In figure 'o' is the centre of a circle. The area of sector OAPB is $\frac{5}{18}$ of the area of the circle find $x$.
To solve this problem, we need to find the angle \( x \) which subtends the arc in the sector OAPB of the circle. We know from the problem statement that the area of sector OAPB is \(\frac{5}{18}\) of the area of the entire circle.
The formula for the area of a sector of a circle is given by:
\(\frac{\theta}{360^\circ} \times \pi r^2\)
where \(\theta\) is the angle in degrees and \( r \) is the radius of the circle.
Since the area of sector OAPB is \(\frac{5}{18}\) of the total area, we have:
\(\frac{x}{360^\circ} \times \pi r^2 = \frac{5}{18} \times \pi r^2\)
By cancelling out \(\pi r^2\) from both sides, we get:
\(\frac{x}{360^\circ} = \frac{5}{18}\)
To find \( x \), we solve the equation:
\(x = \frac{5}{18} \times 360^\circ\)
Calculating the value:
\(x = \frac{5 \times 360^\circ}{18}\)
\(x = \frac{1800^\circ}{18}\)
\(x = 100^\circ\)
Thus, the value of \( x \) is 100 degrees.
This matches the correct answer from the options provided: 100 degrees.
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Important Questions from Mensuration
In a circular garden of radius 15 m, a path of 2 m wide has to be made inside the garden at the rate of ₹ 24 per sq. m. The cost of making the path is: (Take π = \(\frac{22}{7}\) )
A hollow cylinder with outer radius 4 cm and height 2 cm is made up of 1 cm thick metal sheet. What is the volume of metal used? (Take π = \(\frac{22}{7}\))
The sides of a triangular park are 60 m, 297 m and 303 m. Its area is equal to the area of a square-shaped garden. What is the perimeter (in m) of the garden?
The length and breadth of a rectangular field are in the ratio 4 : 3. If the cost of cultivating the field at ₹2 per m 2is ₹600, then the length of the field is:
A cylindrical tank has a capacity of 5632 m3. If the diameter of its base is 8 m, what is the depth of the cylindrical tank? (Use π = \(\frac{22}{7}\))
