Areas of Combination of Plane Figures
Areas of combination of plane figures involve computing the area of regions formed by combining or overlapping standard shapes — circles, semicircles, triangles, rectangles, and sectors. This is covered in Class 10 CBSE Chapter 12 (Areas Related to Circles).
These problems typically ask for the area of a shaded region, which requires adding or subtracting areas of individual shapes. The key skill is identifying which standard shapes form the combined figure and deciding whether to add or subtract their areas.
These questions appear frequently in CBSE board exams (4–5 marks) and require a strong understanding of area formulas for circles, sectors, segments, triangles, and rectangles.
What is Areas of Combination of Plane Figures - Shaded Regions & Solved Examples?
Definition: A combination of plane figures is a region formed by overlapping, joining, or cutting standard 2D shapes. The area of such a region is found by adding or subtracting the areas of the component shapes.
Key Formulas Used:
- Area of circle = πr²
- Area of semicircle = πr²/2
- Area of quadrant = πr²/4
- Area of sector = (θ/360°) × πr²
- Area of segment = Area of sector − Area of triangle
- Area of rectangle = length × breadth
- Area of square = side²
- Area of triangle = ½ × base × height
- Area of equilateral triangle = (√3/4) × side²
Areas of Combination of Plane Figures Formula
Strategies for Finding Combined Areas:
Shaded area = Total area − Unshaded area
OR
Shaded area = Sum of individual shaded parts
Decision Guide:
- If the shaded region is inside a larger shape but outside a smaller shape: Area = Area(large) − Area(small)
- If the shaded region is the union of multiple non-overlapping parts: Area = Sum of all parts
- If shapes overlap: Area = A₁ + A₂ − Area(overlap)
Derivation and Proof
Common Patterns in Combined Figure Problems:
- Circle inscribed in a square: The diameter of the circle = side of the square. Area between them = side² − πr² = side² − π(side/2)².
- Square inscribed in a circle: The diagonal of the square = diameter of circle. Side = d/√2. Area between = πr² − side².
- Semicircles on sides of a right triangle: Areas of semicircles drawn on the three sides relate to each other via Pythagoras theorem.
- Flower/petal patterns: Created by overlapping quadrants drawn from corners of a square. Use symmetry to simplify.
Useful relationship for semicircles on a right triangle:
- Let the right triangle have legs a, b and hypotenuse c.
- Semicircle areas: S₁ = πa²/8, S₂ = πb²/8, S₃ = πc²/8
- By Pythagoras: a² + b² = c², so S₁ + S₂ = S₃.
- This means the two lune (crescent) areas equal the area of the triangle.
Types and Properties
Common Types of Combined Figure Problems:
- Type 1: Shape inside a shape — Circle in square, square in circle, triangle in circle. Shaded area = outer − inner.
- Type 2: Semicircles on diameter — Semicircles drawn on the diameter or sides of a rectangle/triangle.
- Type 3: Overlapping circles/arcs — Two circles overlapping, or two quadrants overlapping to form a lens/petal shape.
- Type 4: Sectors and segments — Shaded region between a chord and arc, or between two concentric arcs.
- Type 5: Ring (annulus) — Area between two concentric circles = π(R² − r²).
- Type 6: Design patterns — Flower petals, crescents, and decorative designs made from arcs and straight lines.
Solved Examples
Example 1: Circle Inscribed in a Square
Problem: A circle is inscribed in a square of side 14 cm. Find the area of the remaining portion (shaded region).
Solution:
Given: Side of square = 14 cm → radius of inscribed circle = 7 cm
Area of square:
- = 14² = 196 cm²
Area of circle:
- = πr² = (22/7) × 49 = 154 cm²
Shaded area:
- = 196 − 154 = 42 cm²
Answer: Area of shaded region = 42 cm²
Example 2: Four Quadrants in a Square
Problem: In a square of side 14 cm, four quadrants of circles of radius 7 cm are drawn at the four corners. Find the area of the remaining portion.
