Question

The radius of a circle with centre 'P' is 10 cm. If chord AB of the circle subtends a right angle at P, find area of minor sector by using the following activity. (\(\pi = 3.14\)) 

Activity : 
r = 10 cm, \(\theta\) = 90\(^\circ\), \(\pi\) = 3.14. 
A(P-AXB) = \(\frac{\theta}{360} \times \boxed{\phantom{\pi r^2}}\) = \(\frac{\boxed{\phantom{90}}}{360} \times 3.14 \times 10^2\) = \(\frac{1}{4} \times \boxed{\phantom{314}}\) <br>
A(P-AXB) = \(\boxed{\phantom{78.5}}\) sq. cm.
 

Hint:

For a right angle at the centre (\(90^\circ\)), the sector is always a quarter of the circle. You can quickly calculate the area as \(\frac{1}{4}\pi r^2\). This is a useful shortcut.

Answer 1

Step 1: Understanding the Concept:
The area of a sector of a circle is a fraction of the total area of the circle, determined by the central angle of the sector. The question requires filling the blanks in the activity to calculate this area.

Step 2: Key Formula or Approach:
The area of a sector with central angle \(\theta\) and radius \(r\) is given by the formula: \[ \text{Area of Sector} = \frac{\theta}{360^\circ} \times \pi r^2 \]

Step 3: Detailed Explanation:
Here is the completed activity with the blanks filled in.
r = 10 cm, \(\theta\) = 90\(^\circ\), \(\pi\) = 3.14.
A(P-AXB) = \(\frac{\theta}{360} \times \boxed{\pi r^2}\) = \(\frac{\boxed{90}}{360} \times 3.14 \times 10^2\)
(Calculation: \(\frac{90}{360} = \frac{1}{4}\) and \(10^2 = 100\))
= \(\frac{1}{4} \times \boxed{314}\) (since \(3.14 \times 100 = 314\))
(Calculation: \(\frac{1}{4} \times 314 = 78.5\))
A(P-AXB) = \(\boxed{78.5}\) sq. cm.

Step 4: Final Answer:
The area of the minor sector (P-AXB) is 78.5 sq. cm.