[bhars18]

Originally Posted by arcade79

hey there seems to b some confusion 2 me.. u said line perpendicular and not perpendicular bisector... so the prob started..

oops... I didn't look into that... sorry. I meant the bisector but missed the term somehow.

and what i said is that there can b 2 perpendiculars on a chord with same length..

totally agreed

i tried this with a circle of radius 2.5 cm and chord of 2 cm.. check that out.. u will find that perpendicular bisector of chord wud b gr8r than 2 cm...

I am unable to get what you are aiming at.

Regards,
Bharathi

[arcade79]

Now try this... I m sorry for getting off the track but cudnt resist posting this...


Ques: What is that which starts with "E", ends with "E" but usually contains only a letter.

Happy gray cell burning

arcade

[bhars18]

Envelope !!!

Most times, contains a letter

That was really nice one

Regards,
Bharathi.

[arcade79]

Incredible.... !!

That was simply supeb.. believe me u r the first one whom i asked this and got the answer... CORRECT.....

Well were u knowing this b4hand... :wink: :

arcade

[bhars18]

I knew a similar one...

An eight letter word carrying all the letters !!! So no problems in guessing

You had given the clue of the E's.

Regards,
Bharathi

[bhars18]

Let ABC be an equilateral triangle and D be a point outside it. The distances AD, BD and CD are 10, 10 and 18 units respectively. Find the side of the equi. triangle.

Regards,
Bharathi.

[arcade79]

Is this the answer?

Consider a triangle ABC

let D be a point such that AD=BD=10 and CD=18.
Let CD cut AB at O.
Now focus on Triangles AOC and BOC.. It shouldnt be difficult to prove that they both are congruent triangles....
Thus angle AOC=angle BOC=90 degrees

Let CO=x and OD=18-x. Let AO=y
As angle BAC=60 deg hence CO=AO tan60 ie x=y.sqrt(3)

In traingle AOD, AD^2= OD^2+ AO^2

or 100 = (18-x)^2 + y^2
Replace x by y.sqrt(3)

Thus y =9.96 cm and hence AC = AO cos 60 = 19.92 or 20 cm

Also y can b 5.6 cm then AC = 11.2 cm...

Regards

Arcade

[bhars18]

Originally Posted by arcade79

Is this the answer?

Hats off !!! You're right !

My approach was a bit different... Would like to share with you

Originally Posted by arcade79

Consider a triangle ABC
let D be a point such that AD=BD=10 and CD=18.
Let CD cut AB at O.
Now focus on Triangles AOC and BOC.. It shouldnt be difficult to prove that they both are congruent triangles....
Thus angle AOC=angle BOC=90 degrees

Precisely... that was it!

Now, CO is the altitude of triangle ABC and OD is the median ( and altitude ) of the triangle ABD.

If a is the side of the equilateral triangle, then
CO = alt = [sqrt(3)/2] * a
OD = 18 - alt
In triangle, ABD,

AD^2 + BD^2 = 2 (OD^2 + OB^2)

OB=a/2 (since equilateral)

This gives the solution a=9*sqrt(3) (+/-) sqrt(19) which comes to 19.9 / 11.2

Regards,
Bharathi.

[arcade79]

I wont consider that to b a different approach...

see u wrote altitude to b (sqrt3/2)*a.. but thats nothg else but sin 60*a.. so i started with tan 60.. u with sin 60.. so its just which ratio one chose...

thus the prob soln goes almost on the parallel tracks...

waiting for more.....

arcade

[arcade79]

Okie now try this...

This is in line with the questions being asked in XLRI... I luv these questions...

For the question below, there is a series of numbers that follows some definite order. Following the series are five sets of two numbers each. Decide which of the sets would be the next two numbers of the series.

7 10 8 13 16 8 19 ...

(A) 22 8
(B) 8 22
(C) 20 21
(D) 22 25
(E) 8 25

Regards

Arcade