Three circles of equal radius are drawn tangential to each other with their centres at the vertices of an equi

Question

Three circles of equal radius are drawn tangential to each other with their centres at the vertices of an equilateral triangleABC.Another equilateral trianglePQR is drawn such that each of its side is tangent to two of the three circles,then he side of trianglePQR is
(a) sqrt(3)/2 times radius of three circles
(b) (1+sqrt(3)) times side of triangle ABC
(c) 5 times radius of three circles
(d) None of these

Answer 1

B

distance BC is equal to 2r

Let S and T be the points on line QR corresponding to perpendiculars drawn from B and C respectively.

ST = 2r

r/QB = sin 30

QB = 2r

QS/QB = cos 30
QS = 1/2 x QB x sqrt(3)

QS = r x sqrt(3)

QR = length of side = 2r(sqrt(3)+1) therefore answer b

Answer 2

The answer is b. The part between where the sides of PQR touches the two circles is as long as a side of ABC, and the parts that stick out on both sides can be shown to be sqrt(3)/2 times a side of ABC by some trigonometry. There are 2 of these pieces so we get 1+2(sqrt(3)/2)=1+sqrt(3) times a side of ABC

Answer 3

for a complete circle the section is pi x radius squared. The determine you have (thoroughly) defined has 3 sectors of their circles that are a million / 6 of this, or a million/2 a circle minus the component to the triangle. This section is comparable to a million/2 the backside situations the top. the backside equals the radius however the top is sin 60 stages of this or (root 3) / 2. consequently component to triangle is the radius squared x (root 3) / 4 the section you desire is: radius squared x ( pi - (root 3) / 2 ) / 2

Answer 4

B, wont explain why but heres this...
http://img226.imageshack.us/img226/2765/maththingkl1.jpg

EDIT:also, you gotta look at this question, i knew it was B before i thought about it because why else would you need to draw triangle ABC(side=diameter of circle) if its not in the answer