Geometry and Measurement Sec 2 E-Mathematics

Mensuration

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Key Concepts

  • Mensuration is the part of mathematics and science that deals with measurement of shapes and solids.
  • In this topic, you need to know how to find:
    • Volume of solids
    • Surface area of solids
    • Units of volume and how to convert between them
    • Measurements of composite solids made from two or more simple solids joined together

1. Volume

  • Volume is the amount of space occupied by a 3-dimensional object.
  • Volume is measured in cubic units because it involves length × width × height.
  • Common units:
    • cm³
    • m³
    • mm³
  • For liquids, volume may also be measured in:
    • mL
    • L

2. Surface Area

  • Surface area is the total area of all the outer surfaces of a solid.
  • It is measured in square units such as:
    • cm²
    • m²
    • mm²

3. Pyramid

  • A pyramid is a solid with:
    • one base that is a polygon
    • triangular faces that meet at a single point called the apex
  • The most common type is a square-based pyramid.

Volume of a pyramid

  • Formula:

    Volume of pyramid=13×base area×vertical height \text{Volume of pyramid} = \frac{1}{3} \times \text{base area} \times \text{vertical height}

  • Important:

    • Use the area of the base
    • Use the vertical height, not the slant height

Surface area of a pyramid

  • Surface area = base area + area of all triangular faces
  • For a square-based pyramid:
    • find area of square base
    • find area of each triangular face
    • add them together

4. Cone

  • A cone is a solid with:
    • one circular base
    • one curved surface
    • one vertex called the apex
  • It can be thought of as a pyramid with a circular base.

Volume of a cone

  • Formula:

    Volume of cone=13πr2h \text{Volume of cone} = \frac{1}{3}\pi r^2 h

    where:

    • rr = radius of base
    • hh = vertical height

Surface area of a cone

  • A cone has:

    • one circular base
    • one curved surface

  • Total surface area:

    Surface area=πr2+πrl \text{Surface area} = \pi r^2 + \pi r l

    where:

    • rr = radius
    • ll = slant height

  • Curved surface area only:

    πrl \pi r l

  • If the cone’s slant height is not given, use Pythagoras’ theorem:

    l=r2+h2 l = \sqrt{r^2 + h^2}

5. Sphere

  • A sphere is a perfectly round 3D solid in which every point on the surface is the same distance from the centre.
  • Examples: basketball, marble, planet Earth (approximately).

Volume of a sphere

  • Formula:

    Volume of sphere=43πr3 \text{Volume of sphere} = \frac{4}{3}\pi r^3

Surface area of a sphere

  • Formula:

    Surface area of sphere=4πr2 \text{Surface area of sphere} = 4\pi r^2

6. Composite Solids

  • A composite solid is a solid made by joining two or more simple solids together.
  • Examples:
    • hemisphere on top of a cylinder
    • cone attached to a cylinder
    • two cuboids joined together

Finding volume of composite solids

  • Split the solid into simpler known shapes
  • Find the volume of each part
  • Add them together

If there is a hollow or cut-out section:

  • find the total volume first
  • subtract the missing part

Finding surface area of composite solids

  • Find the area of all exposed surfaces only
  • Do not include surfaces that are joined inside and cannot be seen from outside

7. Converting Between Units of Volume

  • Volume units are cubic, so conversion involves cubing the linear conversion.

Length conversions

  • 1 cm=10 mm1 \text{ cm} = 10 \text{ mm}
  • 1 m=100 cm1 \text{ m} = 100 \text{ cm}
  • 1 m=1000 mm1 \text{ m} = 1000 \text{ mm}

Volume conversions

Because volume is cubic:

  • 1 cm3=1000 mm31 \text{ cm}^3 = 1000 \text{ mm}^3
  • 1 m3=1 000 000 cm31 \text{ m}^3 = 1\,000\,000 \text{ cm}^3
  • 1 m3=1 000 000 000 mm31 \text{ m}^3 = 1\,000\,000\,000 \text{ mm}^3

Liquid volume conversions

  • 1 cm3=1 mL1 \text{ cm}^3 = 1 \text{ mL}
  • 1000 cm3=1000 mL=1 L1000 \text{ cm}^3 = 1000 \text{ mL} = 1 \text{ L}
  • 1 L=1000 mL1 \text{ L} = 1000 \text{ mL}
  • 1 m3=1000 L1 \text{ m}^3 = 1000 \text{ L}

8. Choosing the Correct Formula

  • For pyramids and cones, volume always has a factor of:

    13 \frac{1}{3}

  • For surface area, check whether the question asks for:

    • total surface area
    • curved surface area only
    • exposed surface area

  • Always identify:

    • radius or diameter
    • vertical height or slant height
    • whether a base is included

