Problem 17 · 2012 Math Kangaroo
Hard
Geometry & Measurement
areapythagorean-triple
The diagram shows a right-angled triangle with side lengths 5, 12 and 13. What is the length of the radius of the inscribed semi-circle?
Show answer
Answer: B — \(\frac{10}{3}\)
Show hints
Hint 1 of 2
The semicircle's flat side lies on the leg 12 and it touches the slanted side 13.
Still stuck? Show hint 2 →
Hint 2 of 2
Its centre is r from both that leg-line and the hypotenuse; set those equal.
Show solution
Approach: centre equidistant from the two sides it touches
- Place the right angle at the origin with legs 12 (along the x-axis) and 5 (along the y-axis); the hypotenuse line is 5x + 12y = 60.
- The semicircle's centre lies at (r, 0): it is r from the y-axis leg and its distance to the hypotenuse, |5r − 60|/13, must also equal r.
- Solve 60 − 5r = 13r, giving 18r = 60, so r = 10/3 (B).
Mark:
· log in to save