Area & Circumference of Circles worksheet with problems involving calculations for full circles, fractions of circles, composite shapes, and missing lengths.

Let’s solve each part step by step. We’ll use these formulas:

- Circumference of a circle = π × diameter (or 2 × π × radius)
- Area of a circle = π × radius²
- For fractions of circles (like semicircles or quarter-circles), we take the fraction of the full area and add straight edges for perimeter.
- For composite shapes (circles + rectangles), break them into parts, calculate each, then add.

We’ll use π ≈ 3.14 unless told otherwise.

---

Section 1: Area and Circumference of Circles

#### a) Radius = 3 cm
Area = π × r² = 3.14 × 3² = 3.14 × 9 = 28.26 cm²
Circumference = 2 × π × r = 2 × 3.14 × 3 = 18.84 cm

#### b) Diameter = 7 cm → Radius = 3.5 cm
Area = 3.14 × (3.5)² = 3.14 × 12.25 = 38.465 cm²
Circumference = π × d = 3.14 × 7 = 21.98 cm

#### c) Radius = 2 cm
Area = 3.14 × 4 = 12.56 cm²
Circumference = 2 × 3.14 × 2 = 12.56 cm

#### d) Diameter = 5 cm → Radius = 2.5 cm
Area = 3.14 × (2.5)² = 3.14 × 6.25 = 19.625 cm²
Circumference = 3.14 × 5 = 15.7 cm

#### e) Diameter = 3 cm → Radius = 1.5 cm
Area = 3.14 × (1.5)² = 3.14 × 2.25 = 7.065 cm²
Circumference = 3.14 × 3 = 9.42 cm

#### f) Radius = 1 cm
Area = 3.14 × 1 = 3.14 cm²
Circumference = 2 × 3.14 × 1 = 6.28 cm

#### g) Radius = 6.5 cm
Area = 3.14 × (6.5)² = 3.14 × 42.25 = 132.665 cm²
Circumference = 2 × 3.14 × 6.5 = 40.82 cm

#### h) Diameter = 4.5 cm → Radius = 2.25 cm
Area = 3.14 × (2.25)² = 3.14 × 5.0625 = 15.89625 cm²
Circumference = 3.14 × 4.5 = 14.13 cm

---

Section 2: Area and Perimeter of Fractional Circle Shapes

Perimeter includes curved part + any straight edges.

#### a) Semicircle, diameter = 10 cm → radius = 5 cm
Curved part = half circumference = (π × 10)/2 = 15.7 cm
Straight edge = 10 cm
→ Perimeter = 15.7 + 10 = 25.7 cm
Area = half circle = (π × 5²)/2 = (78.5)/2 = 39.25 cm²

#### b) Quarter-circle, radius = 6 cm
Curved part = ¼ × 2πr = ¼ × 2×3.14×6 = ¼ × 37.68 = 9.42 cm
Two straight edges = 6 + 6 = 12 cm
→ Perimeter = 9.42 + 12 = 21.42 cm
Area = ¼ × π × 6² = ¼ × 113.04 = 28.26 cm²

#### c) Three-quarters circle, radius = 5 cm
Curved part = ¾ × 2πr = ¾ × 31.4 = 23.55 cm
Two straight edges = 5 + 5 = 10 cm
→ Perimeter = 23.55 + 10 = 33.55 cm
Area = ¾ × π × 25 = ¾ × 78.5 = 58.875 cm²

#### d) Quarter-circle, radius = 8 cm
Curved part = ¼ × 2π×8 = ¼ × 50.24 = 12.56 cm
Two straight edges = 8 + 8 = 16 cm
→ Perimeter = 12.56 + 16 = 28.56 cm
Area = ¼ × π × 64 = ¼ × 200.96 = 50.24 cm²

#### e) Three-quarters circle, radius = 2 cm
Curved part = ¾ × 2π×2 = ¾ × 12.56 = 9.42 cm
Two straight edges = 2 + 2 = 4 cm
→ Perimeter = 9.42 + 4 = 13.42 cm
Area = ¾ × π × 4 = ¾ × 12.56 = 9.42 cm²

#### f) Semicircle, diameter = 12 cm → radius = 6 cm
Curved part = ½ × π × 12 = 18.84 cm
Straight edge = 12 cm
→ Perimeter = 18.84 + 12 = 30.84 cm
Area = ½ × π × 36 = 56.52 cm²

#### g) Quarter-circle, radius = 5.5 cm
Curved part = ¼ × 2π×5.5 = ¼ × 34.54 = 8.635 cm
Two straight edges = 5.5 + 5.5 = 11 cm
→ Perimeter = 8.635 + 11 = 19.635 cm
Area = ¼ × π × (5.5)² = ¼ × π × 30.25 = ¼ × 94.985 = 23.74625 cm²

