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Mensuration - Area and Perimeter of 2D Shapes

Grade 12IGCSE

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

🔑Concepts

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Understanding Perimeter as the total boundary length and Area as the surface space within a 2D shape.

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Identification of standard polygons: Triangles, Quadrilaterals (Rectangle, Square, Parallelogram, Rhombus, Trapezium).

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Properties of Circles: Circumference, Diameter, Radius, and the use of Pi (π).

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Sector and Arc calculations based on the fractional part of a full circle (angle θ/360).

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Compound shapes: Calculating areas by dividing complex figures into simpler rectangles, triangles, or semi-circles.

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Shaded regions: Determining area by subtracting the area of internal shapes from external shapes.

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Unit conversion: Understanding the squared relationship (e.g., 1 cm2=100 mm21\text{ cm}^2 = 100\text{ mm}^2 and 1 m2=10,000 cm21\text{ m}^2 = 10,000\text{ cm}^2).

📐Formulae

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

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

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

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

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

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

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

Area of Sector=θ360×πr2\text{Area of Sector} = \frac{\theta}{360} \times \pi r^2

💡Examples

Problem 1:

Calculate the area of a trapezium where the parallel sides are 12 cm and 18 cm, and the perpendicular height is 7 cm.

Solution:

A=12(12+18)×7=12(30)×7=15×7=105 cm2A = \frac{1}{2}(12 + 18) \times 7 = \frac{1}{2}(30) \times 7 = 15 \times 7 = 105\text{ cm}^2

Explanation:

Identify the parallel sides 'a' and 'b' and the height 'h'. Substitute them into the trapezium formula: Area = 1/2(sum of parallel sides) × height.

Problem 2:

Find the perimeter of a sector with a radius of 10 cm and a central angle of 72∘72^{\circ}. (Take π=3.142\pi = 3.142)

Solution:

Arc Length=72360×2×3.142×10=12.568 cm\text{Arc Length} = \frac{72}{360} \times 2 \times 3.142 \times 10 = 12.568\text{ cm}. Perimeter=Arc Length+2r=12.568+20=32.568 cm\text{Perimeter} = \text{Arc Length} + 2r = 12.568 + 20 = 32.568\text{ cm}.

Explanation:

To find the perimeter of a sector, you must calculate the arc length first and then add the two radii that form the 'v' shape of the sector.

Problem 3:

A circular hole of radius 3 cm is cut out of a square piece of metal with side length 10 cm. Find the area of the remaining metal.

Solution:

Area of Square=10×10=100 cm2\text{Area of Square} = 10 \times 10 = 100\text{ cm}^2. Area of Circle=π×32=9π≈28.27 cm2\text{Area of Circle} = \pi \times 3^2 = 9\pi \approx 28.27\text{ cm}^2. Remaining Area=100−28.27=71.73 cm2\text{Remaining Area} = 100 - 28.27 = 71.73\text{ cm}^2.

Explanation:

This is a compound area problem involving subtraction. Calculate the total area of the outer shape (square) and subtract the area of the shape removed (circle).