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Mensuration - Perimeter and Area of 2D Shapes

Grade 9IGCSE

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

🔑Concepts

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Definition of Perimeter: The total length of the boundary of a closed 2D figure.

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Definition of Area: The measure of the surface region enclosed by a 2D shape.

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Units of Measurement: Perimeter is measured in linear units (cm, m, km) while Area is measured in square units (cm², m², km²).

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Conversion of Units: For area, remember that 1 cm² = 100 mm² and 1 m² = 10,000 cm².

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Compound Shapes: Complex shapes can be split into simpler shapes (rectangles, triangles, circles) to calculate total area or perimeter.

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Arc Length and Sector Area: Fractions of a circle's circumference and area based on the central angle θ.

📐Formulae

Rectangle Area=l×w\text{Rectangle Area} = l \times w

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

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

Trapezium Area=12(a+b)h\text{Trapezium Area} = \frac{1}{2}(a + b)h

Circle Circumference=2πr or πd\text{Circle Circumference} = 2\pi r \text{ or } \pi d

Circle Area=πr2\text{Circle Area} = \pi r^2

Arc Length=θ360×2πr\text{Arc Length} = \frac{\theta}{360} \times 2\pi r

Sector Area=θ360×πr2\text{Sector Area} = \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\frac{1}{2}(8 + 12) \times 5 = \frac{1}{2}(20) \times 5 = 10 \times 5 = 50 \text{ cm}^2

Explanation:

Apply the trapezium area formula A=12(a+b)hA = \frac{1}{2}(a+b)h, where aa and bb are the parallel sides and hh is the height.

Problem 2:

A sector of a circle has a radius of 6 cm and a central angle of 60°. Find the length of the arc. (Use π=3.142\pi = 3.142)

Solution:

Arc Length = 60360×2×3.142×6=16×37.704=6.284 cm\frac{60}{360} \times 2 \times 3.142 \times 6 = \frac{1}{6} \times 37.704 = 6.284 \text{ cm}

Explanation:

Use the arc length formula θ360×2πr\frac{\theta}{360} \times 2\pi r. Substitute θ=60\theta = 60 and r=6r = 6.

Problem 3:

Find the perimeter of a semi-circle with a radius of 7 cm. (Take π=227\pi = \frac{22}{7})

Solution:

Perimeter = (πr)+d=(227×7)+(2×7)=22+14=36 cm(\pi r) + d = (\frac{22}{7} \times 7) + (2 \times 7) = 22 + 14 = 36 \text{ cm}

Explanation:

The perimeter of a semi-circle consists of the curved arc (half of 2πr2\pi r) plus the straight diameter (2r2r).