Quarter Circle: Perimeter

1. The curved arc = (1/4) × 2πr = πr/2
2. Two straight sides (both equal to r) = r + r = 2r

$$P_{\text{quarter circle}} = \frac{\pi r}{2} + 2r$$

Quarter Circle: Area

$$A_{\text{quarter circle}} = \frac{\pi r^2}{4}$$

Worked Examples

Example 1 — Circumference (Basic)

Find the circumference of a circle with diameter 14 cm. (Take π = 22/7)

Step 1: The diameter is given and 14 is a multiple of 7, so use π = 22/7 and C = πd.

$$C = \pi d = \frac{22}{7} \times 14$$

Step 2: Cancel: 14 ÷ 7 = 2

$$C = 22 \times 2 = 44 \text{ cm}$$

Answer: 44 cm

Example 2 — Semicircle Perimeter and Area

A semicircle has a diameter of 20 cm. Find its perimeter and area. (Take π = 3.14)

Step 1: Find the radius.

$$r = 20 \div 2 = 10 \text{ cm}$$

Step 2: Perimeter of semicircle = curved part + straight part

$$P = \pi r + 2r = 3.14 \times 10 + 2 \times 10$$ $$P = 31.4 + 20 = 51.4 \text{ cm}$$

Step 3: Area of semicircle

$$A = \frac{\pi r^2}{2} = \frac{3.14 \times 10 \times 10}{2} = \frac{314}{2} = 157 \text{ cm}^2$$

Answer: Perimeter = 51.4 cm, Area = 157 cm²

Example 3 — Composite Figure (Rectangle + Semicircle)

The figure below shows a rectangle with a semicircle attached to one end. The rectangle is 16 cm long and 8 cm wide. Find the area and perimeter of the whole figure. (Take π = 3.14)

        ________
       |        |
       |        |  8 cm
  8 cm |        |
       |________|
           ↑
       semicircle on this end (diameter = 8 cm)
       (total length = 16 cm)

Imagine a rectangle 16 cm × 8 cm, with a semicircle on one short end (so the semicircle’s diameter = 8 cm).

Step 1: Find the radius of the semicircle.

$$r = 8 \div 2 = 4 \text{ cm}$$

Step 2: Area of rectangle

$$A_{\text{rect}} = 16 \times 8 = 128 \text{ cm}^2$$

Step 3: Area of semicircle

$$A_{\text{semi}} = \frac{\pi r^2}{2} = \frac{3.14 \times 4 \times 4}{2} = \frac{50.24}{2} = 25.12 \text{ cm}^2$$

Step 4: Total area

$$A_{\text{total}} = 128 + 25.12 = 153.12 \text{ cm}^2$$

Step 5: Perimeter — identify all outer edges:

• Two long sides of rectangle: 16 cm + 16 cm = 32 cm
• One short side of rectangle (the end WITHOUT the semicircle): 8 cm
• The curved edge of the semicircle (NOT the diameter — that is shared with the rectangle): πr = 3.14 × 4 = 12.56 cm

Note: Do NOT include the diameter of the semicircle in the perimeter, because it is the inside join — it is already the edge of the rectangle.

$$P = 32 + 8 + 12.56 = 52.56 \text{ cm}$$

Answer: Area = 153.12 cm², Perimeter = 52.56 cm

Example 4 — Shaded Region (Circle inside Square)

A square has side 14 cm. A circle is drawn inside it, touching all four sides. Find the area of the shaded region (the corners outside the circle). (Take π = 22/7)

Step 1: The circle fits exactly inside the square, so the diameter = 14 cm.

$$r = 14 \div 2 = 7 \text{ cm}$$

Step 2: Area of square

$$A_{\text{sq}} = 14 \times 14 = 196 \text{ cm}^2$$

Step 3: Area of circle

$$A_{\text{circle}} = \pi r^2 = \frac{22}{7} \times 7 \times 7 = \frac{22}{7} \times 49 = 22 \times 7 = 154 \text{ cm}^2$$

Step 4: Shaded area = Square − Circle

$$A_{\text{shaded}} = 196 - 154 = 42 \text{ cm}^2$$

Answer: 42 cm²

Example 5 — Quarter Circle Composite

The figure shows a square of side 10 cm with a quarter circle cut from one corner. Find the area of the remaining shape. (Take π = 3.14)

