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We are given a circle with a chord \( XY = 16 \, \text{cm} \). The perpendicular bisector of the chord cuts \( XY \) at \( N \) and the circle at \( P \). We also know that \( NP = 2 \, \text{cm} \).

Let's break down the problem:

### Step 1: Relationship between radius, chord, and bisector
- \( N \) is the midpoint of the chord \( XY \), so:
  \[
  XN = \frac{XY}{2} = \frac{16}{2} = 8 \, \text{cm}.
  \]
- The perpendicular bisector passes through the center \( O \) of the circle, making \( ON \) perpendicular to \( XY \).

We are asked to find the radius \( r \) of the circle. \( O \) is the center of the circle, so we need to calculate \( OP \), which is the radius.

### Step 2: Right triangle in the circle
- The triangle \( ONP \) is a right triangle because \( ON \) is perpendicular to \( XY \).
- From Pythagoras’ theorem:
  \[
  OP^2 = ON^2 + NP^2
  \]
  where \( OP = r \) is the radius we need to find, \( ON \) is the distance from the center to the midpoint of the chord, and \( NP = 2 \, \text{cm} \) is given.

### Step 3: Finding \( ON \)
To find \( ON \), we use the right triangle \( ONX \). By Pythagoras’ theorem again:
\[
ON^2 + XN^2 = r^2
\]
Substitute \( XN = 8 \, \text{cm} \):
\[
ON^2 + 8^2 = r^2
\]
\[
ON^2 + 64 = r^2
\]
\[
ON^2 = r^2 - 64
\]

### Step 4: Applying to \( \triangle ONP \)
Substituting into the equation for \( OP^2 \):
\[
OP^2 = ON^2 + NP^2
\]
\[
r^2 = (r^2 - 64) + 2^2
\]
\[
r^2 = r^2 - 64 + 4
\]
\[
r^2 = r^2 - 60
\]
This simplifies to \( r^2 = 60 \), so:
\[
r = \sqrt{60} \approx 7.75 \, \text{cm}.
\]

### Final Answer:
The radius of the circle is approximately \( 7.75 \, \text{cm} \).

---

Would you like a more detailed breakdown or have any questions? Here are some related questions to explore:

1. How do you find the length of a chord if the radius and the distance from the center to the chord are known?
2. What is the relationship between a chord, its perpendicular bisector, and the radius in a circle?
3. How can you derive the Pythagorean theorem for a right triangle inscribed in a circle?
4. How does the perpendicular bisector theorem help in solving circle-related geometry problems?
5. What are the properties of cyclic quadrilaterals formed by chords and bisectors?

**Tip:** When dealing with circles, always look for opportunities to apply Pythagoras’ theorem, as many circle problems involve right triangles formed by radii and chords.

The perpendicular bisector of a chord XY cuts XY at N and the circle at P. Given that XY = 16 cm and NP = 2 cm, calculate the radius of the circle.

Learn how to calculate the radius of a circle given a chord length of 16 cm and a perpendicular bisector that cuts the chord. This geometry problem covers key concepts like Pythagoras' Theorem and the Perpendicular Bisector Theorem. Suitable for students in Grades 8-10.

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