Area of Semicircle
A semicircle is exactly half of a circle, formed when a diameter divides a circle into two equal parts. The area of a semicircle is half the area of the full circle.
This concept is part of Chapter 12 — Areas Related to Circles in the CBSE Class 10 syllabus. Problems on semicircles appear in combination-of-figures questions where students must add or subtract semicircular regions from rectangles, squares, or other shapes.
Since the area of a full circle is πr², the area of a semicircle follows directly as πr²/2. The perimeter of a semicircle includes the curved part (half the circumference) plus the diameter.
What is Area of Semicircle?
Definition: A semicircle is the region enclosed between a diameter of a circle and one of the two arcs cut off by that diameter.
- A semicircle subtends an angle of 180° at the centre.
- The diameter is the straight edge (base) of the semicircle.
- The curved edge is half the circumference of the full circle.
- Every semicircle is congruent to the other semicircle formed by the same diameter.
Area of Semicircle Formula
Area of Semicircle:
Area = πr² / 2
Where:
- π = 22/7 or 3.14159...
- r = radius of the circle
Perimeter of Semicircle:
Perimeter = πr + 2r = r(π + 2)
Where:
- πr = length of the curved arc (half circumference = 2πr/2)
- 2r = diameter (the straight edge)
In terms of diameter d:
- Area = πd²/8 (since r = d/2)
- Perimeter = πd/2 + d
Derivation and Proof
Deriving the area of a semicircle:
- The area of a full circle with radius r is πr².
- A semicircle is exactly half of a full circle (the diameter divides the circle into two equal parts).
- Therefore, Area of semicircle = πr² / 2 = πr²/2.
Deriving the perimeter of a semicircle:
- The circumference of a full circle is 2πr.
- The curved part of a semicircle is half the circumference = 2πr/2 = πr.
- The straight part is the diameter = 2r.
- Total perimeter = curved part + straight part = πr + 2r = r(π + 2).
Important: The perimeter of a semicircle is NOT half the circumference of the circle. It is half the circumference PLUS the diameter.
Types and Properties
Semicircle problems in Class 10 fall into these categories:
Type 1: Direct Area Calculation
- Given radius or diameter, find the area of the semicircle.
Type 2: Direct Perimeter Calculation
- Given radius or diameter, find the perimeter of the semicircle.
Type 3: Combination of Figures
- Semicircles drawn on sides of a rectangle, square, or triangle.
- Find the area of the shaded region by adding/subtracting areas.
Type 4: Semicircle Inside/Outside a Shape
- Semicircle inscribed in a rectangle or constructed on the hypotenuse of a right triangle.
Type 5: Multiple Semicircles
- Two or more semicircles drawn on different diameters.
- Find total area or the area enclosed between them.
Methods
Steps to find the area of a semicircle:
- Identify the radius (r). If the diameter (d) is given, compute r = d/2.
- Apply the formula: Area = πr²/2.
- Use π = 22/7 or 3.14 as specified in the problem.
- Simplify and write the answer with correct units (cm², m², etc.).
Steps to find the perimeter of a semicircle:
- Identify the radius r.
- Calculate the curved part: πr.
- Add the diameter: 2r.
- Perimeter = πr + 2r.
Common Mistakes:
- Forgetting to add the diameter when calculating perimeter.
- Using the full circle area formula instead of dividing by 2.
- Confusing radius and diameter — always check which is given.
Solved Examples
Example 1: Finding Area Given Radius
Problem: Find the area of a semicircle with radius 7 cm. (Use π = 22/7)
Solution:
Given:
- r = 7 cm
Using Area = πr²/2:
- Area = (22/7) × 7² / 2
- Area = (22/7) × 49 / 2
- Area = 22 × 7 / 2
- Area = 154 / 2 = 77 cm²
Answer: The area of the semicircle is 77 cm².
Example 2: Finding Area Given Diameter
Problem: Find the area of a semicircle with diameter 28 cm. (Use π = 22/7)
Solution:
Given:
- d = 28 cm, so r = 28/2 = 14 cm
Using Area = πr²/2:
- Area = (22/7) × 14² / 2
- Area = (22/7) × 196 / 2
- Area = 22 × 28 / 2
- Area = 616 / 2 = 308 cm²
Answer: The area is 308 cm².
Example 3: Finding Perimeter of a Semicircle
Problem: Find the perimeter of a semicircle with radius 10.5 cm. (Use π = 22/7)
Solution:
Given:
- r = 10.5 cm
Perimeter = πr + 2r:
- Curved part = (22/7) × 10.5 = (22/7) × 21/2 = 22 × 3/2 = 33 cm
- Straight part (diameter) = 2 × 10.5 = 21 cm
- Perimeter = 33 + 21 = 54 cm
Answer: The perimeter is 54 cm.