Step 1: Area of full square

$$A_{\text{sq}} = 10 \times 10 = 100 \text{ cm}^2$$

Step 2: The quarter circle has radius = 10 cm (its two straight sides run along the sides of the square).

$$A_{\text{quarter}} = \frac{\pi r^2}{4} = \frac{3.14 \times 10 \times 10}{4} = \frac{314}{4} = 78.5 \text{ cm}^2$$

Step 3: Remaining area = Square − Quarter circle

$$A = 100 - 78.5 = 21.5 \text{ cm}^2$$

Answer: 21.5 cm²

Composite Figures — Strategy

When a figure combines circles (or parts of circles) with rectangles, triangles, or other shapes:

1. Break it apart. Identify each separate shape.
2. Decide: add or subtract?
   • If a shape is added on → add its area/perimeter portion.
   • If a shape is cut out / removed → subtract its area.
3. Be careful with perimeters. Only count edges that are on the outside boundary. Shared inner edges are NOT part of the perimeter.
4. Label everything before calculating.

Shaded Region Problems — Strategy

$$A_{\text{shaded}} = A_{\text{outer shape}} - A_{\text{inner shape(s)}}$$

Steps:

1. Find the area of the larger outer shape.
2. Find the area of the smaller inner shape(s) that is/are NOT shaded.
3. Subtract.

Common Exam Traps and Mistakes

Read this section carefully — these mistakes cost marks every year.

Trap 1: Forgetting the diameter in the semicircle perimeter

The perimeter of a semicircle is πr + 2r, NOT just πr.

Students who write P = πr are forgetting the straight flat edge. The diameter is a boundary edge — always include it.

Trap 2: Using diameter instead of radius in the area formula

The area formula is A = πr². If the question gives you a diameter of 10 cm, you must use r = 5 cm:

• WRONG: A = π × 10² = 314 cm² (this is 4× too large!)
• RIGHT: A = π × 5² = 78.5 cm²

Trap 3: Including interior edges in the perimeter

When a semicircle sits on top of a rectangle, the flat edge of the semicircle is shared with the rectangle. It is an interior join — do NOT count it twice in the perimeter.

Trap 4: Using the wrong value of π

• 22/7 gives exact fractions when radius is a multiple of 7. Use 3.14 in other cases, or follow the question.
• Never mix 3.14 and 22/7 in the same question.

Trap 5: Forgetting to halve the diameter to get the radius

Always write down r = d ÷ 2 as your first step when given the diameter. This one small step prevents a very common error.

Trap 6: Wrong formula for arc length vs. area

• Circumference/arc involves r to the power 1: C = 2πr
• Area involves r squared: A = πr²

If your answer for area involves just “r” (not r²), something is wrong.

Trap 7: Quarter-circle perimeter — forgetting both straight sides

The perimeter of a quarter circle = πr/2 + 2r (two radius sides, not one).

Quick Reference Table

Values of π:

• π ≈ 3.14 (general use)
• π ≈ 22/7 (when r or d is a multiple of 7)

Key relationship: d = 2r, so r = d/2

Practice Questions

Multiple Choice Questions (MCQ)

Q1. A circle has a radius of 7 cm. What is its circumference? (Take π = 22/7)

A) 22 cm B) 44 cm C) 154 cm D) 88 cm

Q2. A semicircle has a radius of 5 cm. What is its perimeter? (Take π = 3.14)

A) 15.7 cm B) 25.7 cm C) 31.4 cm D) 39.25 cm

Q3. A circle has a diameter of 10 cm. What is its area? (Take π = 3.14)

A) 31.4 cm² B) 78.5 cm² C) 314 cm² D) 157 cm²

Q4. A quarter circle has a radius of 14 cm. What is its area? (Take π = 22/7)

A) 44 cm² B) 154 cm² C) 308 cm² D) 616 cm²

Open-Ended Questions

Q5. A circular pond has a diameter of 21 m. Find the circumference and area of the pond. (Take π = 22/7)

Q6. A window is shaped like a rectangle with a semicircle on top. The rectangle is 60 cm wide and 80 cm tall. The semicircle sits on the top edge of the rectangle (diameter = 60 cm). Find the total area of the window. (Take π = 3.14)

