To solve the problem, we need to analyze the two statements provided and verify their correctness step by step.
### Step 1: Understand the Statements
- **Statement 1:** The circumradius of the triangle formed by the lines \( y = 3x \), \( x = 0 \), and \( y = 5 \) is \( \frac{5\sqrt{5}}{3\sqrt{2}} \).
- **Statement 2:** In a right-angled triangle, the radius of the circumcircle is half of the hypotenuse.
### Step 2: Analyze Statement 2
Statement 2 is a well-known property of right-angled triangles. The circumradius \( R \) of a right-angled triangle is given by:
\[
R = \frac{c}{2}
\]
where \( c \) is the length of the hypotenuse. Therefore, Statement 2 is true.
**Hint for Step 2:** Recall the property of circumradius in right-angled triangles.
### Step 3: Analyze Statement 1
To verify Statement 1, we need to find the vertices of the triangle formed by the lines \( y = 3x \), \( x = 0 \), and \( y = 5 \).
1. **Find Intersection Points (Vertices of the Triangle):**
- **Intersection of \( y = 3x \) and \( x = 0 \):**
- At \( x = 0 \), \( y = 3(0) = 0 \). So, point A is \( (0, 0) \).
- **Intersection of \( y = 3x \) and \( y = 5 \):**
- Set \( 3x = 5 \) → \( x = \frac{5}{3} \).
- So, point B is \( \left(\frac{5}{3}, 5\right) \).
- **Intersection of \( x = 0 \) and \( y = 5 \):**
- At \( x = 0 \), \( y = 5 \). So, point C is \( (0, 5) \).
2. **Vertices of the Triangle:**
- A: \( (0, 0) \)
- B: \( \left(\frac{5}{3}, 5\right) \)
- C: \( (0, 5) \)
### Step 4: Calculate the Length of the Hypotenuse
The triangle ABC is a right-angled triangle at point A.
1. **Calculate the lengths of sides AB and AC:**
- **Length of AB:**
\[
AB = \sqrt{\left(\frac{5}{3} - 0\right)^2 + (5 - 0)^2} = \sqrt{\left(\frac{5}{3}\right)^2 + 5^2} = \sqrt{\frac{25}{9} + 25} = \sqrt{\frac{25 + 225}{9}} = \sqrt{\frac{250}{9}} = \frac{5\sqrt{10}}{3}
\]
- **Length of AC:**
\[
AC = \sqrt{(0 - 0)^2 + (5 - 0)^2} = 5
\]
2. **Calculate the length of BC:**
- **Length of BC:**
\[
BC = \sqrt{\left(\frac{5}{3} - 0\right)^2 + (5 - 5)^2} = \frac{5}{3}
\]
### Step 5: Calculate the Circumradius
Since ABC is a right triangle, the circumradius \( R \) is given by:
\[
R = \frac{c}{2}
\]
where \( c \) is the hypotenuse (which is AB in this case).
1. **Calculate the circumradius:**
\[
R = \frac{AB}{2} = \frac{\frac{5\sqrt{10}}{3}}{2} = \frac{5\sqrt{10}}{6}
\]
### Step 6: Compare with Statement 1
We need to verify if \( \frac{5\sqrt{10}}{6} \) equals \( \frac{5\sqrt{5}}{3\sqrt{2}} \).
1. **Simplify \( \frac{5\sqrt{5}}{3\sqrt{2}} \):**
\[
\frac{5\sqrt{5}}{3\sqrt{2}} = \frac{5\sqrt{5} \cdot \sqrt{2}}{3 \cdot 2} = \frac{5\sqrt{10}}{6}
\]
Since both expressions are equal, Statement 1 is also true.
### Conclusion
Both statements are true, and Statement 2 correctly explains Statement 1.
**Final Answer:** Both Statement 1 and Statement 2 are true.
