IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

Mr.IITIAN007

An equilateral triangle of side 'a' units is inscribed in a circle of radius R .A new circle is inscribed in the triangle .Again a new equilateral triangle is inscribed in the newly inscribed circle.This goes on for n times.
q1.Then the sum of the areas of all circles is :----------
(a) 3

R² { 1- ( 1/3ⁿ) }
(b) 2

R² { 1- (1/ 2ⁿ) }
(c) n

R²{ 1-( 1/ a³) }
(d) None of these

q2.Limit of sum of all the areas of all circles as n tends to infinity, is-----------
(a) 4

R²
(b) 2

R²
(c)

R^a
(d) none of these

q3.Sum of areas of all triangles is :-----------------------
(a) 8nR²
(b) 2nR²
(c) a Rⁿ
(d) none of these

q4.Limit of sum of all the areas of the triangles as n tends to infinity ,is ------------
(a) 2R²
(b) 4R²
(C) 3R^{2
(d) none of these}

Mr.IITIAN007

If something is not visible about the problem then that something is 'pie'.

govindsuku

the ans to the first ......tht i am gettin is pie R^2( 1 - (1/4)^n)/9

akella.as

the answe to first question is option (b)

akella.as

q2 answer 2*22/7*R*R
FOR Q3 AND Q4 answe is option d

akella.as

are my answers correct

cooldude

circle is circumcentre of the equilateral triangle.hence R=a/root3.next one is incentre so it is a/2root3...........
sum of areas of circles is pi{(a square /3)+(a square /12)+........}
it is sum of a GP and we get 4/3 pi R squared[1-(1/4)^n].
as n tends to infinity it becomes4/3 pi R squared.
sum of triangles,(root3)/4Xa squared(1+1/4+1/16.........)
we get,root3XR squared[1-(1/4)^n]
limit is root 3XR squared.hence answer for all is d.
i may be wrong because none of my answers are matching with anything other than none of these.correct me if i am wrong but do rate me if i am correct.

iitianwannabe

cooldude...im getting exactly wht ur getting...
d) in all
mriitian...is this answer right??

Moderator

Please ask one question at a time for the experts to answer.

~ moderator

Mr.IITIAN007

Hey akella,cooldude,govindsuku and iitianwannabe I don't know the correct answers but I agree with your logics especially iitianwannabe.That is why I posted the questions here .MODERATOR this is a single question which has many parts.Can't you see...................................!!!!!!!!!!!!!!!!!

fused_bulb

C , THIS IS A COMPETITIVE WORLD

U DONT NEED TO WORK SO HARD FOR SUCH PROBLEMS .

JUST PUT N=1 IN FIRST Q. TO GET ANSWER AS ( b ) ( obviously pie R^2 ).

apply limits to this part to get the second answer .

similarly for the triangles .

iitkgp_bipin

1. Simply apply geometry , you will find that the radius of circles are R , R/2 , R/4 , .......

Hence sum of areas of circles = pi.R² + pi.(R/2)² + pi.(R/4)² + ....... n terms

= pi.R²{1 - (1/2)ⁿ}/(1 - 1/2) = 2pi.R²{1 - 1/2ⁿ}

2. As n tends to infinity , this area tends to 2pi.R².

3. Sides of triangle also decrease in same order as that of radius of circles.

a , a/2 , a/4 , .............

Sum of areas = (3^1/2/2)a²(1 - 1/2ⁿ)

4. As n tends to infinity , this area tends to (3^1/2/2)a².

Mr.IITIAN007 , I would request you to be polite with Moderator. She has been working hard for the fruitful running of the forum.

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