Example 4: Semicircle on Side of a Square
Problem: A semicircle is drawn on one side of a square of side 14 cm. Find the area of the figure formed. (Use π = 22/7)
Solution:
Given:
- Side of square = 14 cm
- Diameter of semicircle = 14 cm, so r = 7 cm
Steps:
- Area of square = 14² = 196 cm²
- Area of semicircle = πr²/2 = (22/7) × 49/2 = 77 cm²
- Total area = 196 + 77 = 273 cm²
Answer: The total area is 273 cm².
Example 5: Two Semicircles on a Diameter
Problem: A circle has diameter 28 cm. Two semicircles are drawn inside it on two halves of a diameter. Find the total area of the two semicircles. (Use π = 22/7)
Solution:
Given:
- Diameter of large circle = 28 cm
- Each smaller semicircle has diameter = 14 cm, so r = 7 cm
Steps:
- Area of one small semicircle = πr²/2 = (22/7) × 49/2 = 77 cm²
- Area of two small semicircles = 2 × 77 = 154 cm²
Verification: Two semicircles with diameter 14 cm each combine to form a full circle with radius 7 cm. Area = πr² = (22/7) × 49 = 154 cm². Confirmed.
Answer: The total area is 154 cm².
Example 6: Shaded Region Between Circle and Semicircles
Problem: A circle of radius 14 cm has a diameter AB. Two semicircles are drawn inside the circle taking AO and OB as diameters (O is the centre). Find the area of the shaded region between the large circle and the two small semicircles. (Use π = 22/7)
Solution:
Given:
- Radius of large circle = 14 cm, so diameter = 28 cm
- AO = OB = 14 cm (each is a diameter for smaller semicircles)
- Radius of each smaller semicircle = 7 cm
Steps:
- Area of large circle = π(14)² = (22/7) × 196 = 616 cm²
- Area of large semicircle (one half) = 616/2 = 308 cm²
- Area of two small semicircles = 2 × (22/7) × 49/2 = 154 cm²
- Shaded area = Area of large circle − Area of two small semicircles = 616 − 154 = 462 cm²
Answer: The shaded area is 462 cm².
Example 7: Semicircular Garden Path
Problem: A semicircular garden has radius 21 m. A path of width 3.5 m runs along the curved boundary. Find the area of the path. (Use π = 22/7)
Solution:
Given:
- Outer radius R = 21 m
- Width of path = 3.5 m
- Inner radius r = 21 − 3.5 = 17.5 m
Steps:
- Area of outer semicircle = πR²/2 = (22/7) × 441/2 = (22 × 63)/2 = 693 m²
- Area of inner semicircle = πr²/2 = (22/7) × 306.25/2 = (22/7) × 153.125 = 481.25 m²
- Area of path = 693 − 481.25 = 211.75 m²
Answer: The area of the path is 211.75 m².
Example 8: Semicircles on Sides of Right Triangle
Problem: Semicircles are drawn on each side of a right triangle with sides 6 cm, 8 cm, and 10 cm (hypotenuse). Prove that the sum of the areas of the two smaller semicircles equals the area of the semicircle on the hypotenuse minus the area of the triangle.
Solution:
Given:
- Sides: a = 6 cm, b = 8 cm, c = 10 cm
Steps:
- Area of semicircle on side a: A₁ = π(3)²/2 = 9π/2
- Area of semicircle on side b: A₂ = π(4)²/2 = 16π/2 = 8π
- Area of semicircle on hypotenuse: A₃ = π(5)²/2 = 25π/2
- A₁ + A₂ = 9π/2 + 8π = 9π/2 + 16π/2 = 25π/2
- Area of triangle = (1/2) × 6 × 8 = 24 cm²
- A₃ − Area of triangle = 25π/2 − 24
- A₁ + A₂ = 25π/2 = A₃
The sum of the two smaller semicircle areas equals the area of the largest semicircle. This is a direct consequence of the Pythagoras Theorem: a² + b² = c² implies πa²/8 + πb²/8 = πc²/8.
Answer: Proved. The result follows from the Pythagoras Theorem.
Example 9: Finding Radius from Area
Problem: The area of a semicircle is 308 cm². Find its radius and perimeter. (Use π = 22/7)
Solution:
Given:
- Area of semicircle = 308 cm²
Finding radius:
- πr²/2 = 308
- πr² = 616
- (22/7) × r² = 616
- r² = 616 × 7/22 = 196
- r = 14 cm
Finding perimeter:
- Perimeter = πr + 2r = (22/7) × 14 + 28 = 44 + 28 = 72 cm
Answer: Radius = 14 cm, Perimeter = 72 cm.
