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Maths708Updated 7 Sept 202618 pages

National 5 Maths Applications - Unit 3 Overview

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Want to ace your geometry tests? This unit covers everything you need to know about calculating perimeters, areas, and volumes of circles and 3D shapes. Master these formulas and you'll be solving complex problems with confidence!
National 5 Apps of Maths - Unit 3 Notes – page 1

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Circle Perimeter Basics

Ever wondered how to measure the distance around a circle? The circumference formula C = πD is your best mate here. Remember that D is the diameter, and if you're given the radius instead, just double it first.

Let's break it down with real examples. For a circle with diameter 10cm: C = π × 10 = 31.4cm. Easy! But what if you only know the radius is 25cm? First find the diameter 25×2=50cm25 × 2 = 50cm, then apply the formula: C = π × 50 = 157cm.

Semi-circles need special attention. Calculate the full circle's circumference first, then halve it. For a garden with radius 2.1m, the diameter is 4.2m, giving a full circumference of 13.18m. The semi-circle's curved edge is 6.59m, and the total perimeter including the straight edge is 10.79m.

Quick Tip: Always find the full circumference first for semi-circles, then divide by 2 for just the curved part!

National 5 Apps of Maths - Unit 3 Notes – page 2

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Past Paper Perimeter Problems

Badge design problems pop up frequently in exams, so let's tackle them head-on! When you see a shape combining rectangles and semi-circles, break it down step by step.

For Zainab's badge with a 10cm diameter semi-circle and rectangular base, start with the curved perimeter. The semi-circle's circumference is π × 10 ÷ 2 = 15.7cm. Don't forget the straight edges of the rectangle!

The total perimeter includes three straight sides (10cm + 15cm + 15cm = 40cm) plus the curved semi-circle (15.7cm). Adding these gives 55.7cm of gold edging needed.

Exam Strategy: Always identify which edges form the actual perimeter - some internal edges don't count!

National 5 Apps of Maths - Unit 3 Notes – page 3

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Area Formulas You Need

Time to move from perimeters to areas! You'll need several area formulas in your toolkit: rectangles A=l×wA = l × w, triangles A=½×b×hA = ½ × b × h, and the crucial circle area formula A = πr².

Circle areas are straightforward once you've got the radius. For r = 3cm: A = π × 3² = 28.27cm². If you're given the diameter (like 18mm), halve it first to get r = 9mm, then A = π × 9² = 254.34mm².

Composite shapes require adding areas together. Break complex shapes into familiar parts like rectangles and triangles. A rectangle 6×7=42cm26 × 7 = 42cm² plus a triangle ½×7×4=14cm2½ × 7 × 4 = 14cm² gives a total area of 56cm².

Remember: Area is always measured in square units (cm², m², etc.) - never forget those little ²s!

National 5 Apps of Maths - Unit 3 Notes – page 4

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More Area Practice

Let's reinforce those area calculations with different shapes. Besides circles and rectangles, you might encounter rhombuses and kites (both use A = ½ × d₁ × d₂) or parallelograms (A = base × height).

The key to mastering circle areas is getting comfortable with A = πr². Whether the radius is 3cm (giving 28.27cm²) or you need to find it from an 18cm diameter (r = 9cm, giving 254.34cm²), the process stays the same.

Complex shapes become manageable when you split them systematically. Identify each basic shape, calculate its area separately, then add them up. A 10cm × 7cm rectangle (70cm²) combined with a triangle creates larger composite areas.

Pro Tip: Always write down which formula you're using - it helps avoid mixing up perimeter and area calculations!

National 5 Apps of Maths - Unit 3 Notes – page 5

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Volume of Cuboids

Volume calculations start with the simple formula V = l × b × h for cuboids. But here's where unit conversions become crucial - remember that 1cm³ = 1ml and 1000ml = 1L.

Let's solve a practical problem: will a 5-litre jug overflow a tank? Calculate the tank's volume first: V = 30 × 9 × 19 = 5130cm³. Converting to litres: 5130 ÷ 1000 = 5.13L. Since 5.13L > 5L, the tank won't overflow.

