 # how do i work out the area of a circle without a calculator?

## Recommended Posts

pen and paper, basic arithmetic would be my choice ##### Share on other sites

use approxamations?

pi = 3 etc?

##### Share on other sites

2pi = 3.142 x 2 thus the 1st bit is 6.284

whats the Radius? lets assume its 9cms so whats 9 Squared? thats 81

81 x 6.284 = 509 cms

Hmmm... I think I just worked out a Sphere?

nevermind, I said my maths was crap! ##### Share on other sites

I think a good approximate for pi is 3.1416 since the next after the five there's a 9. Memorizing pi is probably the most idiotic thing to do, but for some strange reason I've memorized up to 35 decimals. I wonder what's the current world record...

##### Share on other sites

Why dko you need to approximate anyway? the area of a circle is pi*r^2, you shpould always keep the pi in, unless for some reason you need to know where it approximately lies on the real number line, but generally there's no need to know.

##### Share on other sites

Why do you need to approximate, eh? How about pi being an irrational number. Btw, World Record: Goto, Hiroyuki | Japan | 42195 digits, World Rec since -95

##### Share on other sites
Why do you need to approximate' date=' eh? How about pi being an irrational number. Btw, World Record: Goto, Hiroyuki | Japan | 42195 digits, World Rec since -95[/quote']

You should never approximate unless you have to, the decimals are only one representation of the real numbers.

##### Share on other sites

If you didn't notice, the thread is about Ice wanting to know about counting the area of a circle without a calculator; therefore, you need to approximate quite a bit.

##### Share on other sites

If your asked to find the raidus of a circle of say radius 9m, then you should alewyas express your answer as 81*pi m^2, of course if for some reason (I relaly can't think of any reaosn though) you need to know wherre it approximatley lies on the real number line then you would approximate pi.

##### Share on other sites
If you didn't notice, the thread is about Ice wanting to know about counting the area of a circle without a calculator; therefore, you need to approximate quite a bit.

If he can't multiply numbers just because they're on the wrong side of a decimal place I'd have concerns about what his maths teacher is doing all day.

##### Share on other sites

when i have to calculate the area of a circle without a calculator, I usually leave it in terms of pi.

pi*r^2=A

r=2

2^2=4

A=4pi

Ask whoever is telling you to do this if they want an approximate or exact value. If they want exact you must leave it in terms of pi. However, if you are just doing this on your own, pick one.

##### Share on other sites

I don't think I've used a calculator to work out areas of things approximately for quite some time now. Leaving it in terms of pi is a lot better.

##### Share on other sites

He could of corse leave it with Pi without turning it into numbers. This can be useful where there is a more complex question with more than 1 circle. I remember doing quesitions where pi canceled out, this is the most likely solution if it crops up on a non calculator paper.

##### Share on other sites

Hey there,

I recently did found myself wanting to work out the area of a circle by hand, and here was what I came up with...

Tools:

A compass, a pen/pencil/stylus, a straightedge, paper or otherwise neat drawing surface, and oodles of free time which may or may not drive you into a crushing torpor.

Method:

Take the compass and make for yourself a nice circle of some arbitrary radius (make it big enough to really see everything -- about hand-sized). Be sure to fix this radius in the compass (perhaps you've a nice one with a screw-lock or something else handy). Do not swallow your compass.

Now taking your compass, carefully stick the pointy end on an arbitrary point somewhere on the circle (perhaps your favorite spot?). Trace-out your second circle here and be sure to note how it should cross directly over the center of the first circle you made. It also should intersect the first circle at exactly two points. Choose your favorite of these two and stick the center of a third circle (same radius you've been using again) on that point. It should intersect at two points on the original circle, and you can just pick up your compass and move around that way again. After doing this carefully for a while, you'll notice that you've made a total of five lovely circles--one mother circle in the middle and four clones radiating around her.

Take a step back and examine what you've done. You should definitely notice how these four circles intersect each other outside and inside the mother circle in two places for every pair of circles. There should be four intersections outside the momma circle, and four inside (all at the exact center of the momma). Connect these intersection points by making them straight through the center (normal to the mother circle).

At this time, you should have exactly five circles and two straight lines drawn (via-straightedge). These lines should be perpendicular to each other and make a cross.