#### h) Three-quarters circle, radius = 2.5 cm
Curved part = ¾ × 2π×2.5 = ¾ × 15.7 = 11.775 cm
Two straight edges = 2.5 + 2.5 = 5 cm
→ Perimeter = 11.775 + 5 = 16.775 cm
Area = ¾ × π × 6.25 = ¾ × 19.625 = 14.71875 cm²

---

Section 3: Composite Shapes (Circles + Rectangles)

#### a) Rectangle 6cm x 2cm + semicircle on top (diameter = 6cm → radius = 3cm)
Rectangle area = 6 × 2 = 12 cm²
Semicircle area = ½ × π × 9 = 14.13 cm²
Total area = 12 + 14.13 = 26.13 cm²

#### b) Rectangle 5cm x 1cm + two semicircles (left and right, diameter = 5cm → radius = 2.5cm)
Actually, this is a rectangle with two half-circles → makes one full circle!
Circle area = π × (2.5)² = 19.625 cm²
Rectangle area = 5 × 1 = 5 cm²
Total area = 19.625 + 5 = 24.625 cm²

Wait — looking again: The shape has a vertical dashed line labeled 5cm, and horizontal 1cm at top. Actually, it looks like a rectangle 5cm tall and 1cm wide, with two semicircles on left and right? But that would make total width = 1 + 2.5 + 2.5 = 6cm? Hmm.

Actually, re-examining: It's probably a rectangle 5cm high and 1cm wide, with two semicircles attached to the sides — but if they’re attached to the 5cm sides, then diameter = 5cm → radius = 2.5cm. So yes, two semicircles = one full circle of radius 2.5cm.

So area = rectangle (5×1=5) + circle (π×2.5²=19.625) = 24.625 cm²

But let me check diagram description: “b) 1cm on top, 5cm vertical dashed” — likely means the rectangle is 1cm wide and 5cm tall, and semicircles are on the left and right ends (so diameter = 5cm). Yes.

#### c) Rectangle 6cm x 5cm + quarter-circle on top-right corner (radius = 5cm)
Rectangle area = 6 × 5 = 30 cm²
Quarter-circle area = ¼ × π × 25 = 19.625 cm²
Total area = 30 + 19.625 = 49.625 cm²

Wait — actually, looking at the shape: it’s a rectangle 6cm wide and 5cm tall, with a quarter-circle cut out? Or added? The curve is on the top-right, bulging outward — so added.

Yes, so total area = rectangle + quarter-circle = 30 + 19.625 = 49.625 cm²

#### d) Two quarter-circles and a square? Let’s see:
It shows a square 4cm x 4cm, with a quarter-circle on bottom-left and another on top-right? Actually, it looks like two quarter-circles forming a half-circle, plus a rectangle?

Wait — better interpretation: The shape is made of a 4x4 square, with a quarter-circle attached to the bottom-left (radius 4cm) and another quarter-circle attached to the top-right (also radius 4cm). But that might overlap? No — actually, looking at the dashed lines, it seems like the entire shape is composed of:

- A central square 4x4
- Plus a quarter-circle on the bottom-left (outside the square)
- Plus a quarter-circle on the top-right (outside the square)

But that doesn’t match the drawing. Alternatively, it might be a rectangle 8cm wide and 4cm tall, with two quarter-circles on the ends? Wait — no.

Actually, standard interpretation: This is a "stadium" shape but rotated? Or perhaps it’s two quarter-circles and a rectangle in between.

Looking carefully: There’s a vertical dashed line splitting it, and horizontal dashed line. The shape has:

- Bottom-left: quarter-circle radius 4cm
- Top-right: quarter-circle radius 4cm
- And a rectangle connecting them? Actually, the total shape is symmetric.

Better way: The shape can be seen as a rectangle 4cm x 8cm? No.

Wait — actually, it’s equivalent to a full circle of radius 4cm plus a rectangle 4cm x 4cm? Let me think.

No — here’s correct breakdown:

The shape consists of:
- A rectangle 4cm (width) x 4cm (height) in the center? Not quite.

Actually, from the diagram: It appears to be two quarter-circles (each radius 4cm) placed such that their straight edges form a sort of L-shape, but connected by a rectangle? I think I need to reinterpret.

Alternative approach: The shape is made by taking a 4x4 square, and attaching a quarter-circle to the bottom side and another to the right side — but that would create an irregular shape.

