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.

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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

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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

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

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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

Monday.

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

days.

5. A normal Century has 5 odd days and leap century has 6 odd

days.

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.

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Arithmetic series

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

Tn= a + (n-1)d

A.M*HM=GM^2

geometric progression

Tn=ar^(n-1)

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

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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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