This is the solution to the problem:
To solve the problem above, you have to find the radius of each of the circles. To find the radius of the larger circle inside the triangle...
AC=16 as shown above.
Therefore AD=16-X
AF=AD
AD=16-X
CB=12
Therefore EB=12-X
EB=BD
BD=12-X
BD+AD=AB=20
12-X+16-X=20
28-2X=20
2X=8
X=4
The radius of the larger circle is 4 units. So the area of it is 4x4xpi= 16pi
To find the radius of the smaller circle:
To find the radius of the smaller circle, tangent to the triangle, you have to form a sector.
The radius of the sector is 10, as the diameter (BC) is 20.
AB=BC=10 (The radius of the whole semi-circle)
AC=16, therefore EC=8
Sides EC and BC can form the sides of a right angled triangle, along with BE.
BE^2+EC^2=BC^2
BE^2+64=100
BE^2=36
BE=6
BF=10 as it is the radius on the whole sector.
BF-BE=EF (diameter of the small circle)
10-6=EF
EF=4
Therefore, if the diameter is 4, the radius must be 2.
If the radius is 2, Area=4pi
The question asks for the ratio of the two circles areas, not just the area of each.
Ratio = Area of the larger circle/Area of the smaller circle
= 16pi/4pi
=4
So the larger circle inside the triangle, is 4 times bigger than the circle tangent to both AC and the semi-circle.
