Posted by: SATYASRINIVAS | June 4, 2007


Apollonius theorem ; In a triangle ABC , if AD
is the median to the side BC , then
AB^2 + AC^2 = 2(AD^2 + BD^2) or 2(AD^2 + DC^2) .
Note BD=DC=BC/2.


Let the sides be A,B,C.

Semiperimeter = (A+B+C)/2

Area =(S*(S-A)*(S-B)*(S-C))^0.5

Inradius = Area/ Semiperimeter

Circumradius= A.B.C/ 4.Area


if n is number of sides.
For any regular polynomial, each angle =((n-2)*180)/n and
Total sum of the angles = (n-2)*180 or Each angle* n


Wrong Weight*(100 + Profit %)/100 = Correct Weight


1. 1st January 0001 and every 400 years after that is a Monday.
That is 1st Jan of year 0001, 401, 801,1201,1601,2001 is
2. An year divisible by 4 is a leap year, but century years are not
leap years unless they are divisible by 400. So 2000 is a Leap
Year but 1900, 1800, 1700 are not not leap year.
3. Odd days is the remainder obtained when the number of days
is divided by 7. Example: If it is Sunday today , after 50 days it will
be: 50/7 gives remained 1. Add 1 day to Sunday to get answer
as Monday.
4. A non leap year has 1 odd day and a leap year has 2 odd
5. A normal Century has 5 odd days and leap century has 6 odd
6. January, March, May, July, Aug, Oct and December have 3 odd
days each. April, June, Sept and Nov have 2 odd days each. Feb
has 0 odd days if it is not a leap year and has 1 odd day if it is a
leap year.
6. Number of leap years between any two given years is equal to
the quotient when the difference between the two given years is
divided by 4. Example Number of Leap years between 1947 to
1901 = 1947-1901= 46. 46 /4 = 11.5. So number of leap years=11.


What was the day on 15 Aug 1947 ?

Odd year for 1601-1900 : 5+5+5=15 or 1 odd day
1947-1901= 46. No of odd days =46 or 4
Number of leap years = 46/4=11. No. of odd days =11 or 4
Odd days till July end : 3+0+3+2+3+2+3= 16 or 2
Day is August = 15 or 1
Total Odd days =1+4+4+2+1= 12 or 5 odd days.
So 15th Aug 1947 was Friday.


Arithmetic series

Sn= n(2a +(n-1)d) / 2
Tn= a + (n-1)d


geometric progression


Sn= a[ (1-r^n)/ (1-r) ] where r<1

Sn= a[ (r^n-1)/ (r-1) ] where r>1

S(infinity)= a/ (1-r) when r<1


Right angle triangle:The square of the shortest side is less than the product of the
sum and the difference of the longest and the third side.


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