NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 11: Area Related To Circles - Exercise 11.1

Question 6

Step 1: The diameters of the two smaller parks are given as 16 m and 12 m.

Radius is half of diameter, so:

• Radius of first park = \(\tfrac{16}{2} = 8\, \text{m}\)
• Radius of second park = \(\tfrac{12}{2} = 6\, \text{m}\)

Step 2: Area of a circle is given by the formula:

\( A = \pi r^2 \)

Step 3: Calculate the area of both small parks:

• First park area = \( \pi (8^2) = \pi (64) = 64\pi\, \text{m}^2 \)
• Second park area = \( \pi (6^2) = \pi (36) = 36\pi\, \text{m}^2 \)

Step 4: Add the two areas:

Total area = \(64\pi + 36\pi = 100\pi\, \text{m}^2\)

Step 5: Let the radius of the new (bigger) park be \(R\). Its area is:

\( A = \pi R^2 \)

Since the new park’s area must equal the sum of the smaller parks:

\( \pi R^2 = 100\pi \)

Step 6: Cancel \(\pi\) on both sides:

\( R^2 = 100 \)

Step 7: Take square root:

\( R = \sqrt{100} = 10\, \text{m} \)

Final Answer: The radius of the new park is 10 m. Hence, option (A).