Example 10: Semicircular Window Area
Problem: A window is in the shape of a rectangle of length 50 cm and breadth 30 cm, with a semicircle on one of the longer sides. Find the total area of the window. (Use π = 3.14)
Solution:
Given:
- Rectangle: 50 cm × 30 cm
- Semicircle diameter = 50 cm, so r = 25 cm
Steps:
- Area of rectangle = 50 × 30 = 1500 cm²
- Area of semicircle = πr²/2 = 3.14 × 625/2 = 3.14 × 312.5 = 981.25 cm²
- Total area = 1500 + 981.25 = 2481.25 cm²
Answer: The total area of the window is 2481.25 cm².
Real-World Applications
Architecture and Construction:
- Semicircular arches, windows, and doorways are common in classical and modern architecture.
- Calculating material required for semicircular structures uses the area formula.
Engineering:
- Cross-sections of tunnels and bridges are often semicircular.
- Water flow through semicircular channels is computed using the area.
Landscaping:
- Semicircular garden beds, fountains, and driveways require area calculations for materials and planting.
Sports:
- The D-shaped area near the goal in football is a semicircular region.
- Track curves at stadium ends are semicircular.
Key Points to Remember
- Area of semicircle = πr²/2.
- Perimeter of semicircle = πr + 2r (curved arc + diameter).
- In terms of diameter: Area = πd²/8.
- A semicircle subtends an angle of 180° at the centre.
- The perimeter of a semicircle is NOT half the circumference — it includes the diameter.
- When semicircles are drawn on sides of a right triangle, the sum of areas on the two legs equals the area on the hypotenuse (by Pythagoras).
- Always check whether the problem gives radius or diameter before substituting.
- For shaded region problems, identify which areas to add and which to subtract.
- Use π = 22/7 when the radius is a multiple of 7 for cleaner calculations.
- Use π = 3.14 when explicitly stated in the problem.
Practice Problems
- Find the area of a semicircle with radius 21 cm. (Use π = 22/7)
- Find the perimeter of a semicircle with diameter 42 cm. (Use π = 22/7)
- A semicircular plot has area 1232 m². Find its radius. (Use π = 22/7)
- Semicircles are drawn on each side of a rectangle with length 14 cm and breadth 7 cm. Find the total area of the four semicircles.
- A circular disc of radius 14 cm is cut into two equal halves. Find the perimeter of each half.
- Find the area of the shaded region when a semicircle of diameter 7 cm is removed from a square of side 7 cm. (Use π = 22/7)
- A path 2 m wide surrounds a semicircular garden of radius 10 m. Find the area of the path. (Use π = 3.14)
- Three semicircles are drawn on the three sides of a right triangle with legs 5 cm and 12 cm and hypotenuse 13 cm. Find the total area of all three semicircles. (Use π = 3.14)
Frequently Asked Questions
Q1. What is the formula for the area of a semicircle?
The area of a semicircle is πr²/2, where r is the radius. Since a semicircle is half of a full circle, its area is half of πr².
Q2. What is the perimeter of a semicircle?
The perimeter of a semicircle is πr + 2r. This includes the curved arc (πr, which is half the circumference) and the straight edge (the diameter, 2r).
Q3. Is the perimeter of a semicircle half the circumference?
No. The perimeter of a semicircle = πr + 2r. The curved part alone is πr (half the circumference), but the perimeter also includes the diameter (2r) as the straight boundary.
Q4. How do you find the area of a semicircle when the diameter is given?
First find the radius: r = d/2. Then apply Area = πr²/2. Alternatively, use Area = πd²/8 directly.
Q5. What angle does a semicircle subtend at the centre?
A semicircle subtends an angle of 180° at the centre. It is a sector with central angle 180°, which is half of the full 360°.
Q6. How is the semicircle related to the Pythagoras Theorem?
If semicircles are drawn on all three sides of a right triangle, the area of the semicircle on the hypotenuse equals the sum of areas on the other two sides. This follows from c² = a² + b², since area of semicircle on a side of length s = π(s/2)²/2 = πs²/8.
Q7. What is the difference between a semicircle and a segment?
A semicircle is the region between a diameter and one of the arcs — it is exactly half the circle. A segment is the region between any chord and its arc. A semicircle is a special case of a segment where the chord is the diameter.
Q8. Can a semicircle have different orientations?
Yes. A semicircle can open upwards, downwards, left, or right depending on the position of the diameter. The area and perimeter formulas remain the same regardless of orientation.
Q9. How do you find the radius if the area of a semicircle is given?
Use πr²/2 = A. Then r² = 2A/π. Take the positive square root: r = √(2A/π).
Q10. What is the area of a semicircle with radius 7 cm?
Area = πr²/2 = (22/7) × 49/2 = 77 cm².
