Circle worksheets in PDF | Printable Circle Worksheets - Free Printable

Let’s solve each problem one by one. We’ll use the formula for the area of a circle:
Area = π × r², where r is the radius.

We’ll also remember:
- A semicircle is half a circle → divide area by 2.
- A quarter circle is 1/4 of a full circle → divide area by 4.
- For shapes made of multiple parts, add or subtract areas as needed.
- Use π ≈ 3.1416 (or your calculator’s π button) and round to 2 decimal places.

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Problem 1: Semicircle with diameter 8 cm

Radius = 8 ÷ 2 = 4 cm

Full circle area = π × 4² = π × 16 ≈ 50.2655
Semicircle = 50.2655 ÷ 2 ≈ 25.13 cm²

✔ Final Answer for #1: 25.13

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Problem 2: Quarter circle with radius 4.7 mm

Area = (π × 4.7²) ÷ 4
4.7² = 22.09
π × 22.09 ≈ 69.398
Divide by 4 → 69.398 ÷ 4 ≈ 17.35 mm²

✔ Final Answer for #2: 17.35

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Problem 3: Circle with a slice missing — looks like 3/4 of a circle? Wait — let’s check.

The shape shows a circle with a 90° wedge removed → so it’s 3/4 of a full circle.

Radius = 11 m

Full circle area = π × 11² = π × 121 ≈ 380.1327
3/4 of that = 380.1327 × 0.75 ≈ 285.10 m²

✔ Final Answer for #3: 285.10

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Problem 4: Two red quarter circles forming a semicircle? Let’s see.

Each red part is a quarter circle. Together they make a semicircle.

Total height = 16 cm → that’s the diameter of each quarter circle? Actually, looking at the diagram: the vertical line spans both top and bottom quarters → so each quarter has radius = 16 ÷ 2 = 8 cm

So two quarter circles = one semicircle with radius 8 cm.

Area = (π × 8²) ÷ 2 = (π × 64) ÷ 2 = 32π ≈ 100.53 cm²

✔ Final Answer for #4: 100.53

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Problem 5: Two purple quarter circles — same idea?

They are arranged to form a semicircle again? Yes — two quarter circles = one semicircle.

Radius = 12 cm

Area = (π × 12²) ÷ 2 = (π × 144) ÷ 2 = 72π ≈ 226.19 cm²

✔ Final Answer for #5: 226.19

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Problem 6: Three yellow quarter circles? Wait — looks like three-quarters of a circle.

Actually, counting the pieces: there are 3 quarter-circles → total = 3/4 of a full circle.

Height given = 4 m → that’s the diameter? Or radius? Looking at the diagram: the vertical arrow spans from top to bottom of the shape — which includes two radii (top quarter and bottom quarter). So diameter = 4 m → radius = 2 m.

Wait — actually, in this diagram, the 4 m is labeled next to the entire height — which covers two quarter-circles stacked vertically → so each quarter-circle has radius = 4 ÷ 2 = 2 m

But we have THREE quarter-circles → total area = 3 × (quarter circle area)

Quarter circle area = (π × 2²) ÷ 4 = (π × 4) ÷ 4 = π ≈ 3.1416

Three of them = 3 × 3.1416 ≈ 9.42 m²

Alternatively: 3/4 of a full circle with radius 2 m:

Full circle = π × 4 = 12.5664
3/4 = 9.4248 → rounds to 9.42

✔ Final Answer for #6: 9.42

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Problem 7: Square with a white circle inside — shaded area is square minus circle.

Square side = 12 cm → area = 12 × 12 = 144 cm²

Circle fits perfectly inside → diameter = 12 cm → radius = 6 cm

Circle area = π × 6² = π × 36 ≈ 113.097

Shaded area = 144 - 113.097 ≈ 30.90 cm²

✔ Final Answer for #7: 30.90

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Problem 8: Shape looks like a quarter circle cut out from a square? Wait — no.

Actually, it’s a right triangle with a curved side — looks like a quarter circle sector but shaded outside? Let me think.

Looking closely: It’s a square corner (right angle), and the curve is concave inward — meaning the shaded region is the area between the straight lines and the arc — so it’s the area of the square corner MINUS the quarter circle.