Important Definitions

  • Mensuration: the branch of mathematics dealing with measurement of lengths, areas and volumes.
  • Volume: the amount of space taken up by a 3D object.
  • Surface area: the total area of all the outside surfaces of a solid.
  • Base: the surface on which a solid stands, or the main face used in a formula.
  • Apex: the pointed top of a pyramid or cone.
  • Radius: the distance from the centre of a circle or sphere to its edge or surface.
  • Diameter: a straight line passing through the centre of a circle or sphere, equal to twice the radius.
  • Vertical height: the perpendicular distance from the base to the top or apex.
  • Slant height: the length measured along the sloping face of a cone or pyramid.
  • Curved surface area: the area of the curved outer surface of a solid, not including the base.
  • Sphere: a round 3D solid with all points on the surface equidistant from its centre.
  • Composite solid: a solid formed by combining two or more simple solids.
  • Exposed surface: a surface that can be seen from the outside.
  • Cubic unit: a unit used for volume, such as cm³ or m³.
  • Square unit: a unit used for area, such as cm² or m².
  • Hemisphere: half of a sphere.

Worked Examples

Example 1: Volume and surface area of a cone

A cone has radius 3 cm3\text{ cm} and vertical height 4 cm4\text{ cm}. Find:

  1. its volume
  2. its total surface area

Step 1: Write down the given information

  • r=3 cmr = 3\text{ cm}
  • h=4 cmh = 4\text{ cm}

To find surface area, we need slant height ll.

Step 2: Find the slant height

l=r2+h2 l = \sqrt{r^2 + h^2}
l=32+42=9+16=25=5 cm l = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\text{ cm}

Step 3: Find the volume

V=13πr2h V = \frac{1}{3}\pi r^2 h
V=13π(3)2(4) V = \frac{1}{3}\pi (3)^2(4)
V=13π(9)(4) V = \frac{1}{3}\pi (9)(4)
V=12π cm3 V = 12\pi \text{ cm}^3

Using the π key on your calculator:

V≈12π≈37.70 cm3 V \approx 12\pi \approx 37.70\text{ cm}^3

Step 4: Find the total surface area

Surface area=πr2+πrl \text{Surface area} = \pi r^2 + \pi rl
=π(3)2+π(3)(5) = \pi(3)^2 + \pi(3)(5)
=9π+15π = 9\pi + 15\pi
=24π cm2 = 24\pi \text{ cm}^2

Using the π key on your calculator:

24π≈75.40 cm2 24\pi \approx 75.40\text{ cm}^2

Final answers

  • Volume = 12π cm312\pi\text{ cm}^3 or 37.70 cm337.70\text{ cm}^3
  • Total surface area = 24π cm224\pi\text{ cm}^2 or 75.40 cm275.40\text{ cm}^2

Example 2: Volume and surface area of a sphere

A sphere has diameter 10 cm10\text{ cm}. Find:

  1. its radius
  2. its volume
  3. its surface area

Step 1: Find the radius

r=d2=102=5 cm r = \frac{d}{2} = \frac{10}{2} = 5\text{ cm}

Step 2: Find the volume

V=43πr3 V = \frac{4}{3}\pi r^3
V=43π(5)3 V = \frac{4}{3}\pi (5)^3
V=43π(125) V = \frac{4}{3}\pi (125)
V=5003π cm3 V = \frac{500}{3}\pi \text{ cm}^3

Using the π key on your calculator:

V≈500π3≈523.60 cm3 V \approx \frac{500\pi}{3} \approx 523.60\text{ cm}^3

Step 3: Find the surface area

A=4πr2 A = 4\pi r^2
A=4π(5)2 A = 4\pi(5)^2
A=4π(25) A = 4\pi(25)
A=100π cm2 A = 100\pi \text{ cm}^2

Using the π key on your calculator:

A=100π≈314.16 cm2 A = 100\pi \approx 314.16\text{ cm}^2

Final answers

  • Radius = 5 cm5\text{ cm}
  • Volume = 5003π cm3\frac{500}{3}\pi\text{ cm}^3 or 523.60 cm3523.60\text{ cm}^3
  • Surface area = 100π cm2100\pi\text{ cm}^2 or 314.16 cm2314.16\text{ cm}^2

Example 3: Composite solid

A solid is made of a cylinder with a hemisphere on top.