Solution:
Area of square: = 14² = 196 cm²
Area of 4 quadrants:
- Each quadrant = πr²/4 = (22/7) × 49/4 = 38.5 cm²
- Total = 4 × 38.5 = 154 cm²
Note: 4 quadrants of radius 7 cm = 1 full circle of radius 7 cm = 154 cm²
Shaded area:
- = 196 − 154 = 42 cm²
Answer: Area of remaining portion = 42 cm²
Example 3: Semicircles on the Sides of a Square
Problem: Semicircles are drawn on each side of a square with side 10 cm as diameter. Find the area enclosed by all four semicircles (the flower-petal region outside the square).
Solution:
Radius of each semicircle: r = 10/2 = 5 cm
Area of 4 semicircles:
- = 4 × πr²/2 = 4 × π(25)/2 = 50π cm²
Area of square: = 100 cm²
Petal area outside square:
- = Area of 4 semicircles − Area of square
- = 50π − 100
- = 50(22/7) − 100
- = 157.14 − 100 = 57.14 cm²
Answer: Area of petal region = (50π − 100) cm² ≈ 57.14 cm²
Example 4: Two Semicircles Inside a Rectangle
Problem: A rectangle has dimensions 20 cm × 10 cm. Two semicircles of diameter 10 cm are drawn inside it on the shorter sides. Find the area of the shaded region (rectangle minus two semicircles).
Solution:
Given: l = 20 cm, b = 10 cm, r = 5 cm
Area of rectangle:
- = 20 × 10 = 200 cm²
Area of 2 semicircles = 1 circle:
- = πr² = π(25) = 25π = 78.57 cm²
Shaded area:
- = 200 − 78.57 = 121.43 cm²
Answer: Shaded area = (200 − 25π) cm² ≈ 121.43 cm²
Example 5: Segment of a Circle
Problem: Find the area of the shaded segment in a circle of radius 14 cm, where the central angle of the sector is 90°.
Solution:
Area of sector (90°):
- = (90/360) × πr² = ¼ × (22/7) × 196 = 154 cm²
Area of triangle (right-angled, both sides = 14 cm):
- = ½ × 14 × 14 = 98 cm²
Area of segment:
- = Area of sector − Area of triangle
- = 154 − 98 = 56 cm²
Answer: Area of segment = 56 cm²
Example 6: Equilateral Triangle with Circular Arcs
Problem: An equilateral triangle of side 12 cm has three sectors of radius 6 cm drawn at each vertex (each sector has angle 60°). Find the area of the remaining portion of the triangle.
Solution:
Area of equilateral triangle:
- = (√3/4) × 12² = (√3/4) × 144 = 36√3 = 62.35 cm²
Area of 3 sectors (each 60°):
- = 3 × (60/360) × πr² = 3 × (1/6) × (22/7) × 36
- = (1/2) × (22/7) × 36 = 396/7 = 56.57 cm²
Note: 3 sectors of 60° each = ½ circle of radius 6 cm.
Remaining area:
- = 62.35 − 56.57 = 5.78 cm²
Answer: Area of remaining portion = (36√3 − 396/7) cm² ≈ 5.78 cm²
Example 7: Two Overlapping Circles
Problem: Two equal circles of radius 7 cm intersect such that each passes through the centre of the other. Find the area of the overlapping region.
Solution:
Step 1: The distance between centres = radius = 7 cm.
Step 2: The overlapping region consists of two identical circular segments, each with central angle 120° (since the triangle formed by the two centres and one intersection point is equilateral → each angle at centre = 60°, but the segment subtends 2 × 60° = 120°).
Area of one sector (120°):
- = (120/360) × πr² = (1/3) × (22/7) × 49 = 154/3 cm²
Area of equilateral triangle (side = 7):
- = (√3/4) × 49 = 49√3/4 cm²
Area of one segment:
- = 154/3 − 49√3/4 cm²
Total overlapping area = 2 segments:
- = 2(154/3 − 49√3/4)
- = 308/3 − 49√3/2
- = 102.67 − 42.43 = 60.24 cm²
Answer: Overlapping area ≈ 60.24 cm²
Example 8: Running Track (Annular Region)
Problem: A running track consists of two straight portions and two semicircular ends. The inner semicircles have radius 35 m and the track width is 7 m. Find the area of the track.
