ABC is a right angled triangle with ∠ABC=90∘ and side lengths AB=24 and BC=7. A semicircle is inscribed in ABC, such that the diameter is on AC and it is tangent to AB and BC. If the radius of the semicircle is an improper fraction of the form a/b, where a and b are positive, coprime integers, what is the value of a+b?

Question

ABC is a right angled triangle with ∠ABC=90∘ and side lengths AB=24 and BC=7. A semicircle is inscribed in ABC, such that the diameter is on AC and it is tangent to AB and BC. If the radius of the semicircle is an improper fraction of the form a/b, where a and b are positive, coprime integers, what is the value of a+b?

Answer 1

To find the radius of the semicircle, we can use the information given about the right-angled triangle ABC and the properties of circles inscribed in right-angled triangles.

First, let's draw a diagram to visualize the triangle and the inscribed semicircle:

A
|\
| \
| \
| \
r|____\ B
̣̣ ̣ ̣ ̣ ̣ ̣ ̣ ̣ ̣ ̣ ̣ ̣ ̣
C |

We are given that AB = 24 and BC = 7.

Since the semicircle is inscribed in triangle ABC, the diameter of the semicircle must be the hypotenuse of the right-angled triangle ABC, which is AC.

Using the Pythagorean theorem, we can find the length of AC:

AC² = AB² + BC²
AC² = 24² + 7²
AC² = 576 + 49
AC² = 625
AC = √625
AC = 25

Now, the diameter of the semicircle is 25. Since the radius of a circle is half the length of its diameter, the radius of the semicircle is:

r = AC/2
r = 25/2
r = 12.5

Therefore, the radius of the semicircle is 12.5, which can be expressed as an improper fraction 25/2. The sum of the numerator and denominator is 25 + 2 = 27.

Hence, the value of a + b is 27.