Mathematics Class 10
🏠 Home Log in
TOPIC 28 OF 35

Area Combinations of Plane Figures and Exercises

🎓 Class 10 Mathematics CBSE Theory Ch 11 — Areas Related to Circles ⏱ ~30 min
🌐 Language: [gtranslate]

This MCQ module is based on: Area Combinations of Plane Figures and Exercises

This mathematics assessment will be based on: Area Combinations of Plane Figures and Exercises
Targeting Class 10 level in Mensuration, with Intermediate difficulty.

Upload images, PDFs, or Word documents to include their content in assessment generation.

Combinations of Plane Figures

Many practical shapes are built up by joining or overlapping simple figures: a flower bed made of circle and square, a path of semicircles along a rectangle, a logo with circular cut-outs. To find their areas, we break the shape into standard pieces, find each piece's area, and add or subtract as appropriate.

Strategy
  1. Identify the basic shapes (rectangle, triangle, circle, sector, semicircle).
  2. Write down each dimension carefully; convert units if required.
  3. Decide: do pieces add (non-overlapping union) or subtract (hole, unshaded region)?
  4. Compute using standard formulas. Keep π as 22/7 or 3.14 as directed.

Worked Problem A — Semicircles on sides of a square

A square of side 14 cm has a semicircle drawn on each side as diameter, outward. Find the total area of the figure.

14 cm
Fig. — Square with four outward semicircles on its sides

Solution. Square area = 14 × 14 = 196 cm². Each semicircle has diameter 14, radius 7. Area of one = (1/2)π(7²) = (1/2)(22/7)(49) = 77 cm². Four semicircles = 308 cm². Total area = 196 + 308 = 504 cm².

Worked Problem B — Shaded region

In the figure (a quadrant of radius 14 cm with a semicircle drawn on the radius as diameter), find the area of the shaded region.

Solution. Quadrant area = (1/4)π(14²) = (1/4)(22/7)(196) = 154 cm². Inside the quadrant, semicircle on one radius has diameter 14, radius 7, area = (1/2)(22/7)(49) = 77 cm². Triangle formed by the two radii and chord = (1/2)(14)(14) = 98 cm². Using the standard calculation, shaded area = 98 cm² in a typical arrangement. (Depends on exact arrangement — check figure.)

Activity — Design a flower bed

Draw a rectangle of dimensions 20 m × 10 m for a park. At two opposite shorter sides, attach a semicircle outward. Compute the total area of the "stadium" shape. Then compute its perimeter.

Rectangle area = 200 m². Two semicircles (diameter 10) = one full circle = π(5²) = 25π ≈ 78.54 m². Total area ≈ 278.54 m². Perimeter = two long sides + circumference of one circle = 40 + 10π ≈ 71.42 m.

Exercise 11.2

(Unless stated otherwise, take π = 22/7.)

Q1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is open at the top. Ignoring the thickness of the plastic sheet, determine the area of sheet required. (Not in this exercise — moving to circles-related.)

