Area & Volume

Junior Cycle Higher Level Mathematics revision notes with diagrams, key terms and self-check questions.

4 min readHigher LevelBy Studytok
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Area measures flat surface coverage, while volume measures the three-dimensional space inside a solid. In the Junior Cycle Higher Level exam, you will calculate boundaries, composite areas, surface areas, and volumes for cylinders, prisms, spheres, and combined shapes.

1. Units of Measure, Conversions, and Scale

Always check your units before starting any calculation. If dimensions are mixed, convert them all to the requested unit first.

QuantityConversions
Length1 m=100 cm1\text{ m} = 100\text{ cm}, 1 cm=10 mm1\text{ cm} = 10\text{ mm}
Area1 m2=(100)2 cm2=10 000 cm21\text{ m}^2 = (100)^2\text{ cm}^2 = 10\,000\text{ cm}^2
Volume1 m3=(100)3 cm3=1 000 000 cm31\text{ m}^3 = (100)^3\text{ cm}^3 = 1\,000\,000\text{ cm}^3
Capacity1 litre=1000 cm31\text{ litre} = 1000\text{ cm}^3, 1 ml=1 cm31\text{ ml} = 1\text{ cm}^3

Remember that area scale factors square the linear factor, while volume factors cube it. On a scaled drawing with a ratio of 1:n1:n, each unit on paper represents nn units in real life. Divide the real length by nn to find the drawing length.

2. Perimeter and Area of Plane Figures

Perimeter is the total boundary distance around a shape, while area measures the flat space inside.

  • Rectangle: Area=l×w\text{Area} = l \times w; Perimeter=2l+2w\text{Perimeter} = 2l + 2w
  • Triangle: Area=12bh\text{Area} = \frac{1}{2}bh, where height hh must meet base bb at 90∘90^\circ
  • Disc (Circle): Area=πr2\text{Area} = \pi r^2; Circumference=2πr\text{Circumference} = 2\pi r

When working with a semicircle, its perimeter includes both the curved arc and the flat diameter: P=πr+2rP = \pi r + 2r. For any sector with angle θ\theta, the arc length is θ360∘×2πr\frac{\theta}{360^\circ} \times 2\pi r and its sector area is θ360∘×πr2\frac{\theta}{360^\circ} \times \pi r^2. For composite 2D shapes, split the diagram into familiar triangles, rectangles, or discs, then add or subtract their individual areas.

3. Nets of Prisms and Cylinders

A net is a flat two-dimensional pattern that folds up to make a three-dimensional solid without any overlapping faces.

  • Cuboid: Unfolds into six rectangles, made of three matching pairs.
  • Right polygonal prism: Has two matching polygon bases joined by rectangular sides perpendicular to the bases. A triangular prism has two triangles and three rectangles, while a hexagonal prism has two hexagons and six rectangles.
  • Cylinder: Unfolds into two matching circular bases and one large rectangle.

The cylinder's curved wall becomes the body of the rectangle. Its height is the cylinder height hh, and its width matches the circumference of the circular base, 2πr2\pi r.

A cylinder unfolds into two circular bases of radius r and a rectangle of width 2πr and height h.
A cylinder unfolds into two circular bases of radius r and a rectangle of width 2πr and height h.

4. Volume of Solids and Combinations

Prism volume = base area × perpendicular distance between bases. In V = ½bhl, h is the triangle’s perpendicular height; l is the distance between triangular ends.

A right triangular prism has a shaded triangular end with base b and perpendicular height h; length l runs between its matching ends.
A right triangular prism has a shaded triangular end with base b and perpendicular height h; length l runs between its matching ends.
  • Cuboid: V=l×w×hV = l \times w \times h
  • Cylinder: V=πr2hV = \pi r^2 h
  • Triangular Prism: V=(12bh)×lV = (\frac{1}{2}bh) \times l
  • Sphere: V=43πr3V = \frac{4}{3}\pi r^3
  • Hemisphere: Halve the sphere formula to get V=23πr3V = \frac{2}{3}\pi r^3.

For composite solids, calculate each component separately. For example, a silo made of a cylinder with a hemispherical cap has volume V=πr2h+23πr3V = \pi r^2 h + \frac{2}{3}\pi r^3. If an object has a hole drilled out, subtract the missing volume from the main solid.

5. Curved and Total Surface Area

Take care to distinguish between curved surface area and total surface area.

SolidCurved Surface Area (CSA)Total Surface Area (TSA)
Cylinder2πrh2\pi rh2πrh+2πr22\pi rh + 2\pi r^2
Sphere4πr24\pi r^24πr24\pi r^2
Solid Hemisphere2πr22\pi r^22πr2+πr2=3πr22\pi r^2 + \pi r^2 = 3\pi r^2
Triangular PrismNone — all faces are flat.2(Triangle Area)+Sum of 3 Rectangles2(\text{Triangle Area}) + \text{Sum of } 3 \text{ Rectangles}

A cylinder’s outside curved area is 2πrh. An ideal open-top cylinder, ignoring thickness, has area 2πrh + πr². When shapes are joined together into a composite solid, ignore touching interior faces because they are no longer on the outside.

A hemispherical cap sits flush on a cylinder. A translucent view reveals their shared circular face inside the combined solid.
A hemispherical cap sits flush on a cylinder. A translucent view reveals their shared circular face inside the combined solid.

Key terms

Perimeter
The total outer boundary length enclosing a two-dimensional shape.
Cross-Section
The two-dimensional shape formed by a flat cut through a solid.
Prism
A solid with matching, parallel polygonal ends and the same cross-section along its length.
Net
A 2D pattern of connected faces that folds along edges to build a 3D solid.
Curved Surface Area
The area of the curved exterior surfaces of a shape, excluding its flat ends.
Total Surface Area
The complete exterior area of all faces of a solid combined, both flat and curved.

Check yourself

  1. Convert an area of 0.4 m² into cm².

    0.4×10 000=4000 cm20.4 \times 10\,000 = 4000\text{ cm}^2.

  2. What is the total boundary perimeter of a semicircle with radius 7 cm, using π = 22/7?

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

  3. Find the volume of a sphere of radius 3 cm in terms of π.

    V=43π(33)=43×27π=36π cm3V = \frac{4}{3}\pi(3^3) = \frac{4}{3} \times 27\pi = 36\pi\text{ cm}^3.

  4. A cylinder has radius 4 cm and height 10 cm. Find its curved surface area in terms of π.

    CSA=2πrh=2π(4)(10)=80π cm2\text{CSA} = 2\pi rh = 2\pi(4)(10) = 80\pi\text{ cm}^2.

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