Three identical circles are inscribed within an equilateral

I guessed this by using an approximate value of each side of the triangle...If u see the figure then it can be visualized that each side of the equilateral triangle have a length slightly greater than the sum of the 4 radius=4*3 =12. Therefore perimeter must be greater than 12+12+12=36 approx...

which is choice A.

Just.... How do you prove that angle BAE (with E the center of the circle on the left hand side) is equal to 30 degree ???

This approach would give 100% accurate answer if only one option were more than 36, but in our case both A and B are more than 36. So chances are 50/50 to pick the right answer. Still good way of thinking for educated guess.


The big triangle is equilateral, so the angels are 60 degrees each and BAE is half of it, so 30 degrees.

I reformulate : How do you know BAE is half ?

How else? Look at the attachment below: Right triangles ACE and ABE are congruent (they share the same hypotenuse, CE=BE=radius, hence the third sides are also equal). So <CAE=<BAE=<CAB/2=30.

Hope it's clear.

Of course' ! Thx... Kudos !!!

I have a more straightforward way to solve it

If we take the three centers of the inner circles we can draw an isoceles triangle with a total perimeter of 27 (9 X 3).

There is a relationship between increase in in the side of an isoceles triangle and the perimeter of it. In that case we are increasing the above mentioned triangle in three. as it is the length of the ratio the inner circles. In that case we will increase the perimeter from 27 to (9 + 3 ) = 12 X 3 = 36. The only answer that has an aproximate value to 36 is A.

Please comment.

I dont think I understand what you just did there.
The triangle formed by joining the centers, is not isoceles, it is equilateral and also its perimeter is 6*3=18 and not 27. And finally not sure where you get that 36 from. The answers above are the exact answers not approximations

Also I'd add that the answer choices A and B are both more than 36: \(A\approx{49}\) and \(B\approx{40}\).

As for 36, we can think in the following way: the side of the triangle must be more than \(4r=12\), so the perimeter must be more than \(3*12=36\) (see utin's post above). This approach would give 100% accurate answer if only one option were more than 36, but in our case both A and B are more than 36. So chances are 50/50 to pick the right answer. Still good way of thinking for educated guess.

Hope it's clear.

Makes sense. Educated guess can be really helpful if one has no idea of how to proceed with the problem.

In this problem, by guessing one is looking for something which is 36+

A is about 49
B is about 40

Most of the ppl guessing this one will end up with B .......... it helps to know the correct way to solve question

Thanks Bunuel.

Nice explanation Bunuel. Honestly, I could'nt find an easy way so I just ballparked.

So we have that each side is more than 2(2r) >12 so three sides >36. The only answer choice that makes sense is A.

Hope it helps
Cheers!
J

This question can be solved in 3 easy step
1. calculate the area of 3 inscribed circle
2. calculate the approximate side of equilateral triangle from answers and compute area of equilateral triangle
3. compare the answer correctly.

Solution
Area of 3 inscribed circle = 3*3.14*9 =84. So area of triangle must be greater than 84 as circle are inscribed in triangle

a. 18(1+√3) =>6(2.7) =>16 A=(256*√3)/4=64√3 (approximately greater than 100 ) = 108
b. 9(1+2√3) =>3(4.46) => 13.3 =13 A=(169*√3)/4= 42√3(approximately 70) =74
c. 18√3 => 6√3 => 10.3 =10 . No need of calculation
d. 9(1+√3) => 3(2.7) => 8 .No need of calculation
e. 6(1+√3) => 2(2.7) => 6 .No need of calculation

SO answer is A.

They are asking perimeter of the big triangle. Would this method still hold true. ?

great explanation as always, thanks Bunuel

picked B, but purely guess.
I could eliminate C, D, and E.
Suppose we draw a diagonal for each circle so that to form 3 equilateral triangles with sides 6
now we have 6 equal sides of these small equilateral triangles, which is 36. But it has to be slightly more than 36, since we have 3 equal parts that are not covered between these 3 small triangles.

C,D,and E are <36. Between A and B...picked B.

bunnuel's explanation is good...have to analyze it more...

bumped into it another time...still had difficulties in understanding the process to solve it...nevertheless, I could eliminate C, D, and E, and after some more thinking, B as well..

so we have 3 circles, diameter of each is 6. thus, the length of each side should be more than 12. now we have 12+ * 3 = 36+.

A. 18(1+√3) = 18+18√3 = 18+slightly less than 36, or overall slightly less than 54. of these two, picked A.
B. 9(1+2√3) = 9+ 18√3, so 9+slightly less than 36, or overall -> slightly less than 45.
C. 18√3 = less than 36, so out.
D. 9(1+√3) = 9+9√3 = 8+18- => slightly less than 26, so out.
E. 6(1+√3) = 6 + 6√3 = that's slightly less than 18, so out.

(1st)From the Center of the Bottom 2 Circles:

Draw a Line Connecting the 2 Centers = 2 * (r) = 2 * (3) = 6


(2nd)Concept: Radius draw to a Tangent Line will create a 90 Degree Angle at the Point of Tangency

Drop a Radius from Each of the 2 Bottoms Circles' Centers ----> to the Bottom Side of the Equilateral Triangle

Now have a Rectangle with:

Length = 6 across the Bottom of the Triangle
&
Width = 3 = radius up to the Center of Each Circle


(3rd) Drawn an Angle Bisector from the Center of Each Circle to Each of the 2 Bottom Vertices of the Triangle


This creates 2 Congruent Triangles that are:

30-60-90 Degree Triangles

and

Side Opposite the 30 Degree Angle = radius of 3

and

Side Opposite the 60 Degree Angle = 1/2 of the Remaining Portion of the Bottom Side of the Equilateral Triangle =

3 * sqrt(3)



(4th)Calculate Perimeter of Equilateral Triangle


The Bottom Side of the Equilateral Triangle can now be found:

3 * sqrt(3) + 6 + 3 * sqrt(3) =

6 + 6*sqrt(3)


Perimeter = 3 * (Side Length) = 3 * [ 6 + 6*sqrt(3) ] =

18 + 18*sqrt(3)


-A-


EDIT: Should have known....just see Bunuel's Explanation lol