Semicircles

As shown in the diagram, we start by joining the centres of the two semi circles. Let the radius of small circle be r cm. Then,we construct a right angled triangle with the following measures:

1. hypotenuse = (r+18)cm
2. side 1= (36-r) cm
3. side 2 =18 cm

then using Pythagoras theorem:

(r+18)^2 = (36-r)^2 + (18)^2

r^2 + 2(18)r +18^2 = [ (36)^2 -2(36)r +r^2] +(18)^2

r^2 + 36r +18^2 = (36)^2 - 72r +r^2 +18^2

cancelling like terms,

36r = 36^2 - 72r

108r = 36 X 36

r = 12

The radius of the large semicircle, $R$ , is equal to $D/2=18$ . Let the radius of the small semicircle be denoted by $r$ .

The radii of both semicircles would meet at the point of intersection of the semicircles and form a straight angle. This line segment will form the hypotenuse of our right triangle, and is equal to $18+r$ .

Using the Pythagorean theorem, we construct our right triangle in the bottom left-hand corner: $(36-r)^2+18^2=(18+r)^2$ .

Expansion yields: $324+1296-72r+r^2=324+36r+r^2$ .

Simplifying and isolating for $r$ reveals that $r=12$ .

It's a quite simple problem.

It can be done with the help of the Pythagoras theorem.

Let the radius of smaller circle be x.

Let the centre of the yellow circle be O and the centre of the blue circle be O'.

Let the left corner of the of the square in which the blue circle is touching be A.

Therefore OAO' is a right triangle where

OA= 36-x

AO'= 18(radius of the blue circle)

OO' = x + 18 ( it will be a straight line since the circles are tangent to each other. The radius x and the radius 18 will both be perpendicular to the tangent and therefore form a straight line.)

Now by just applying the Pythagoras theorem.

(36-x)^2 + 18^2 = (x+18)^2

And x comes out to be 12.

I made an image so that u could understand what I am trying to say better. But it wasn't uploading. So I hope u found this useful.

Put the origin at the bottom left hand of the box. Then the equation of the blue circle is $(x-18)^{2}+y^{2}=18$ and the equation of the yellow circle is $x^{2}+(y-36+r)^{2} = r^{2}$ where r is the radius of the yellow circle. Next we can solve for the value of r that yields only 1 solution (since the circles are tangent); if we restrict 0<r<36, then this value is unique, with all other values of r yielding 0 or 2 solutions. (r=36 also yields a tangency)

solution loade d

I only guesstimated. Small circle covers 2/3 of the side. The side is 36 cm long. Divide by 3 and get 12.the circle length/diameter is 24, so half of that is 12 cm.

Let A is the centre of the yellow circle, B is the centre of the blue circle, C is the point of tangency and D is the point at the bottom left of the square. => A, B, C are aligned. Let x is the radius of the yellow circle. => AD²+BD²=AB²=(AC+BC)² => (36-x)²+18²=(x+18)² => x²-72x+36²+18²=x²+36x+18² => 108x=1296 => x=12cm => The area of the yellow semi-circle is S=½πx²=72π (cm²).