Geometry Problem: Finding the Radius of the Smaller Circle Inscribed Between a Semicircle and a Right-Angle Triangle

Consider a right-angle triangle ABC with side lengths 3 units, 4 units, and 5 units. A semicircle is drawn on side BC such that it touches both AC and AB. Additionally, there is a smaller circle inscribed between the semicircle and the triangle. The objective of this article is to determine the radius of this smaller circle.

Step 1: Identify the Triangle Dimensions

First, we identify the dimensions of the right-angle triangle ABC as follows:

AB 3 (height) AC 4 (base) BC 5 (hypotenuse)

Step 2: Calculate the Radius of the Semicircle

Since the semicircle is drawn on side BC, its radius R is half the length of BC:

R BC/2 5/2 2.5

Step 3: Calculate the Area of Triangle ABC

The area A of triangle ABC can be computed using the formula:

A 1/2 × base × height 1/2 × 4 × 3 6

Step 4: Calculate the Inradius of the Inscribed Circle of Triangle ABC

The inradius r of a triangle can be calculated using the formula:

r A / s, where s is the semi-perimeter of the triangle:

s AB AC BC/2 3 4 5/2 6

Substituting the values:

r 6 / 6 1

Step 5: Determine the Radius of the Smaller Circle

The smaller circle is inscribed between the semicircle and triangle ABC. The radius r_s of the smaller circle can be found using the difference between the radius of the semicircle and the inradius of the triangle:

r_s R - r 2.5 - 1 1.5

Thus, the radius of the smaller circle is 1.5 units.

Additional Geometry Calculation

The bisector of angle A meets BC at D, dividing BC in the ratio 4:3. Parallel lines DE and DF form similar triangles ADE and BDF to ABC. Using the given ratios and geometric properties, we can determine:

CE/AE 4/3 AE 12/7 DF/AF 12/7 DA √(144/49) 12√2/7 Distance from G to A 120.414/7 0.7097 GJ/DE 1 - 0.707 0.293 GJ 0.502 JI^2 0.502^2 - 0.251^2 0.0623 Radius of the small circle EF √0.0623 0.249 cm

In the second geometry problem, assuming that A is the right angle, the center of the semicircle is on the hypotenuse BC, and the radius of the semicircle is 12/7. Let AC be 4, and the center of the small circle be O. The calculation involves finding the center E of the small circle and its radius r. Using trigonometric properties, we can solve for r to be:

r (24 ± √320) / 21