(Actual Q1) Find the area of the shaded region in the figure: a circle of radius 7 cm is inscribed in a square, and a flower-like shape is shaded between them (area outside the circle inside the square).
Square side = 14. Square area = 196. Circle area = (22/7)(49) = 154. Shaded (corners) = 196 − 154 = 42 cm².
Q2. Find the area of the shaded region: two concentric circles of radii 7 cm and 14 cm with sector angle 40° at the centre (annular sector).
Big sector − small sector = (40/360)(22/7)[14² − 7²] = (1/9)(22/7)(147) = (22)(147)/63 = 154/3 cm² ≈ 51.33 cm².
Q3. Find the area of the shaded region in a square of side 14 cm with semicircles drawn on two opposite sides inward.
Square = 196. Two inward semicircles overlap in the middle. For a lens-less configuration (e.g., two non-overlapping semicircles drawn on two non-adjacent sides with diameter 14), each semicircle area = 77. Shaded (square minus two inscribed semicircles, if non-overlapping) = 196 − 154 = 42 cm².
Q4. Find the area of the shaded region: an equilateral triangle of side 12 cm with a circle inscribed in it, shaded = triangle − circle.
Triangle = (√3/4)(144) = 36√3. Inradius r = side/(2√3) = 12/(2√3) = 2√3. Circle area = π(12) = 12π. Shaded = 36√3 − 12π ≈ 62.35 − 37.71 ≈ 24.64 cm².
Q5. From each corner of a square of side 4 cm, a quadrant of radius 1 cm is cut. Additionally, a circle of diameter 2 cm is cut from the centre. Find the area of the remaining portion.
Square = 16. Four quadrants = 1 full circle of r = 1 ⇒ π. Centre circle r = 1 ⇒ π. Total cut = 2π. Remaining = 16 − 2π ≈ 16 − 6.28 ≈ 9.72 cm², or 16 − 44/7 = 68/7 ≈ 9.71 cm² (taking π = 22/7).
Q6. In a circular table-cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle. Find the area of the design (area outside triangle but inside circle).
Circumradius R = 32 ⇒ side a = R√3 = 32√3. Triangle area = (√3/4)(32√3)² = (√3/4)(3072) = 768√3. Circle area = π(1024) = 1024π. Design = 1024π − 768√3 ≈ 3217.14 − 1330.14 ≈ 1887 cm².
Q7. In the figure, ABCD is a square of side 14 cm. With centres A, B, C, D, four quadrants of radius 7 cm are drawn inside the square. Find the area of the shaded region (square minus four quadrants).
Four quadrants of radius 7 = 1 full circle = (22/7)(49) = 154. Square = 196. Shaded = 196 − 154 = 42 cm².
Q8. The figure depicts a racing track with inner straight portions 106 m each and two semicircular ends of inner diameter 60 m. Width of track is 10 m. Find (i) the distance around the track along its inner edge, (ii) the area of the track.
Inner radius r = 30, outer R = 40. (i) Inner perimeter = 2(106) + 2·π·30 = 212 + 60π = 212 + 188.57 ≈ 400.57 m. (ii) Track area = 2(straight rect 106 × 10) + (annular ring: π(R² − r²)) = 2120 + π(1600 − 900) = 2120 + 700π ≈ 2120 + 2200 = 4320 m².
Q9. In the figure, AB and CD are two diameters of a circle (perpendicular) and OD is the diameter of a smaller circle. If OA = 7 cm, find the area of the shaded region.
Big circle r = 7, area = 49π. Small circle r = 3.5, area = 12.25π. Shaded arrangement depends on figure. For the standard NCERT shade: Area = (area of semicircle on AB) − (area of small circle) + (triangle ACD area). Computed answer is typically 66.5 cm² (with π = 22/7).
Q10. The area of an equilateral triangle ABC is 17320.5 cm². With each vertex as centre, a circle is drawn with radius equal to half the length of the side of the triangle. Find the area of the shaded region (triangle minus three circular sectors). (Use π = 3.14, √3 = 1.73205.)
Area = (√3/4)a² = 17320.5 ⇒ a² = 17320.5 × 4 / 1.73205 = 40000 ⇒ a = 200 cm. Radius r = 100. Each angle of triangle = 60°. Sum of three sectors = 3 × (60/360)π(100)² = (1/2)π(10000) = 5000π = 15700 cm². Shaded = 17320.5 − 15700 = 1620.5 cm².
Q11. On a square handkerchief, nine circular designs each of radius 7 cm are made (3 × 3 grid). Find the area of the remaining portion of the handkerchief.
Side of square = 3 × 14 = 42 cm. Area = 1764. 9 circles: 9 × (22/7)(49) = 9 × 154 = 1386. Remaining = 1764 − 1386 = 378 cm².
Q12. In the figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the shaded region (quadrant minus triangle OBD).
Quadrant = (1/4)(22/7)(12.25) = 9.625 cm². Triangle OBD = (1/2)(3.5)(2) = 3.5 cm². Shaded = 9.625 − 3.5 = 6.125 cm².
Q13. In the figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region (quadrant minus square). (Use π = 3.14.)
Diagonal OB = 20√2, this is the radius of the quadrant. Quadrant = (1/4)(3.14)(800) = 628. Square = 400. Shaded = 628 − 400 = 228 cm².
Q14. AB and CD are arcs of two concentric circles of radii 21 cm and 7 cm, centre O. If ∠AOB = 30°, find the area of the shaded region (annular sector).
Area = (30/360)(22/7)(21² − 7²) = (1/12)(22/7)(441 − 49) = (1/12)(22/7)(392) = (22 × 392)/(12 × 7) = (8624)/84 = 308/3 cm² ≈ 102.67 cm².

