Exercise 9.2 — Circle Areas
Area of circle, circular path or ring and area of sector.
Exercise 9.2 – Area of a Circle, Ring & Sector: Shaded Region Problems
Exercise 9.2 from Chapter 9, "Areas of Plane Figures," of Class 8 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus) extends the area concepts learned for straight-edged shapes to circles, rings (circular paths), sectors, and semicircles. This is one of the most practically useful exercises in the chapter, with real-world applications ranging from cutting circular buttons from sheet metal to calculating how much grass a tethered horse can graze.
This exercise assumes familiarity with the area formulas for rectangles, squares, and trapeziums covered in Exercise 9.1, since most "shaded region" problems here combine circular area formulas with these straight-edged shapes.
Area of a Circle — Where Does πR² Come From?
Rather than simply stating the formula, this section explains it through a hands-on activity: a circle is cut into many thin equal sectors, which are then rearranged alternately (pointing up and down) to form a shape that looks like a parallelogram.
- The base of this parallelogram is half the circle's circumference, i.e., ½ × 2πR = πR.
- The height of this parallelogram is equal to the circle's radius, R.
- Using the parallelogram area formula (base × height), the area becomes πR × R = πR².
Area of a circle with radius R = πR²
Area of a Circular Ring (Path)
A circular ring (or circular path) is the region between two concentric circles — an outer circle of radius R and an inner circle of radius r. Its area is found by subtracting the area of the smaller circle from the area of the larger one.
Area of ring = πR² − πr² = π(R² − r²) = π(R + r)(R − r)
The final form, π(R + r)(R − r), uses the factorisation of a difference of squares — a useful algebraic identity that often simplifies calculations involving rings.
Area of a Sector, Semicircle & Quadrant
A sector is the region of a circle bounded by two radii and the arc between them — like a "pizza slice." If the sector's angle is x° and the radius is r, then both the arc length and the sector's area are simply fractions of the full circle's circumference and area, in proportion to x° out of the full 360°.
Length of arc, l = (x ÷ 360) × 2πrPerimeter of sector = l + 2rArea of sector = (x ÷ 360) × πr²
Two special cases of the sector formula are extremely common in problems:
| Special Case | Sector Angle | Area Formula |
|---|---|---|
| Semicircle | 180° | ½ × πr² |
| Quadrant (quarter circle) | 90° | ¼ × πr² |
Exercise 9.2 – Solved Questions, Step by Step
A rectangular acrylic sheet measuring 36 cm × 25 cm has 56 circular buttons, each of diameter 3.5 cm, cut out from it.
Area of sheet = 36 × 25 = 900 sq.cmThe radius of each button is 3.5 ÷ 2 = 1.75 cm. Using π = 22/7, the area of each button is πr². Multiplying by 56 and simplifying the fractions carefully gives the combined area of all buttons.
Total area of 56 buttons = 56 × (22/7) × (3.5/2)² = 539 sq.cmArea of remaining sheet = 900 − 539 = 361 sq.cmA circle is inscribed in a square of side 28 cm — meaning the circle fits exactly inside the square, touching all four sides. The circle's diameter must therefore equal the square's side length.
Diameter of circle = side of square = 28 cm ⇒ Radius = 28 ÷ 2 = 14 cmArea of circle = πr² = (22/7) × 14² = (22/7) × 14 × 14 = 616 sq.cm(i) Square with Four Semicircles on Its Sides
Semicircles of diameter 'd' are constructed on each side of a square of side 'd'. The total length across the figure (two radii plus the square's side) is given as 42 cm.
r + d + r = 42 ⇒ (d/2) + d + (d/2) = 42 ⇒ 2d = 42 ⇒ d = 21 cmSo the radius of each semicircle is r = d/2 = 21/2 cm. The shaded region is made up of 4 semicircles, which together equal the area of 2 full circles.
Area of shaded region = 4 × ½ × πr² = 2πr² = 2 × (22/7) × (21/2)² = 693 sq.cm(ii) Outer Circle with Two Inner Semicircles Removed
Two semicircles of diameter 10.5 m are constructed inside a circle of diameter 21 m. The shaded region is the area of the outer circle minus the combined area of the two inner semicircles.
Radius of outer circle, R = 21/2 m ⇒ Area = π(21/2)² = (441/4)πRadius of inner semicircles, r = 10.5/2 = 21/4 m ⇒ Sum of 2 semicircle areas = πr² = (441/16)πArea of shaded region = (441/4)π − (441/16)π = 259.875 sq.mThis figure has two big semicircles (radius 42 cm each) and four small semicircles (radius 21 cm each, since their diameter equals 42 cm).
Area of each big semicircle = ½ × πR² = ½ × (22/7) × 42² = 2772 sq.cmArea of each small semicircle = ½ × πr² = ½ × (22/7) × 21² = (11/7) × 21² sq.cmThe S-shape is formed by adding and subtracting these semicircles in an alternating pattern (large minus small, plus small, plus small, large minus small):
Area of shaded region = [2772 − (11/7)(21)²] + (11/7)(21)² + (11/7)(21)² + [2772 − (11/7)(21)²] = 5544 sq.cmThis figure consists of four half circles and two quarter circles, with OA = OB = OC = OD = 14 cm (these act as the radii of the quarter circles).