Solution:
Given: Inner radius R₁ = 35 m, Outer radius R₂ = 35 + 7 = 42 m
Area of two semicircular ends = one full annulus:
- = π(R₂² − R₁²) = (22/7)(42² − 35²)
- = (22/7)(1764 − 1225) = (22/7)(539) = 22 × 77 = 1694 cm²
If straight length = L, area of two rectangular strips:
- = 2 × L × 7 = 14L
For just the semicircular portions:
Answer: Area of curved track portions = 1694 m²
Real-World Applications
Real-life applications:
- Architecture: Calculating areas of decorative windows (circular arcs in rectangular frames), floor tiles, and garden designs.
- Agriculture: Finding the area irrigated by a rotating sprinkler (sector area) or the area of circular fields within rectangular plots.
- Sports: Calculating the area of running tracks, cricket grounds, and archery targets.
- Art and design: Rangoli patterns, tile designs, and stained glass windows involve combined geometric figures.
- Manufacturing: Cutting circular or semicircular pieces from rectangular sheets — the remaining material (waste) is a combined figure.
Key Points to Remember
- Combined figure area = total area − unwanted area OR sum of parts.
- Area of a ring (annulus) = π(R² − r²).
- Area of a sector = (θ/360) × πr².
- Area of a segment = area of sector − area of triangle.
- 4 quadrants of same radius = 1 full circle.
- 2 semicircles of same radius = 1 full circle.
- For equilateral triangle: area = (√3/4) × a² and each angle = 60°.
- Always identify which shapes form the boundary of the shaded region.
- Sketch the figure and label all dimensions before computing.
- Check your answer: shaded area must be less than the area of the bounding shape.
Practice Problems
- A square of side 20 cm has a circle inscribed in it. Find the area of the shaded region (square minus circle).
- A circle of radius 10 cm has an equilateral triangle inscribed in it. Find the area of the region between the circle and the triangle.
- Two circles of radii 5 cm and 3 cm overlap. If the distance between their centres is 4 cm, find the area common to both circles. (Use the segment formula.)
- A rectangular sheet 44 cm × 20 cm has 4 circles of radius 5 cm stamped out from it. Find the remaining area.
- In a circle of radius 21 cm, a sector of 60° is cut out. Find the area of the remaining part of the circle.
- Semicircles are drawn on each side of a right triangle with sides 6 cm, 8 cm, and 10 cm. Show that the sum of the two lune areas equals the area of the triangle.
Frequently Asked Questions
Q1. How do you find the area of a shaded region?
Identify the shapes that form the boundary. If the shaded region is between two shapes: area = outer shape area − inner shape area. If it consists of multiple non-overlapping parts, add their areas.
Q2. What is the area of a ring (annulus)?
A ring is the region between two concentric circles with radii R (outer) and r (inner). Area = π(R² − r²) = π(R + r)(R − r).
Q3. How do you find the area of a segment?
Area of segment = Area of sector − Area of triangle. The sector is defined by the central angle, and the triangle is formed by the two radii and the chord.
Q4. What is the difference between a sector and a segment?
A sector is the region between two radii and the arc (like a pizza slice). A segment is the region between a chord and the arc (sector minus the triangle).
Q5. Can combined figure areas be negative?
No. Area is always positive. If your calculation gives a negative result, you likely subtracted the larger area from the smaller one — reverse the order.
Q6. How do you find the area of a flower petal pattern?
Use symmetry. A flower with 4 petals in a square is typically: area of 4 semicircles − area of square (petals outside) or area of square − area of 4 quadrants (space between petals inside). Identify the pattern and use the correct formula.
Q7. What is a lune?
A lune is a crescent-shaped region between two circular arcs. In the classic problem with semicircles on a right triangle's sides, the two lune areas on the legs exactly equal the area of the right triangle.
Q8. When should I use 22/7 vs 3.14 for π?
Use the value specified in the problem. If not specified: use 22/7 when dimensions are multiples of 7 (simplifies calculation), and 3.14 otherwise. In CBSE board exams, the problem usually specifies which to use.
