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# Circle Properties

## Circle Properties

Shortcut tricks on Circle Properties are one of the most important topics in exams. Time is the main factor in competitive exams. If you know how to manage time then you will surely do great in your exam. Most of us skip that part. Few examples on circle shortcuts is given in this page below. These shortcut tricks cover all sorts of tricks on Circle. Visitors please read carefully all shortcut examples. You can understand shortcut tricks on Circle Properties by these examples.

First of all do a practice set on math of any exam. Choose any twenty math problems and write it down on a page. Solve first ten math problems according to basic math formula. You also need to keep track of timing. After finish write down total time taken by you to solve those ten maths. Now read our examples on circle shortcut tricks and practice few questions. After doing this go back to the remaining ten questions and solve those using shortcut methods. Again keep track of timing. The timing will be surely improved this time. But this is not enough. If you need to improve your timing more then you need to practice more.

### Few Important things to Remember

Math section in a competitive exam is the most important part of the exam. It doesn’t mean that other topics are not so important. You can get a good score only if you get a good score in math section. You can get good score only by practicing more and more. All you need to do is to do math problems correctly within time, and only shortcut tricks can give you that success. But it doesn’t mean that you can’t do math problems without using any shortcut tricks. You may do math problems within time without using any shortcut tricks. You may have that potential.

But, so many other people may not do the same. So Circle shortcut tricks here for those people. We always try to put all shortcut methods of the given topic. But it possible we miss any. We appreciate if you share that with us. Your little help will help others.

### Circle Properties

A circle is a simple form and it is a set of series of all points in a plane that is given and its distance make same from the center. In maths exam papers there are two or three question are given from this chapter.

This type of problem are given in Quantitative Aptitude which is a very essential paper in banking exam. Under below given some more example for your better practice.

Circle Properties

radius of a circle is denoted as = r. ( radius is half of diameter )
The Diameter of a circle is denoted as = AB = 2r

• Area of a circle

Area of a circle is = πr2
Where π = 22 / 7, or π = 3.14285………
Or Area of Circle is = 22 / 7 x r x r.

• Circumference of a Circle

Circumference of a Circle is = 2 π r
OR, 2 x 22 / 7 x r (r = radius).

• Semi-Circle:

Area of semi-circle = πr2 / 2 = 1 / 2πr2
Circumference = 2πr / 2 + 2r = r( π + 2 )
Diameter :

. = 2 √(area /π )

• Area = π ( diameter / 2 )2
• Perimeter = π ( diameter ).
• Diameter = ( Perimeter / π  ).

### Example #1 – Circle Properties

The cost of railing around a ground is Rs.7040/- at Rs.16/- per foot. Find the area of that circular ground.

1. 10280 sq.foot
2. 13770 sq.foot
3. 15400 sq.foot
4. 17440 sq.foot

Show Answer Show How to Solve Open Rough Workspace

How to Solve
Let, the radius of a circular ground is r foot.
2πr = 7040 / 16
= 440 foot.

So, r = 440 x 7 / 2 x 22
= 70 foot.

Area of ground = πr2
= 22 x (70)2 / 7
= 15400 sq.foot.

Rough Workspace

### Example #2 – Circle Properties

The radius of a circle is 35 meters. What would be the circumference of that circle?

1. 220 meter
2. 246 meter
3. 281 meter
4. 300 meter

Show Answer Show How to Solve Open Rough Workspace

How to Solve
We applied the formula 2πr.
Take, π = 22 / 7.

Circumference = ( 2 x 22 / 7 ) x 35
= 220 meters.

Rough Workspace

### Example #3 – Circle Properties

The inner circumference of a 15 meter wide race track is 550 meter. What would be the radius of the outer circle?

1. 102.5 meter
2. 110.5 meter
3. 125.8 meter
4. 141.3 meter

Show Answer Show How to Solve Open Rough Workspace

How to Solve
At first we consider the inner radius be r2.
2πr2 = 550
r2 = 550 x 7 / 2 x 22
= 87.5

So, the outer radius, that is r1 = 15 + 87.5
= 102.5 meter.

Rough Workspace

### Different types of Mensuration methods with Shortcut Tricks

1. 