# 12: Circles & Triangles

Problem 1: Find the radius of the 3 circles in the diagram below

We would attempt to solve the adjoining problem

A few words about interpreting a problem like this.
The diagram shows a circle and an equilateral triangle inside it. Then one more circle is drawn inside it and one external circle too. We need to know in advance that if you drop a perpendicular from the vertex to the base in an equilateral triangle, its length is aSin 60 where the length of the side of the triangle is a. Further, it must be known that that vertical line contains what is known as the Centroid of the triangle and for an equilateral triangle, the centroid is also the centre of the circumcircle as well as the centre of the incircle. We must also know that the Centroid divides the vertical line in the ratio of 2:1. Armed with these three pieces of information, we proceed to solve the problem.

First we  find the length of the two radii, R and r, of the Circumcircle and the Incircle of the Equilateral Triangle shown
 The diagram on the right explains the computation, based on what we mentioned above.   ​Then look below to see the solution for finding out the radius of the third circle.

Problem no. 1A: We add one more circle at the corner of the triangle adjoining the incircle. Find its radius.
 What we always look out for is a right angled triangle inside our problem. Then we can use Sin x, Cos x, Tan x and we can use Pythagoras theorem. Here, we have culled out the Right Angled Triangle at the lower right corner of the diagram and have redrawn it. The radius of the smaller circle is called x and the fact that the radius falling on a tangent has to be perpendicular to the tangent has been used in the dotted rectangle drawn in the second part of the diagram. Then again we have culled out another right angled triangle from the second diagram and this, the third triangle, is where we have used trigonometry to get to our answer. the R and r are known from above and have been shown.  Now, note the method used here. A portion of the original diagram has been redrawn. This is a very important method which you have to get used to and learn.  In this particular problem being an equilateral triangle we know that the angle is 30

Problem no. 2: Two Circles inside a semicircle. Find the radius of the smaller circles if the semicircle has radius R

We have culled out a rectangle from the diagram but look at the dotted line: it is R since it goes from the centre of the semicircle upto its circumference. And this line also contains the r, the radius of the smaller circle, thus the last right angled triangle comes into play which is now very easy for us to compute.

Problem 3: A square inside a circle and another small circle between the square and the outer circle. Find its radius given side of the square as a.

The diagonal is of an isosceles right angled triangle and so the sides of the square being a, the diagonal will be a*sq.rt 2 which is equal to the diameter of the large circle, 2R.
See the vertical drop from the centre to the circumference i.e. R

Problem 4:
​Two circles inside a semicircle. Find their radii given the radius of the semicircle as x

The radius of the large circle, R, inside the semicircle is merely half of the radius of the semicircle i.e. x/2.

To find small circle radius r, we need to cull out a portion of that area which we have shown.

In this diagram R+r and R-r is clear. Please note the line CQ. Since the semicircle and the small circle touch at the point Q, CQ will contain both the centres as well as both the radii, x & r. That is what we have shown and that is where we get our moolah.

We have two equations the first one from the top right angled triangle and the second from the diagonal of the rectangle. Pythagoras theorem is used as in this case we do not know the angles in the triangles and thus cannot use Sin or Cos or Tan.

Problem 5: Find the radius of the small circle at the corner of the diagram having a circle inside a square.

This problem is easier than Problem 4 since we know the angle: 45 degree since it is a square. But still we need to cull out the right angled triangle from the corner and redraw the same to get the answer.

The picture is somewhat hazy but let that be a challenge for the student to decode the answer. Preferably you should solve all these problems yourself.