Q7. The figure shows a large circle of radius 14 cm with a small circle of radius 7 cm cut out from the centre. Find the area of the shaded region (the ring shape). (Take π = 22/7)

Q8. A square piece of paper has a side of 20 cm. Four quarter circles, each with radius 10 cm, are drawn at the four corners. Find the total area of the shaded regions outside the quarter circles but inside the square. (Take π = 3.14)

Q9. A running track consists of a rectangle 100 m long and 40 m wide, with a semicircle on each short end (each semicircle has diameter 40 m). Find the total perimeter of the track. (Take π = 3.14)

Q10. A logo is made from a square card of side 28 cm. A quarter circle of radius 28 cm is drawn from one corner of the square, as shown. Find the area of the remaining part of the card (the part of the square that is not covered by the quarter circle). (Take π = 22/7)

Answers to Practice Questions

Q1. B — 44 cm C = 2πr = 2 × (22/7) × 7 = 2 × 22 = 44 cm

Q2. B — 25.7 cm P = πr + 2r = 3.14 × 5 + 2 × 5 = 15.7 + 10 = 25.7 cm

Q3. B — 78.5 cm² r = 10 ÷ 2 = 5 cm; A = πr² = 3.14 × 5 × 5 = 78.5 cm²

Q4. B — 154 cm² A = πr²/4 = (22/7) × 14 × 14 ÷ 4 = (22/7) × 196 ÷ 4 = (22 × 28) ÷ 4 = 616 ÷ 4 = 154 cm²

Q5.

• Circumference: C = πd = (22/7) × 21 = 66 m
• Area: r = 21 ÷ 2 = 10.5 m; A = πr² = (22/7) × 10.5 × 10.5 = (22/7) × 110.25 = 346.5 m²

Q6.

• r = 60 ÷ 2 = 30 cm
• Area of rectangle = 60 × 80 = 4800 cm²
• Area of semicircle = πr²/2 = 3.14 × 30 × 30 ÷ 2 = 2826 ÷ 2 = 1413 cm²
• Total area = 4800 + 1413 = 6213 cm²

Q7.

• Area of large circle = πR² = (22/7) × 14 × 14 = (22/7) × 196 = 616 cm²
• Area of small circle = πr² = (22/7) × 7 × 7 = (22/7) × 49 = 154 cm²
• Shaded area = 616 − 154 = 462 cm²

Q8.

• Area of square = 20 × 20 = 400 cm²
• Total area of 4 quarter circles = 1 full circle = πr² = 3.14 × 10 × 10 = 314 cm²
• Shaded area = 400 − 314 = 86 cm²

Q9.

• Two straight sides of rectangle: 100 + 100 = 200 m
• Two semicircles (diameter 40 m each) together form one full circle: C = πd = 3.14 × 40 = 125.6 m
• Total perimeter = 200 + 125.6 = 325.6 m

Q10.

• Area of square = 28 × 28 = 784 cm²
• The quarter circle has radius 28 cm and is drawn from one corner, so it lies entirely within the square.
• Area of quarter circle = πr² ÷ 4 = (22/7) × 28 × 28 ÷ 4 = (22/7) × 784 ÷ 4 = (22 × 112) ÷ 4 = 2464 ÷ 4 = 616 cm²
• Remaining area (square − quarter circle) = 784 − 616 = 168 cm²

Note for Q10: A quarter circle of radius equal to the square’s side, drawn from a corner, always fits exactly within the square. The remaining part of the card is the square minus the quarter circle. In the actual exam, always confirm from the diagram which region is shaded.

Summary — Things to Remember for Exam Day

1. d = 2r → always halve the diameter before using the area formula
2. Circumference = πd or 2πr (use whichever form matches the given information)
3. Area = πr² (radius only, squared)
4. Semicircle perimeter = curved part (πr) + flat part (2r) — never forget the diameter
5. Quarter circle perimeter = curved arc (πr/2) + two straight sides (2r)
6. Use 22/7 when r or d is a multiple of 7; use 3.14 otherwise
7. Shaded region = outer area − inner area
8. Composite perimeter: only count edges on the outer boundary
9. Four quarter circles = one full circle (useful shortcut)
10. Write r = ___ and show all working steps for method marks