11.2 Summary

Key Formulas
  1. Length of arc of a sector with central angle θ and radius r: \(\dfrac{\theta}{360}\times 2\pi r\).
  2. Area of sector with central angle θ and radius r: \(\dfrac{\theta}{360}\times \pi r^2\).
  3. Area of segment = Area of corresponding sector − Area of corresponding triangle.
  4. For combined figures, decompose into standard shapes; add or subtract areas as per the geometry.
Looking Ahead
The next chapter lifts us into three dimensions — Surface Areas and Volumes — where you'll combine familiar solids (cubes, cylinders, cones, hemispheres, spheres) to describe and measure compound objects such as capsules, gulab jamun shapes, ice-cream cones and more.
Competency-Based Questions
Q1. A circular pond of radius 5 m is surrounded by a 2 m wide path. Apply the area formulas to find the area of the path alone.
L3 Apply
Outer radius 7, inner 5. Area of path = π(49 − 25) = 24π ≈ 75.43 m².
Q2. Analyse why the area of a segment with central angle 60° in a unit-radius circle is less than 0.1 square units.
L4 Analyse
Sector area = (60/360)π = π/6 ≈ 0.524. Triangle is equilateral with side 1: area = √3/4 ≈ 0.433. Segment = 0.524 − 0.433 = 0.091, which is indeed less than 0.1. The sector and triangle are close in size for small to moderate angles.
Q3. Evaluate: If a tyre rotates 10 revolutions in a minute along a straight path, and its radius is 35 cm, how far does a point on its rim travel in that minute?
L5 Evaluate
The distance equals 10 circumferences = 10 × 2π(35) = 700π ≈ 2200 cm or 22 m. Note this is straight-line distance only if measured along the track; the point itself traces a cycloid curve.
Q4. Design a garden plan for a 20 m × 14 m plot containing a rectangular lawn and two circular flower beds (each of radius 2 m). Compute total lawn area.
L6 Create
Plot area = 280 m². Two circles area = 2π(4) = 8π ≈ 25.13 m². Lawn area = 280 − 25.13 ≈ 254.87 m². Add a flagstone path of, say, 1 m × 20 m to subtract further if wished.
Assertion & Reason
A: The perimeter of a sector with central angle θ° and radius r is 2r + (θ/360)(2πr).
R: The perimeter equals the two radii plus the arc length.
A) Both true; R explains A
B) Both true; R does not explain A
C) A true, R false
D) A false, R true
Answer: A. A sector is bounded by two radii and an arc, and these formulas compute each piece.
A: The area of a semicircle of radius r equals half of πr².
R: A semicircle is a sector with 180° central angle.
A) Both true; R explains A
B) Both true; R does not explain A
C) A true, R false
D) A false, R true
Answer: A. (180/360)πr² = πr²/2.

Frequently Asked Questions

What is the summary of Chapter 11 Areas Related to Circles?
Chapter 11 covers area and perimeter of circular sectors and segments, plus area of combinations of circles and polygons - all using pi, radius and central angles.
Which formula gives the area of a sector?
Area of sector = (theta/360) x pi r squared, where theta is the central angle in degrees and r is the radius.
What is the most common exam mistake in this chapter?
Confusing sector area with segment area, using degrees where radians are expected (rare in Class 10), and forgetting to subtract the triangle for segments.
How do you find the area of the region between two concentric circles?
Area of annulus = pi (R^2 - r^2), where R is the outer and r is the inner radius. The answer is the big circle minus the small one.
Is Chapter 11 important for board exams?
Yes. CBSE Class 10 boards usually feature one 3-4 mark area-combinations question and one sector/segment numerical worth 2-3 marks.
What diagrams appear most in Chapter 11?
Common diagrams include sectors of a circle, segments with chords, circular flower-bed patches, running tracks and clock-face sector regions.
AI Tutor
Mathematics Class 10
Ready
Hi! 👋 I'm Gaura, your AI Tutor for Area Combinations of Plane Figures and Exercises. Take your time studying the lesson — whenever you have a doubt, just ask me! I'm here to help.