Wait — looking at common problems: This is likely a shape formed by a rectangle 4cm x 8cm with two semicircles? No.

Actually, let’s count areas directly:

From the diagram: There is a region that is a quarter-circle in the bottom-left (area = ¼ π r² = ¼ π 16 = 4π), and another quarter-circle in the top-right (same area), and a rectangle in the middle that is 4cm x 4cm? But how are they arranged?

Perhaps it’s easier: The entire shape is equivalent to a full circle of radius 4cm plus a 4x4 square? Let’s calculate:

If you have two quarter-circles, that’s half a circle: area = ½ × π × 16 = 8π ≈ 25.12 cm²

Plus the rectangular part: which is 4cm x 4cm = 16 cm²? But where is the rectangle?

Actually, upon second thought, the shape is composed of:

- A rectangle 4cm wide and 4cm tall (the inner part)
- Plus a quarter-circle attached to the bottom (extending down)
- Plus a quarter-circle attached to the right (extending right)

But then the total area would be rectangle + two quarter-circles = 16 + 2*(¼ π 16) = 16 + 8π ≈ 16 + 25.12 = 41.12 cm²

But let’s verify with another method: The bounding box is 8cm x 8cm? No.

I recall that in some textbooks, this shape is called a "rounded corner" shape, but here it’s two separate quarter-circles.

Actually, looking at the labels: There is a 4cm label on the bottom, 4cm on the right, and dashed lines indicating symmetry.

Best interpretation: The shape is made of a 4x4 square, with a quarter-circle of radius 4cm attached to the bottom side (so extending downward), and another quarter-circle of radius 4cm attached to the right side (extending rightward). So total area = square + two quarter-circles = 16 + 2*(¼ * π * 16) = 16 + 8π ≈ 16 + 25.12 = 41.12 cm²

Yes, that makes sense.

So area = 16 + 8*3.14 = 16 + 25.12 = 41.12 cm²

---

Section 4: Find Missing Lengths

#### a) Area = 100 cm² → find radius
Area = π r² = 100
r² = 100 / π ≈ 100 / 3.14 ≈ 31.847
r ≈ √31.847 ≈ 5.64 cm (let’s keep more digits: sqrt(100/3.14) = sqrt(31.84713375796178) ≈ 5.643)

But perhaps they want exact or rounded? Since others used 3.14, we’ll use that.

r = √(100 / 3.14) = √31.847 ≈ 5.64 cm

#### b) Circumference = 314 cm → find diameter
C = π d → d = C / π = 314 / 3.14 = 100 cm

#### c) Area = 55 cm² → find radius
π r² = 55
r² = 55 / 3.14 ≈ 17.5159
r ≈ √17.5159 ≈ 4.185 cm → round to 4.19 cm? But let’s compute exactly.

55 / 3.14 = 5500 / 314 = 2750 / 157 ≈ 17.51592356687898
sqrt ≈ 4.1855 → so 4.19 cm if rounding to two decimals.

But perhaps keep as is.

#### d) Circumference = 12 cm → find diameter
C = π d → d = 12 / 3.14 ≈ 3.82 cm

Now, let’s compile all answers neatly.

Final Answer:

Section 1:
a) Area: 28.26 cm², Circumference: 18.84 cm
b) Area: 38.465 cm², Circumference: 21.98 cm
c) Area: 12.56 cm², Circumference: 12.56 cm
d) Area: 19.625 cm², Circumference: 15.7 cm
e) Area: 7.065 cm², Circumference: 9.42 cm
f) Area: 3.14 cm², Circumference: 6.28 cm
g) Area: 132.665 cm², Circumference: 40.82 cm
h) Area: 15.89625 cm², Circumference: 14.13 cm

Section 2:
a) Area: 39.25 cm², Perimeter: 25.7 cm
b) Area: 28.26 cm², Perimeter: 21.42 cm
c) Area: 58.875 cm², Perimeter: 33.55 cm
d) Area: 50.24 cm², Perimeter: 28.56 cm
e) Area: 9.42 cm², Perimeter: 13.42 cm
f) Area: 56.52 cm², Perimeter: 30.84 cm
g) Area: 23.74625 cm², Perimeter: 19.635 cm
h) Area: 14.71875 cm², Perimeter: 16.775 cm

Section 3:
a) 26.13 cm²
b) 24.625 cm²
c) 49.625 cm²
d) 41.12 cm²

Section 4:
a) 5.64 cm
b) 100 cm
c) 4.19 cm
d) 3.82 cm

Note: All values calculated using π = 3.14. Some may vary slightly if more precise π is used, but for school level, 3.14 is fine.

Final Answer:
[All answers as above, compiled]