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Mensuration - Perimeter and Area of 2D Shapes (including Trapeziums and Circles)

Grade 8IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Perimeter is the total distance around the boundary of a 2D shape, measured in linear units (mm, cm, m).

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Area is the measure of the surface covered by a 2D shape, measured in square units (mm², cm², m²).

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The constant π\pi (Pi) represents the ratio of a circle's circumference to its diameter, approximately 3.142 or 22/7.

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A Trapezium is a quadrilateral with at least one pair of parallel sides.

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Compound shapes should be decomposed into simpler polygons (like rectangles and triangles) to calculate total area.

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To convert units of area, remember the square factor: 1 cm2=100 mm21 \text{ cm}^2 = 100 \text{ mm}^2 (since 10×10=10010 \times 10 = 100).

📐Formulae

Rectangle: Area=l×w\text{Area} = l \times w, Perimeter=2(l+w)\text{Perimeter} = 2(l + w)

Triangle: Area=12×b×h\text{Area} = \frac{1}{2} \times b \times h

Parallelogram: Area=b×h\text{Area} = b \times h

Trapezium: Area=12(a+b)h\text{Area} = \frac{1}{2}(a + b)h (where aa and bb are parallel sides)

Circle Circumference: C=2πrC = 2\pi r or C=πdC = \pi d

Circle Area: A=πr2A = \pi r^2

Arc Length: L=θ360×2πrL = \frac{\theta}{360} \times 2\pi r

Sector Area: A=θ360×πr2A = \frac{\theta}{360} \times \pi r^2

💡Examples

Problem 1:

Calculate the area of a trapezium where the parallel sides are 8 cm and 12 cm, and the perpendicular height is 5 cm.

Solution:

Area=12(8+12)×5=12(20)×5=10×5=50 cm2\text{Area} = \frac{1}{2}(8 + 12) \times 5 = \frac{1}{2}(20) \times 5 = 10 \times 5 = 50 \text{ cm}^2

Explanation:

Identify the parallel sides a=8a=8 and b=12b=12, and height h=5h=5. Plug these into the trapezium area formula 12(a+b)h\frac{1}{2}(a+b)h.

Problem 2:

Find the circumference and area of a circle with a radius of 7 cm. (Use π=227\pi = \frac{22}{7})

Solution:

C=2×227×7=44 cmC = 2 \times \frac{22}{7} \times 7 = 44 \text{ cm}; A=227×72=227×49=22×7=154 cm2A = \frac{22}{7} \times 7^2 = \frac{22}{7} \times 49 = 22 \times 7 = 154 \text{ cm}^2

Explanation:

Use the formulas C=2πrC = 2\pi r for circumference and A=πr2A = \pi r^2 for area. Substituting r=7r=7 and π=22/7\pi=22/7 allows for easy cancellation.

Problem 3:

A semi-circle has a diameter of 10 cm. Find its total perimeter.

Solution:

Arc length=12×π×10=5π≈15.71 cm\text{Arc length} = \frac{1}{2} \times \pi \times 10 = 5\pi \approx 15.71 \text{ cm}. Total Perimeter=15.71+10=25.71 cm\text{Total Perimeter} = 15.71 + 10 = 25.71 \text{ cm}.

Explanation:

The perimeter of a semi-circle consists of the curved arc (half the circumference) PLUS the straight diameter. Failing to add the diameter is a common mistake.