Three girls Sangeeta, Namrata and Mandip are playing a game by standing on circle of radius 5 m drawn in a park . Sangeeta throws a ball to Namrata, Namrata to Mandip, Mandeep to Sangeeta. If the distance between Sangeeta and Namrata and between Namrata and Mandip is 6 m each, what is the distance between Sangita and Mandip?

Options

A. 7.6 cm
B. 8.6 cm
C. 9.6 cm
D. 4.6 cm

Answer

To solve the problem, we need to find the distance between Sangeeta and Mandip given that Sangeeta and Namrata are 6 m apart, and Namrata and Mandip are also 6 m apart. We will use the properties of a circle and the Pythagorean theorem.

### Step-by-Step Solution:

1. **Identify the Points:**
- Let S be the position of Sangeeta, N be the position of Namrata, and M be the position of Mandip on the circle.

2. **Draw the Circle:**
- Draw a circle with a radius of 5 m. The center of the circle is O.

3. **Use the Given Distances:**
- The distance between Sangeeta and Namrata (SN) = 6 m.
- The distance between Namrata and Mandip (NM) = 6 m.

4. **Apply the Circle Properties:**
- Since S, N, and M are points on the circle, the distance from the center O to each point (S, N, M) is the radius, which is 5 m.

5. **Construct Right Triangles:**
- Draw the perpendicular from the center O to the line segment SM, meeting it at point T. This bisects the segment SM into two equal parts (ST and TM).

6. **Use the Pythagorean Theorem:**
- In triangle OST:
- \( OS^2 = OT^2 + ST^2 \)
- Here, \( OS = 5 \) m (radius), and let \( OT = x \) m, then:
\[ 5^2 = x^2 + ST^2 \quad \text{(1)} \]
\[ 25 = x^2 + ST^2 \]

7. **For Triangle STN:**
- In triangle STN:
- \( SN^2 = ST^2 + TN^2 \)
- Here, \( SN = 6 \) m, and \( TN = 5 - x \) m (since OT = x):
\[ 6^2 = ST^2 + (5 - x)^2 \quad \text{(2)} \]
\[ 36 = ST^2 + (25 - 10x + x^2) \]

8. **Set Up the Equations:**
- From equation (1):
\[ ST^2 = 25 - x^2 \]
- Substitute \( ST^2 \) in equation (2):
\[ 36 = (25 - x^2) + (25 - 10x + x^2) \]
\[ 36 = 50 - 10x \]
\[ 10x = 50 - 36 \]
\[ 10x = 14 \quad \Rightarrow \quad x = 1.4 \]

9. **Find ST:**
- Substitute \( x \) back into equation (1):
\[ ST^2 = 25 - (1.4)^2 \]
\[ ST^2 = 25 - 1.96 = 23.04 \]
\[ ST = \sqrt{23.04} \approx 4.8 \text{ m} \]

10. **Calculate SM:**
- Since \( SM = 2 \times ST \):
\[ SM = 2 \times 4.8 = 9.6 \text{ m} \]

### Final Answer:
The distance between Sangeeta and Mandip (SM) is approximately **9.6 m**.

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