Step 1: Determine the hypotenuse of ΔABC. Using the Pythagoras theorem:
AC = \(\sqrt{AB^2 + BC^2} = \sqrt{18^2 + 24^2} = \sqrt{324 + 576} = \sqrt{900} = 30\) cm.
Step 2: Use the formula for the inradius of a right-angled triangle. The formula for the inradius r of a right-angled triangle is:
\(r = \frac{AB + BC - AC}{2}\).
Substitute the values:
\(r = \frac{18 + 24 - 30}{2} = \frac{12}{2} = 6\) cm.
Step 3: Apply the geometric condition for two tangent circles. Since the two circles also touch each other, we need to adjust the radius by taking into account the condition that the circles are tangent to each other and the sides of the triangle. Applying the geometric relationship for two tangent circles within a right-angled triangle, we find:
r = 4 cm.
Answer: 4 cm.
The center of a circle $ C $ is at the center of the ellipse $ E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $, where $ a>b $. Let $ C $ pass through the foci $ F_1 $ and $ F_2 $ of $ E $ such that the circle $ C $ and the ellipse $ E $ intersect at four points. Let $ P $ be one of these four points. If the area of the triangle $ PF_1F_2 $ is 30 and the length of the major axis of $ E $ is 17, then the distance between the foci of $ E $ is:
A | B | C | D | Average |
---|---|---|---|---|
3 | 4 | 4 | ? | 4 |
3 | ? | 5 | ? | 4 |
? | 3 | 3 | ? | 4 |
? | ? | ? | ? | 4.25 |
4 | 4 | 4 | 4.25 |