Area of each quarter circle = ¼ × πR² = ¼ × (22/7) × 14² = 154 sq.cmArea of each half circle (radius 7 cm) = ½ × πr² = ½ × (22/7) × 7² = 77 sq.cmThe shaded pinwheel pattern is formed by alternately adding and removing these half-circle and quarter-circle regions:
Area of shaded region = 154 − 77 + 77 + 154 − 77 + 77 = 308 sq.cmPoints A, B, C, D are centres of four equal circles that touch each other externally in pairs, and ABCD forms a square of side 7 cm.
Area of square ABCD = side × side = 7 × 7 = 49 sq.cmSince the circles touch each other, each circle's radius is half the square's side: r = 7/2 cm. The shaded region inside the square (not covered by the circles) is the square's area minus four quarter circles (one at each corner).
Area of each quarter circle = ¼ × πr² = ¼ × (22/7) × (7/2)² = 77/8 sq.cmArea of shaded region = 49 − 4 × (77/8) = 49 − 38.5 = 10.5 sq.cmAn equilateral triangle has area 49√3 sq.cm. At each vertex, a circle of radius 7 cm (half the triangle's side, which is 14 cm) is drawn. Since each interior angle of an equilateral triangle is 60°, each circle contributes a sector of angle 60° inside the triangle.
Area of each 60° sector = (60/360) × πr² = (1/6) × (22/7) × 7² = 77/3 sq.cmThe portion of the triangle not covered by any circle is the triangle's area minus the combined area of all three sectors:
Required area = 49√3 − 3 × (77/3) = 49(1.732) − 77 = 84.868 − 77 = 7.868 sq.cm(i) Four Equal Circles, Each of Radius 'a', Touching One Another
When four equal circles of radius 'a' touch each other, joining their centres forms a square of side 2a (since each side equals the sum of two radii from neighbouring circles).
Area of square = (2a) × (2a) = 4a²Each circle contributes one quarter circle of radius 'a' inside this square (at each corner). Four such quarters together make one full circle of radius 'a'.
Sum of 4 quarter circles = πa²Area between the circles = 4a² − πa² = (4 − 22/7)a² = (6/7)a² sq.units(ii) Four Equal Circles at the Corners of a 24 cm Square
Each circle has radius equal to half the square's side: r = 24/2 = 12 cm. The square side (joining the centres) is 24 cm.
Area of square = 24 × 24 = 576 sq.cmSum of 4 quarter circles = πr² = (22/7) × 12² ≈ 452.16 sq.cmArea between the circles = 576 − 452.16 = 123.84 sq.cmA trapezium-shaped cardboard ABCD has AB ∥ CD and ∠BCD = 90°. Given AB = BC = 3.5 cm and DE = 2 cm, a quarter circle is removed from one corner.
The parallel sides of the trapezium are AB = 3.5 cm and CD = DE + EC = 2 + 3.5 = 5.5 cm, with the distance between them being BC = 3.5 cm.
Area of trapezium = ½ × (AB + CD) × BC = ½ × (3.5 + 5.5) × 3.5 = 15.75 sq.cmThe quarter circle removed has radius 3.5 cm (since ∠BCD = 90° allows a 90° quarter circle to fit exactly at corner C).
Area of quarter circle = ¼ × πr² = ¼ × (22/7) × 3.5 × 3.5 = 9.625 sq.cmArea of remaining cardboard = 15.75 − 9.625 = 6.125 sq.cmA horse is tethered at one corner of a rectangular field measuring 70 m × 52 m, by a rope of length 21 m. The area the horse can graze forms a quarter-circle sector (90°) of radius equal to the rope's length, since the field's corner provides a 90° angle and the rope cannot stretch beyond the field's boundary in this case.
Grazing area = (90/360) × πr² = ¼ × (22/7) × 21² = 346.5 sq.mCommon Mistakes to Avoid in Exercise 9.2
- Using diameter instead of radius: The formula πr² always uses the radius. If a question gives the diameter, divide by 2 first before squaring.
- Forgetting the ½ or ¼ for semicircles and quadrants: A semicircle's area is ½πr², and a quadrant's area is ¼πr² — leaving these factors out gives an answer 2 or 4 times too large.
- Mixing up R and r in ring problems: Always assign R to the larger (outer) circle and r to the smaller (inner) circle — reversing them gives a negative area.
- Not simplifying fractions before multiplying: With π = 22/7, look for opportunities to cancel the 7 in the denominator with a multiple of 7 in the radius before multiplying out — this avoids large intermediate numbers and arithmetic slips.
- Forgetting unit conversion for real-world problems: As seen in earlier exercises, always check whether all given measurements are in the same unit before combining them in a formula.
- Assuming a sector's angle without checking the figure: Sector angles are not always 90° or 180° — always verify the angle given in the problem (like the 60° sectors in Question 7) before applying the (x/360) fraction.
What Exercise 9.2 Prepares You For
This exercise rounds out the area toolkit for Chapter 9 by adding circular shapes to the straight-edged shapes covered in Exercise 9.1. Together, these two exercises let you find the area of almost any composite figure — whether it's made of triangles, trapeziums, rectangles, circles, semicircles, or sectors.
These ideas reappear in Class 9 and Class 10 mensuration chapters dealing with surface areas and volumes of solids like cylinders, cones, and spheres — all of which rely on the same circle and sector area formulas, simply applied to curved surfaces instead of flat ones. A strong grasp of πr² and sector proportions here will make those later topics significantly easier.