  • Radius of both parts =7 cm= 7\text{ cm}
  • Height of cylinder =10 cm= 10\text{ cm}

Find:

  1. total volume
  2. total exposed surface area

Step 1: Identify the parts

The solid consists of:

  • one cylinder
  • one hemisphere

Step 2: Find the volume of the cylinder

Vcyl=πr2h V_{\text{cyl}} = \pi r^2 h
Vcyl=π(7)2(10) V_{\text{cyl}} = \pi(7)^2(10)
Vcyl=π(49)(10)=490π cm3 V_{\text{cyl}} = \pi(49)(10) = 490\pi \text{ cm}^3

Step 3: Find the volume of the hemisphere

Volume of sphere:

43πr3 \frac{4}{3}\pi r^3

Volume of hemisphere:

Vhemi=12×43πr3=23πr3 V_{\text{hemi}} = \frac{1}{2}\times \frac{4}{3}\pi r^3 = \frac{2}{3}\pi r^3
Vhemi=23π(7)3 V_{\text{hemi}} = \frac{2}{3}\pi(7)^3
Vhemi=23π(343)=6863π cm3 V_{\text{hemi}} = \frac{2}{3}\pi(343) = \frac{686}{3}\pi \text{ cm}^3

Step 4: Add the volumes

Vtotal=490π+6863π V_{\text{total}} = 490\pi + \frac{686}{3}\pi
=14703π+6863π = \frac{1470}{3}\pi + \frac{686}{3}\pi
=21563π cm3 = \frac{2156}{3}\pi \text{ cm}^3

Using the π key on your calculator:

Vtotal≈2156π3≈2257.76 cm3 V_{\text{total}} \approx \frac{2156\pi}{3} \approx 2257.76\text{ cm}^3

Step 5: Find exposed surface area

Exposed surfaces are:

  • curved surface of hemisphere
  • curved surface of cylinder
  • bottom circular base of cylinder

Do not include:

  • the circular face between hemisphere and cylinder, because it is internal

Curved surface area of hemisphere:

2πr2 2\pi r^2
2π(7)2=98π 2\pi(7)^2 = 98\pi

Curved surface area of cylinder:

2πrh 2\pi rh
2π(7)(10)=140π 2\pi(7)(10) = 140\pi

Bottom base of cylinder:

πr2=π(7)2=49π \pi r^2 = \pi(7)^2 = 49\pi

Step 6: Add exposed areas

Atotal=98π+140π+49π A_{\text{total}} = 98\pi + 140\pi + 49\pi
=287π cm2 = 287\pi \text{ cm}^2

Using the π key on your calculator:

Atotal≈287π≈901.64 cm2 A_{\text{total}} \approx 287\pi \approx 901.64\text{ cm}^2

Final answers

  • Total volume = 21563π cm3\frac{2156}{3}\pi\text{ cm}^3 or 2257.76 cm32257.76\text{ cm}^3
  • Exposed surface area = 287π cm2287\pi\text{ cm}^2 or 901.64 cm2901.64\text{ cm}^2

Common Mistakes to Avoid

  • Using diameter instead of radius in formulas for cones and spheres
  • Using slant height instead of vertical height when finding volume
  • Forgetting the 13\frac{1}{3} in the volume formula for pyramids and cones
  • Using surface area units for volume, or volume units for area
    • volume must be in cm³, m³
    • surface area must be in cm², m²
  • Including hidden internal surfaces when finding surface area of composite solids
  • Forgetting to include the base area when the question asks for total surface area of a cone
  • Including the base when the question asks for curved surface area only
  • Not converting units before substituting into formulas
    • for example, radius in cm and height in m
  • Converting volume units wrongly by multiplying by 10 or 100 instead of cubing the conversion factor
  • Rounding too early in the working, leading to inaccurate final answers
  • Forgetting that:
    • 1 cm3=1 mL1\text{ cm}^3 = 1\text{ mL}
    • 1000 cm3=1 L1000\text{ cm}^3 = 1\text{ L}

Exam Tips

  • Start by writing the correct formula clearly. This often earns method marks.

  • Define the symbols you use:

    • rr = radius
    • hh = vertical height
    • ll = slant height

  • If the diameter is given, immediately write:

    r=d2 r = \frac{d}{2}

  • For composite solids, write:

    • “Total volume = volume of part A + volume of part B”
    • “Exposed surface area excludes internal faces”

  • If you need slant height of a cone, mention:

    • “Using Pythagoras’ theorem”

  • Always write units in the final answer:

    • cm³ for volume
    • cm² for surface area

  • If exact answers are acceptable, leave answers in terms of π\pi first.

  • If a decimal answer is needed, use the π key on your calculator — this gives accurate results for exams. Leave answers in exact form (in terms of π) where possible.

  • Read the question carefully to check whether it asks for:

    • total surface area
    • curved surface area
    • volume
    • capacity

  • For unit conversion questions:

    • convert before calculation if easier
    • or calculate first, then convert carefully using the correct cubic relationship

  • Useful mark-earning phrases:

    • “Area of base = …”
    • “Vertical height = …”
    • “Curved surface area = …”
    • “Internal surface not included”
    • “Hence, total volume is …”

Quick Summary

  • Volume measures the space inside a solid; surface area measures the outside area.

