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

Grade 8Cambridge (IGCSE)

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

🔑Concepts

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The perimeter is the total distance around the outside of a 2D shape. For a circle, this distance is called the circumference. For rectilinear shapes, add all side lengths together.

Diagram of a rectangle showing length and width for perimeter calculation.
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The area of a trapezium is calculated using the average of the two parallel sides multiplied by the perpendicular height: A=12(a+b)hA = \frac{1}{2}(a+b)h. The height must be perpendicular to the parallel bases.

Diagram of a trapezium with parallel sides a and b and height h.
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Circles are defined by their radius (rr) or diameter (dd). The circumference C=2πrC = 2\pi r and the area A=πr2A = \pi r^2. A diameter is twice the radius: d=2rd = 2r.

Circle showing the radius from the center to the edge.
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Sectors are fractions of a circle. The arc length and sector area are proportional to the central angle θ\theta out of 360∘360^{\circ}.

📐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.

Problem 4:

A flower bed is in the shape of a sector of a circle with a radius of 6 m6\text{ m} and a central angle of 120∘120^{\circ}. Calculate the area of the flower bed. (Give your answer in terms of π\pi)

A sector with a 120 degree angle and 6m radius.

Solution:

Area=θ360×πr2\text{Area} = \frac{\theta}{360} \times \pi r^2 Area=120360×π×62\text{Area} = \frac{120}{360} \times \pi \times 6^2 Area=13×36π\text{Area} = \frac{1}{3} \times 36\pi Area=12π m2\text{Area} = 12\pi \text{ m}^2

Explanation:

Substitute the given radius r=6r = 6 and angle θ=120\theta = 120 into the sector area formula. Simplify the fraction 120360\frac{120}{360} to 13\frac{1}{3} and calculate 62=366^2 = 36 to find the final area.

Problem 5:

Calculate the area of the composite shape consisting of a rectangle of 10 cm10\text{ cm} by 4 cm4\text{ cm} with a semi-circle removed from one of the shorter sides.

A rectangle with a semi-circle cut out from the left side.

Solution:

Area of Rectangle=10×4=40 cm2\text{Area of Rectangle} = 10 \times 4 = 40 \text{ cm}^2 Radius of semi-circle=4÷2=2 cm\text{Radius of semi-circle} = 4 \div 2 = 2 \text{ cm} Area of semi-circle=12×π×22=2π≈6.28 cm2\text{Area of semi-circle} = \frac{1}{2} \times \pi \times 2^2 = 2\pi \approx 6.28 \text{ cm}^2 Total Area=40−6.28=33.72 cm2\text{Total Area} = 40 - 6.28 = 33.72 \text{ cm}^2

Explanation:

Find the area of the full rectangle first. The diameter of the semi-circle is equal to the width of the rectangle (4 cm4\text{ cm}), so the radius is 2 cm2\text{ cm}. Calculate the area of the semi-circle and subtract it from the rectangle's area.