllomando8460]

woo hoo 4.88 Pi r^2 area of circle 75/3.14 = 23.88 then sr of 23.88 = 4.88 thanks for the fun

[@Stylux-z1p]

Ok , let's imagine a 2D circular area of 75 mi² A 2D circular area is inherently associated with a center\midpoint (the tower) and a radius Formula of a circular area(circle) is : A = π r² given : A = 75 m² π r² = ? 75 = π r² 75/π = π r²/π --> isolate r² 75/π = r² √[75/π] = ±√[r²] √[75/π] = r 4.886025119029199 ≈ r r ≈ 4.89 miles away from the tower is the connection range of the signal ✔ I think the signal a tower or router is sending/propagating is spherical 3D

[@raya.pawley3563]

Thank you.

[@reiniernn9071]

I would HATE to get these type of question for checking my maths.... Because I do know to much about physics (electronics including senders). I've never seen any cell tower with a nice circulair radiation. More often I see 6 seperate senders placed in a hexagon.model. This generates an area with is not exact circulair. Also these high frequencies as used for cellphones will travle more in straight line than low frequence radio waves (with vertical antenna). As a child , >50 years ago , as a 15 years old hobbyist, I tested the range of a 50 milliWatt 100mhz sender....it was connected to a dipole (directional) antenna for one of the tests. With our equipment it reached out for 5 miles in one direction (8 km) but only 300meter in the perpendiculair direction. This makes the circulair assumption incorrect for any normal examination questions as this one.....unless also is GIVEN that the signal has equal strength in all horizontal directions. Thois is more a question about reading and making guessed assumptions than checking any real math knowledge.

[@karlrschneider]

Very few antennas are truly omnidirectional. This is a dumb question.

[@Surreal_Wizard]

there used to be this radio station in Yellow Pine, Idaho, KRSN, "Now with an effective broadcast radius of four point eighteightsixo'twofiveoneonenine miles!" So I'll go with that as the answer.

[@tupoa360]

Depends on down tilt and fading. Service areas are seldom a circle.

[@marknesselhaus4376]

You beat me to it ;-)

[@tupoa360]

@@marknesselhaus4376 Great minds brother!

[@PraiseDog]

@@tupoa360 Actually not. I would not call my mind great, but my thinking was "There are of course some other factors here that would affect the transmission, everyone here with a brain bigger than a pea realizes this, but the problem obviously expects us to ignore those in this calculation which is testing our knowledge of geometry, not physical properties of radio waves. Tablet Class Math should have been your clue to this.

[@Poult100]

😂 You just know you're avoiding the theoretical mathmatical issue here, guys!

[@lj6079]

​@@PraiseDogwow. You're so clever

[@vallelungak]

4.88 miles

[@garycgibson]

If the tower was located at the head of a deep straight canyon a half mile wide 75 square miles would be 150 miles long (unless the canyon is a quarter mile wide and the max range is 300 minus compensation for curve of the Earth), so the answer then is 150. The land topology matters. I live 4 miles from a cell tower and get no signal because there is a mountain in the way .

[@Hewhowalks-fv5mq]

Fairly simple 😀😃

[@TheAZZA0990]

sq. root of 75/pi = approx. 4.885 miles!! :)

[@dhy5342]

root 75/pi

[@daveussell1222]

Much better than 22/7 is 355/113

[@leetucker9938]

4.88 miles 🙂

[@cato451]

20ish miles

[@dazartingstall6680]

πr² = 75 r² = 75/π r = √(75/π) _(*grabs calculator*)_ r = 4.89 miles (rounded)

[@josephlaura7387]

4.886miles

[@periramana8556]

You are making 1 inch solution a imile long

[@robertanderson7049]

I have checked out at least 3 of your exercises on KZbin. You talk way too much. You run things into the ground. Very boring. The tower broacasts in a circular area that is 75 miles Square. The area of a cirlce is pi times radius squared.

[@billmorris2613]

4.33 miles, assuming the tower is in the center of the square that contains 75 square miles.

[@PatWyckoff]

Why do we assume the area is circular?

[@martinda7446]

Physics. Waves from a radio tower will have an aerial that propagates equally in all directions like a stone dropped in a pond.

[@stevewaters449]

What else would you suggest?

[@georgecattani7199]

8.7miles

[@WoozyPolarBear]

sounds like you may have solved for the diameter, and not the radius.

[@karlrschneider]

150 miles if the service area is 1/2 x 75 miles.