Area of 2D shapes review worksheet | KS3-4 maths | Teachit - Free Printable

Explanation:
We are told that each shape has an area (or shaded area) of 48 cm², and we need to find the missing length labeled a (and for shape 8, also b). We’ll go one by one.

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1. Rectangle
Length = 12 cm, width = a cm
Area = length × width = 12 × a
Set equal to 48:
12a = 48 → a = 48 ÷ 12 = 4

✔ a = 4 cm

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2. Right triangle
Base = a cm, height = 6 cm
Area = (1/2) × base × height = (1/2) × a × 6 = 3a
Set equal to 48:
3a = 48 → a = 48 ÷ 3 = 16

✔ a = 16 cm

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3. Triangle with base = a cm, height = 16 cm
Area = (1/2) × base × height = (1/2) × a × 16 = 8a
Set equal to 48:
8a = 48 → a = 48 ÷ 8 = 6

✔ a = 6 cm

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4. Trapezoid
Parallel sides: 7 cm and 9 cm
Height = a cm
Area = (1/2) × (sum of parallel sides) × height = (1/2) × (7 + 9) × a = (1/2) × 16 × a = 8a
Set equal to 48:
8a = 48 → a = 6

✔ a = 6 cm

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5. Trapezoid
Parallel sides: 10 cm and a cm
Height = 4 cm
Area = (1/2) × (10 + a) × 4 = 2 × (10 + a) = 20 + 2a
Set equal to 48:
20 + 2a = 48
2a = 28
a = 14

✔ a = 14 cm

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6. Rectangle
Length = 3a cm, width = a cm
Area = 3a × a = 3a²
Set equal to 48:
3a² = 48 → a² = 16 → a = √16 = 4
(Lengths are positive, so we take positive root)

✔ a = 4 cm

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7. Square with side a cm, but corners cut off — wait!
Look carefully: The shape is a square of side a cm, with two small squares (one at top-left, one at bottom-right) cut out — but the problem says “shaded area” is 48 cm². However, the diagram shows the *entire* shape (including the cut-out corners?) — but actually, in standard problems like this, if it's drawn as a square with two small squares removed from corners, and the *shaded area* is the main shape (i.e., the big square minus the two small squares), we’d need more info.

But looking again: The shape looks like a square of side a cm, and the two small squares at corners are just markings (right angles), not cutouts — they’re just indicating the corners are right angles. So it’s just a square of side a cm.

So area = a² = 48
→ a = √48 = √(16×3) = 4√3 ≈ 4 × 1.732 = 6.928 → round to 1 decimal place: 6.9

Wait — but the instruction says: *“Calculate the exact missing length … or round your answer to 1 decimal place when necessary.”*
Since √48 is irrational, we give exact form or rounded. But in earlier problems, they used exact integers. Let’s double-check the diagram description.

Actually, re-reading: In shape 7, it's drawn as a square with side labeled "a cm" on top and right side, and two small squares (like tick marks) at top-left and bottom-right — those are just right-angle indicators (like in shapes 1 and 6), not cutouts. So yes, it's a full square.

So area = a² = 48 → a = √48 = 4√3 cm (exact), or ≈ 6.9 cm (to 1 decimal).

The question says “exact missing length … or round when necessary”. Since √48 simplifies to 4√3, that’s exact. But maybe they expect decimal? Let’s hold and check others first.

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8. L-shaped figure
We need to find a and b.
Let’s break into two rectangles:

Option 1: Top horizontal rectangle: width = 4 cm, height = a cm
Bottom horizontal rectangle: width = 10 cm, height = 2 cm

But wait — the vertical side on left is labeled b cm, and the total height is b cm. The top part has height a, and bottom part has height 2, so:
b = a + 2

Now area = area of top rectangle + area of bottom rectangle
Top rectangle: width = 4 cm, height = a cm → area = 4a
Bottom rectangle: width = 10 cm, height = 2 cm → area = 20
Total area = 4a + 20 = 48
→ 4a = 28 → a = 7
Then b = a + 2 = 9

✔ a = 7 cm, b = 9 cm

Double-check: Is there overlap? No — the L-shape is made of two non-overlapping rectangles: top-left 4×7 and bottom 10×2. But wait — the bottom rectangle extends under the top one? Actually, in an L-shape like this, the bottom rectangle is full width 10 cm, and the top rectangle sits on the left part, width 4 cm, so no overlap. Yes, correct.

Area = 4×7 + 10×2 = 28 + 20 = 48 ✔

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Now back to shape 7: Square of side a, area = a² = 48
Exact value: a = √48 = √(16×3) = 4√3 cm
If they want decimal to 1 place: √48 ≈ 6.928 → 6.9

But let’s see if any other shape required rounding — only this one does. Others gave integers.

The instruction says: “round your answer to 1 decimal place when necessary.” So for shape 7, we should give 6.9.

However, sometimes exam boards accept exact surd. But since they said “use π as 3.14”, implying decimals are preferred, and “round when necessary”, and 4√3 is not a decimal, I think they expect 6.9.

Let me verify √48:
6.9² = 47.61
7.0² = 49
So 6.93² = (6.9 + 0.03)² = 47.61 + 2×6.9×0.03 + 0.0009 = 47.61 + 0.414 + 0.0009 = 48.0249 → too high
6.928² ≈ 48.00 → yes, so 6.9 is correct to 1 decimal.

So final answers:

1. a = 4
2. a = 16
3. a = 6
4. a = 6
5. a = 14
6. a = 4
7. a = 6.9
8. a = 7, b = 9

All checked.

Final Answer:
1. 4 cm
2. 16 cm
3. 6 cm
4. 6 cm
5. 14 cm
6. 4 cm
7. 6.9 cm
8. a = 7 cm, b = 9 cm