Koundinya-Mathematics Teachers Hub: 2012

Monday, 19 November 2012

Monday, 3 September 2012

Wednesday, 29 August 2012

Monday, 27 August 2012

we know that integration by parts is done using

${\color{black} \int d(uv)=uv=\int udv+\int vdu}$

Now,

Let u = x and v = 1/x; then we get,

${\color{black} 1=\int x.(\frac{-1}{x^{2}})dx+\int \frac{1}{x}dx}$

${\color{Yellow}\Rightarrow 1=-\int \frac{1}{x}dx+\int \frac{1}{x}dx}$
${\color{black}\Rightarrow 1=0}$

Can you explain this???

Monday, 13 August 2012

Number of regions...

A circle and a parabola are drawn on a sheet of paper. Find the maximum number of regions they divide the paper into.

I can see 7...can you work out and improvise?

Area of octagon...

ABCD is a square with length of a side 1 cm. An octagon is formed by lines joining the vertices of the square to the midpoints of opposite sides. Find the area of the octagon.

Answer appears to be 1/6 sq cm... try to arrive at it

Sunday, 12 August 2012

MATHS@EASE- 4

MATHS@EASE - 3

Friday, 27 July 2012

MATH@EASE - 2

MATH@EASE-1

https://docs.google.com/open?id=0B5wIYAVf3IFaWUYwNVpSOEZkQnM

Saturday, 17 March 2012

Area of triangle 2...

ABC is a right angled triangle with orthocentre at C. Given the circumradius as 'R' and the angle between the medians not passing through C be Ɵ. Find the area of triangle

Solution:
The Area of the triangle is
${\color{black} \frac{4R^{2}tan\theta }{3}}$

Area of Triangle...

Wednesday, 14 March 2012

Here follows a problem which troubles (possibly)..

FIND THE SHORTEST LINE SEGMENT WITH ITS ENDS ON TWO SIDES OF A GIVEN TRIANGLE WHICH BISECTS THE TRIANGLE

i have a solution to this problem i will post the same however i wish someone comes out with a solution

... it's been days and no reply has come...so, here is the solution

consider the figure below

Let PQ be the shortest line segment such that Area of ΔCPQ is equal to Area of the Quadrilateral PQBA.

Let CP = x , CQ = y , PQ = u.

QM and BE are perpendiculars from Q and B on to CA, we can see that ΔCQM ||| ΔCBE, Then
${\color{black} \frac{QM}{CQ}=\frac{BE}{BC}=sin C}$

Hence we will have QM = ysin C and BE = a sin C

${\color{Yellow}\therefore 2\frac{QM X CP}{2}=\frac{2\mathit{xy}sin C}{2}=\frac{CA X BE}{2} = \frac{absin C}{2}}$

PQ bisects ΔABC implies

${\color{black} xy\ {} =\ {} \frac{ab}{2}\ {}or \ {} 2xy \ {}=\ {}ab\ {}\Rightarrow \ {} y\ {}=\ {}\frac{ab}{2x} }$

Using the Cosine formula for triangle CPQ

${\color{Yellow} \mathit{u}^{2}=PQ^{2}=CP^{2}+CQ^{2}-2.CP.CQ.cosC }$

${\color{black} =\ {}\mathit{x}^{2}+\mathit{y}^{2}-2\mathit{xy}cosC\ {}= \ {}\mathit{x}^{2}+\frac{a^{2}b^{2}}{4\mathit{x}^{2}}-abcosC}$

Let us assume this is equal to r. then

${\color{black} r\mathit{x}^{2}\ {}=\ {}\mathit{x}^{4}-(abcosC)\mathit{x}^{2}+\frac{a^{2}b^{2}}{4}}$

Upon simplification , we get
${\color{Yellow} (\mathit{x}^{2}-\frac{(abcosC)+r}{2})^{2}+\frac{a^{2}b^{2}}{4}\ {} =\ {}\frac{(abcosC+r)^{2}}{4} }$

Observation : Minimum value of R.H.S is reached when the first etrm of L.H.S is 0. ie.,
${\color{black} (\mathit{x}^{2}=\frac{(abcosC)+r}{2})^{2} }$

This implies that r = ab(1-cosC) and consequently we get

${\color{Yellow} \mathit{x} = \sqrt{\frac{ab}{2}}\ {}\Rightarrow \ {}\mathit{y}=\sqrt{\frac{ab}{2}} }$

Therefore length of PQ can be found as
${\color{Yellow} =\ {} \sqrt{\frac{(c-a+b)(c+a-b)}{2}} }$

Generalisation:

Find a+b-c, b+c-a, c+a-b, choose the least and next higher of them; for example if a+b-c and b+c-a are these two in that order, then P and Q must be chosen on BC and BA such that
${\color{black} BP,BQ\ {}=\ {} \sqrt{\frac{ca}{2}} }$
and the shortest length PQ that bisects triangle abc will then be
${\color{black} PQ=\ {} \sqrt{\frac{(a+b-c)(b+c-a)}{2}} }$

S Ramanujan problem--for school kids

This is a problem posed by S Ramanujan .. try ... ${\color{blue} 3=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{1+7\sqrt{.........}}}}}}}}$

Here is the solution

${\color{blue} 3=\sqrt{1+8}}$

${\color{blue} \Rightarrow 3=\sqrt{1+2\sqrt{16}}}$

${\color{blue} \Rightarrow 3=\sqrt{1+2\sqrt{1+15}}}$

${\color{blue} \Rightarrow 3=\sqrt{1+2\sqrt{1+3\sqrt{25}}}}$

so on.. we get

${\color{blue} 3=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{1+7\sqrt{.........}}}}}}}}$

Tuesday, 10 January 2012

problem on probability

Four identical dice are rolled once, find the probability of getting atleast three different numbers.

Koundinya-Mathematics Teachers Hub