Finding missing dimensions works backwards from the volume formula. If volume = 105cm³ and two dimensions are 7cm and 6cm, then 105 = 7 × ? × 6. Solving: ? = 105 ÷ 42 = 2.5cm.

Unit Check: Always convert cm³ to ml or litres when dealing with liquids - examiners love testing this!

National 5 Apps of Maths - Unit 3 Notes – page 6

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Combined Area Problems

This page shows how area calculations work with multiple shapes together. When you're dealing with semicircles and rectangles in the same problem, calculate each area separately before adding them up.

The process involves using A = πr² for circular parts and A = l × b for rectangular sections. Breaking down complex shapes into manageable pieces makes even challenging problems feel straightforward.

Total areas come from careful addition of all component parts. Make sure you've identified every section that contributes to the final answer.

Stay Organised: List each shape's area separately before adding - it prevents calculation errors!

National 5 Apps of Maths - Unit 3 Notes – page 7

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Cylinder Volume

Cylinder volume uses the formula V = πr²h, combining the circular base area with height. For a cylinder with radius 6cm and height 10cm: V = π × 6² × 10 = 1130.4cm³.

When given a diameter instead of radius, halve it first. A 10cm diameter gives r = 5cm, so for height 20cm: V = π × 5² × 20 = 1570.8cm³ (rounded to 1 decimal place).

Working backwards from volume to find missing dimensions needs algebraic manipulation. If V = 942cm³ and height = 12cm, then 942 = π × r² × 12. Solving: r² = 942 ÷ (π × 12), giving r ≈ 5cm.

Decimal Places: Pay attention to rounding instructions - exams often specify 1 d.p. or 3 s.f.!

National 5 Apps of Maths - Unit 3 Notes – page 8

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Cone Volume

Cone volume uses V = ⅓πr²h - it's exactly one-third of a cylinder's volume. For a cone with 30cm diameter and 40cm height, first find r = 15cm, then V = ⅓π × 15² × 40 = 9424.8cm³.

Truncated cones (cones with tops cut off) require calculating two volumes and subtracting. Find the full cone volume, then subtract the small cone that was removed to get the water volume.

Unit conversions matter here too. Once you have volume in cm³, divide by 1000 to get litres. A volume of 425.16cm³ equals 0.425L.

Two-Step Problems: Always calculate in cm³ first, then convert to ml or litres as the final step!

National 5 Apps of Maths - Unit 3 Notes – page 9

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Sphere and Hemisphere Volume

Sphere volume uses the formula V = ⅔πr³. For a sphere with 19cm diameter, r = 9.5cm gives V = ⅔π × 9.5³ = 3591.36cm³. The cubed radius makes these calculations larger quickly!

Hemispheres are exactly half a sphere, so use V = ⅓πr³. Notice how the formula changes from ⅔ to ⅓ - it's literally half the sphere formula. For r = 5cm: V = ⅓π × 5³ = 261.8cm³.

The relationship between sphere and hemisphere formulas makes sense: ⅓ is exactly half of ⅔. This connection helps you remember which formula to use.

Formula Connection: Hemisphere volume = ½ × sphere volume, so ⅓πr³ = ½ × ⅔πr³!

National 5 Apps of Maths - Unit 3 Notes – page 10

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Pyramid and Prism Volume

Pyramid volume always uses V = ⅓Ah, where A is the base area. For a pyramid with base area 14.5cm² and height 18cm: V = ⅓ × 14.5 × 18 = 87cm³. The ⅓ factor makes pyramids much smaller than prisms.

Square-based pyramids need you to find the base area first. With 2cm sides, the base area is 4cm², so V = ⅓ × 4 × 24 = 32cm³. Always calculate the base area separately before applying the main formula.

Prism volume is simpler: V = A × length. For a triangular prism with triangle area 10cm² and length 7cm: V = 10 × 7 = 70cm³. No fractions needed here!

Shape Recognition: Pyramids come to a point (use ⅓), prisms have constant cross-sections (no fraction needed)!

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