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Question

If the perimeter of a semicircular park is 360 m, then what is its area? (Take π = 22/7)

This question was previously asked in
CDS II 2021 General Knowledge Previous Year Paper (14-Nov-2021)
The correct answer is

7700 m2

Calculating the Area of a Semicircular Park from its Perimeter

We are asked to find the area of a semicircular park given its perimeter. The perimeter of a semicircle is the sum of the length of its curved arc and the length of its diameter.

Understanding Semicircle Perimeter

A semicircle is half of a circle. Let the radius of the semicircle be \(r\).

  • The length of the curved arc is half the circumference of a full circle: \(\frac{1}{2} \times 2\pi r = \pi r\).
  • The length of the diameter is \(2r\).

Therefore, the perimeter of the semicircular park is the sum of the arc length and the diameter:

Perimeter = Arc length + Diameter

Perimeter = \(\pi r + 2r\)

Perimeter = \(r(\pi + 2)\)

Using the Given Perimeter to Find the Radius

We are given that the perimeter of the semicircular park is 360 m. We are also asked to use \(\pi = 22/7\).

Substitute the given values into the perimeter formula:

\(r(\pi + 2) = 360\)

\(r\left(\frac{22}{7} + 2\right) = 360\)

Now, calculate the value inside the parenthesis:

\(\frac{22}{7} + 2 = \frac{22}{7} + \frac{2 \times 7}{7} = \frac{22}{7} + \frac{14}{7} = \frac{22 + 14}{7} = \frac{36}{7}\)

So, the equation becomes:

\(r \times \frac{36}{7} = 360\)

To find \(r\), multiply both sides by \(\frac{7}{36}\):

\(r = 360 \times \frac{7}{36}\)

\(r = \frac{360}{36} \times 7\)

\(r = 10 \times 7\)

\(r = 70 \text{ m}\)

The radius of the semicircular park is 70 m.

Calculating the Area of the Semicircular Park

The area of a semicircle is half the area of a full circle with the same radius. The area of a full circle is \(\pi r^2\).

Area of semicircle = \(\frac{1}{2} \pi r^2\)

Substitute the values of \(\pi = 22/7\) and \(r = 70 \text{ m}\):

Area = \(\frac{1}{2} \times \frac{22}{7} \times (70)^2\)

Area = \(\frac{1}{2} \times \frac{22}{7} \times 70 \times 70\)

We can simplify this calculation:

Area = \(\frac{1}{2} \times 22 \times \frac{70}{7} \times 70\)

Area = \(\frac{1}{2} \times 22 \times 10 \times 70\)

Area = \(11 \times 10 \times 70\)

Area = \(110 \times 70\)

Area = \(7700\)

The area of the semicircular park is 7700 square meters.

The unit for area is square meters (\(\text{m}^2\)) since the radius is in meters.

Summary of Calculation Steps

  1. Identify the formula for the perimeter of a semicircle: \(P = r(\pi + 2)\).
  2. Substitute the given perimeter (360 m) and value of \(\pi\) (22/7) into the formula.
  3. Solve the equation to find the radius \(r\).
  4. Identify the formula for the area of a semicircle: \(A = \frac{1}{2} \pi r^2\).
  5. Substitute the calculated radius \(r\) and value of \(\pi\) into the area formula.
  6. Calculate the area and state the result with the correct units.

Revision Table: Semicircle Formulas

Measurement Formula (with radius \(r\))
Perimeter of Semicircle \(\pi r + 2r = r(\pi + 2)\)
Area of Semicircle \(\frac{1}{2} \pi r^2\)
Circumference of Full Circle \(2\pi r\)
Area of Full Circle \(\pi r^2\)

Additional Information on Semicircle Geometry

A semicircle is formed by cutting a circle along its diameter. It consists of a curved boundary (the arc) and a straight boundary (the diameter).

  • The center of the semicircle is the midpoint of its diameter, which is also the center of the original circle.
  • All points on the curved arc are equidistant from the center, with the distance being the radius \(r\).
  • The angle subtended by the diameter at any point on the circumference (the curved part) is always 90 degrees (a right angle). This is a property related to Thales's theorem.

Understanding the distinction between the perimeter and the arc length is crucial. The perimeter includes the straight edge (diameter), while the arc length is only the curved part.

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Important Questions from Plane Figures

  1. A rectangular grassy plot is 38 m long and 32 m wide. It has a gravel path of uniform width on the inside having an area of $600 \text{ m}^2$. The width of the path is:
  2. A river, 2 m deep and 45 m wide is flowing at the speed of 3 km per hour. The quantity of water that runs into the sea per minute is:
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    (Use $\pi = \frac{22}{7}$)

  4. If a circle and a square have the same perimeter, then
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