If you will notice, these lines make new intersection points on the circle....you may continue to divide the circle this way (by powers of two) if you repeat the above method by placing your compass on these new points and trace around the momma circle again until all new points on the circle have become home to a new circle center (same radii of course).

Now, while that is interesting certainly, it is only a mere step (necessary of course) in the right direction to answering your original question..."how do i work out the area of a circle without a calculator?"

So here we go....

Please notice your original two lines making the cross intersect the mother circle in four places that are perfectly (at least in your head they are perfect...because this is all an exercise to emphasize the validity of mathematical ideas here) and evenly spaced around the circular perimeter/circumference. Please draw a square inside this circle (a circumscription, you might say) with vertices at those four points.

Now, I hope you've made your circle big enough to really see all of what comes next.

Focus your attention on your favorite quadrant of the circle....I'll use the second just because I'm weird like that. You should see a very nice 45-45-90 triangle there...one fourth of the area of the square (which happens to be the biggest square that can fit).

The legs of this triangle are each a magnitude R of the radius you randomly picked, making the hypotenuse a line of magnitude [(2RR)^(1/2)]. This simplifies to R times the square root of 2.

Now notice the line you drew before (unless you stopped after making the cross...if so, keep dividing the circle's angles in half as described in the beginning) cuts this triangle in half, bisecting the hypotenuse and running through the center of the momma circle. That line should have intersected the mother circle 45 degrees above the quadrant you're looking at....in other words, it should have split the cross's 90 degree angle in half there. Draw a new line from the point where that splitting line intersects the mother circle to one of the two vertices of the 45-45-90 triangle that is NOT at the center of the mother circle.

This should make for a new right triangle...one that has a long leg and a very short leg. You will notice that you could repeat this for all the quadrants and you'd have made for yourself a nice little octagon. By continuing to divide the circle's angles in half and making these new little triangles on top of the hypotenuses of the old right triangles, you can successfully construct polygons with powers of two as their side number (square-->2^2 : octogon-->2^3 : 16-sides-->2^4 etc.).

You know the area of a square is base times height.....in this case R squared divided by two gets our triangle, and 4 times that for the entire polygon.

Area =(( (2^2)(R^2) )/ 2) + ...next polygons...

The base of the triangle sitting on top of the triangle that is part of the big square is just half of the old hypotenuse we just found a second ago....R times the square root of 2.

So side B for this triangle is R times the square root of 2 all divided by 2.

We'll refer to that as B.

The other leg of this new triangle, leg A, is at present unstated. Notice, however, that there is another right triangle that has half the area in the original 45-45-90 triangle. It is kinda flipped with it's hypotenuse as the radius of the circle. Well, we already know that it's hypotenuse is R, and one of it's sides is just half of the 45-45-90's hypotenuse...this is side B of the new triangle, too. Now we can find the other leg (actually it is obvious in this special case....but a couple polygons down the line, it would not have been so obvious because everything becomes a radical mess!). The other leg is the square root of all of R squared minus B squared. Now, this is the missing leg of that triangle, but as for the smallest leg of the new triangle--the one whose hypotenuse and side A are unknown--we can say that R minus the square root of all of R squared minus B squared is that side length....side A for the new triangle (the one piggy-backing on the 45-45-90).

With this knowledge, we can find the hypotenuse of that triangle, and simply multiply A and B and divide by 2 to get the area of the triangle. Multiply by the next power of two (with respect to whatever polygon that hypotenuse is a part of) and you've an area that is even closer to that of a circle...

Area ==( (2)(R^2) ) + ( (2)(R^2)(-1+sqrt(2)) + ...next polygons...

You can repeat this method to get closer and closer to the area of a circle, and if you assume R is one, closer and closer to a hand estimation of the elusive Pi constant.

Hope this helped!

##### Share on other sites
2pi = 3.142 x 2 thus the 1st bit is 6.284

whats the Radius? lets assume its 9cms so whats 9 Squared? thats 81

81 x 6.284 = 509 cms

Hmmm... I think I just worked out a Sphere?

nevermind' date=' I said my maths was crap! [/quote']

You appear to have magically input a '2' into the equation. A = pi r^2, not 2 pi r ^2.

The volume of a sphere is 4/3 pi r^3, the surface area is 4 pi r ^2. 