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Mensuration - Compound shapes

Grade 11IGCSE

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

🔑Concepts

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Decomposition: Breaking down a complex shape into simpler constituent shapes (rectangles, triangles, circles, etc.).

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Subtraction Method: Calculating the area or volume of a larger 'bounding' shape and subtracting the 'empty' or cut-out sections.

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Internal Boundaries: When calculating the perimeter of 2D compound shapes or the surface area of 3D compound solids, internal edges where shapes meet must not be counted.

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Unit Consistency: Ensuring all measurements are in the same units (e.g., all in cm or all in m) before performing calculations.

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Cross-sectional Area: For compound prisms, identifying the uniform cross-section is key to finding the volume (V=Areacross−sectionimeslengthV = Area_{cross-section} imes length).

📐Formulae

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

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

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

Volume of a Cylinder: V=πr2hV = \pi r^2 h

Volume of a Cone: V=13πr2hV = \frac{1}{3} \pi r^2 h

Volume of a Sphere: V=43πr3V = \frac{4}{3} \pi r^3

Surface Area of a Sphere: A=4πr2A = 4\pi r^2

Curved Surface Area of a Cone: A=πrlA = \pi r l (where ll is slant height)

💡Examples

Problem 1:

A compound 2D shape consists of a rectangle with dimensions 10cm by 6cm, with a semi-circle removed from one of the 6cm sides. Find the total area of the remaining shape.

Solution:

Area of Rectangle = 10×6=60 cm210 \times 6 = 60 \text{ cm}^2. Radius of semi-circle = 6/2=3 cm6 / 2 = 3 \text{ cm}. Area of Semi-circle = 12π(32)=4.5π≈14.14 cm2\frac{1}{2} \pi (3^2) = 4.5\pi \approx 14.14 \text{ cm}^2. Total Area = 60−14.14=45.86 cm260 - 14.14 = 45.86 \text{ cm}^2.

Explanation:

This problem uses the subtraction method. We calculate the area of the full rectangle first, then identify the radius of the missing semi-circle (which is half the side length) and subtract it from the total.

Problem 2:

A solid toy is made by joining a cone to the flat face of a hemisphere. Both the cone and the hemisphere have a radius of 5cm. The slant height of the cone is 13cm. Calculate the total surface area of the toy.

Solution:

Curved Surface Area (CSA) of Cone = π×5×13=65π\pi \times 5 \times 13 = 65\pi. CSA of Hemisphere = 12(4π×52)=50π\frac{1}{2}(4 \pi \times 5^2) = 50\pi. Total Surface Area = 65π+50π=115π≈361.28 cm265\pi + 50\pi = 115\pi \approx 361.28 \text{ cm}^2.

Explanation:

When two solids are joined, the faces that touch (the circular bases) are no longer on the 'outside'. Therefore, the total surface area is the sum of the curved surface area of the cone and the curved surface area of the hemisphere only.

Problem 3:

A metal trough is 2m long. Its cross-section is a semi-circle with a diameter of 40cm. Find the volume of the trough in cubic centimeters (cm3cm^3).

Solution:

Convert length to cm: 2m=200cm2\text{m} = 200\text{cm}. Radius r=20cmr = 20\text{cm}. Area of cross-section (semi-circle) = 12π(202)=200π cm2\frac{1}{2} \pi (20^2) = 200\pi \text{ cm}^2. Volume = 200π×200=40,000π≈125,663.71 cm3200\pi \times 200 = 40,000\pi \approx 125,663.71 \text{ cm}^3.

Explanation:

The trough is a prism with a semi-circular cross-section. We first ensure units are consistent (converting meters to centimeters), calculate the area of the semi-circle, and multiply by the length of the trough.