Wait — actually, the shape shown is bounded by two straight lines (each 110 km long) meeting at right angle, and a circular arc connecting their ends — and the shaded region is INSIDE the arc? No — the shading is on the “corner” side — actually, re-examining:

It appears to be a quarter circle itself — because the two sides are equal (110 km) and meet at 90°, and the curve connects them — so it’s a quarter circle with radius 110 km.

Yes! That makes sense.

Area = (π × 110²) ÷ 4
110² = 12100
π × 12100 ≈ 38013.27
÷ 4 = 9503.32 km²

✔ Final Answer for #8: 9503.32

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Problem 9: Square with diagonal 15 mm — shaded regions are two segments? Let’s analyze.

First, the square has a diagonal of 15 mm. In a square, diagonal d = s√2 → so side s = d / √2 = 15 / √2 ≈ 10.6066 mm

Area of square = s² = (15/√2)² = 225 / 2 = 112.5 mm²

Now, inside the square, there’s a white lens-shaped region formed by two quarter circles? Actually, looking at the diagram: it seems like two quarter circles drawn from opposite corners, overlapping in the middle — creating a "lens" in the center that’s white, and the rest is shaded.

Actually, standard problem: when you draw two quarter circles from adjacent corners of a square, each with radius = side length, they overlap — but here the diagonal is given, not the side.

Wait — perhaps easier: the white region is the intersection of two semicircles? Or maybe it's two quarter circles centered at opposite corners?

Actually, common problem: if you have a square, and you draw a quarter circle from each of two opposite corners, each with radius = side length, then the overlapping region is the white part.

But here, the diagonal is 15 mm — so let’s find side first.

Side s = 15 / √2 ≈ 10.6066 mm

Each quarter circle has radius = s = 10.6066 mm

Area of one quarter circle = (π × s²)/4 = (π × 112.5)/4 ≈ (353.429)/4 ≈ 88.357 mm²

Two such quarter circles = 2 × 88.357 = 176.714 mm²

But they overlap in the white region — and together they cover the whole square plus the white region twice? Actually, the union of the two quarter circles covers the entire square, and the overlapping part (white) is counted twice.

So: Area of two quarter circles = Area of square + Area of white region

→ White region = 2 × (quarter circle) - square = 176.714 - 112.5 = 64.214 mm²

Then shaded area = Total square - white region = 112.5 - 64.214 = 48.286 mm²

Wait — but looking at the diagram: the shaded regions are the four corner triangles? Actually, no — in the image, the shaded parts are the two gray regions on the left and right, and the white is the lens in the middle.

Actually, another way: sometimes this is called the "vesica piscis" inside a square.

Standard formula: for a square of side s, with two quarter circles from opposite corners, the area of the overlapping region (white) is:

White area = 2 × (area of sector) - area of square

Sector is 90° sector → area = (π s²)/4

So white = 2*(π s²/4) - s² = (π s²/2) - s² = s² (π/2 - 1)

Then shaded = square - white = s² - [s² (π/2 - 1)] = s² [1 - (π/2 - 1)] = s² [2 - π/2]

Let’s compute numerically.

s² = 112.5

π/2 ≈ 1.5708

So white = 112.5 × (1.5708 - 1) = 112.5 × 0.5708 ≈ 64.215

Shaded = 112.5 - 64.215 = 48.285 mm²

Rounded to 2 decimals: 48.29 mm²

But wait — let me double-check with exact values.

s² = (15/√2)^2 = 225/2 = 112.5 exactly.

White area = s² (π/2 - 1) = 112.5 × (π/2 - 1)

Compute: π/2 ≈ 1.57079632679

Minus 1 = 0.57079632679

Times 112.5 ≈ 64.2145867639

Shaded = 112.5 - 64.2145867639 = 48.2854132361 → rounds to 48.29

✔ Final Answer for #9: 48.29

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## Final Answers:

1) 25.13
2) 17.35
3) 285.10
4) 100.53
5) 226.19
6) 9.42
7) 30.90
8) 9503.32
9) 48.29