  • Volume units are cubic units like cm³; surface area units are square units like cm².

  • Volume of a pyramid:

    13×base area×height \frac{1}{3}\times \text{base area} \times \text{height}

  • Surface area of a pyramid = base area + area of triangular faces.

  • Volume of a cone:

    13πr2h \frac{1}{3}\pi r^2 h

  • Total surface area of a cone:

    πr2+πrl \pi r^2 + \pi rl

  • Volume of a sphere:

    43πr3 \frac{4}{3}\pi r^3

  • Surface area of a sphere:

    4πr2 4\pi r^2

  • For a cone, slant height is found by:

    l=r2+h2 l = \sqrt{r^2 + h^2}

  • For composite solids, split into simple solids, then add or subtract volumes and count only exposed surfaces.

  • Key conversions:

    • 1 cm3=1000 mm31\text{ cm}^3 = 1000\text{ mm}^3
    • 1 cm3=1 mL1\text{ cm}^3 = 1\text{ mL}
    • 1000 cm3=1 L1000\text{ cm}^3 = 1\text{ L}
    • 1 m3=1000 L1\text{ m}^3 = 1000\text{ L}

  • Always check whether the question gives radius or diameter, and whether it wants exact form or decimal form.

✏️ 29 practice questions available

30 questions from school exam papers

Q1

A new structure shown in the diagram below, has been built. It is made up of hemispherical bottom with radius of 60 m and a right conical top of radius 60 m and height 80 m. Calculate the volume of the structure.

A composite 3D shape consisting of a cone on top of a hemisphere. The cone has a height of 80 m and a base radius of 60 m (marked with a dashed vertical line for height and a horizontal line for radius). The hemisphere has a radius of 60 m. The cone and hemisphere share a common circular base of radius 60 m.
📊 Diagram: A composite 3D shape consisting of a cone on top of a hemisphere. The cone has a height of 80 m and a base radius of 60 m (marked with a dashed vertical line for height and a horizontal line for radius). The hemisphere has a radius of 60 m. The cone and hemisphere share a common circular base of radius 60 m.
3 marks
Math_Sec2NA_SA2_2023_ACS_Barker 2023
Q2

The diagram shows a solid wooden toy made up of a circular cone and a hemisphere. The toy is 6 cm wide and the height of the cone is 4 cm tall. Calculate the volume of the wooden toy.

A composite solid consisting of a cone on top of a hemisphere. The width (diameter) of the entire toy is labeled as 6 cm. The height of the cone portion is labeled as 4 cm. The hemisphere sits at the base of the cone, and dashed lines indicate the internal dimensions and the circular base shared between the cone and hemisphere.
📊 Diagram: A composite solid consisting of a cone on top of a hemisphere. The width (diameter) of the entire toy is labeled as 6 cm. The height of the cone portion is labeled as 4 cm. The hemisphere sits at the base of the cone, and dashed lines indicate the internal dimensions and the circular base shared between the cone and hemisphere.
3 marks
Math_Sec2NA_SA2_2023_Bartley_Sec 2023
Q3

Calculate the total surface area of the wooden toy.

Same composite solid as in question 13a - a cone sitting on top of a hemisphere, with diameter 6 cm and cone height 4 cm.
📊 Diagram: Same composite solid as in question 13a - a cone sitting on top of a hemisphere, with diameter 6 cm and cone height 4 cm.
3 marks
Math_Sec2NA_SA2_2023_Bartley_Sec 2023
Q4

The diagram shows a solid wooden toy made up of a circular cone and a hemisphere. The toy is 6 cm wide and the height of the cone is 4 cm tall. Calculate the volume of the wooden toy.

A right square pyramid with a square base of side 12 cm. The apex is labeled A. The slant height AY is marked as 10 cm. The diagram shows x and y as internal measurements (likely indicating the perpendicular height and half the base). A dashed line from A shows the height to the center of the base.
📊 Diagram: A right square pyramid with a square base of side 12 cm. The apex is labeled A. The slant height AY is marked as 10 cm. The diagram shows x and y as internal measurements (likely indicating the perpendicular height and half the base). A dashed line from A shows the height to the center of the base.
3 marks
Math_Sec2NA_SA2_2023_Bartley_Sec 2023
Q5

Calculate the total surface area of the wooden toy.

Same pyramid as in part (a) - right square pyramid with square base of side 12 cm, slant height 10 cm
📊 Diagram: Same pyramid as in part (a) - right square pyramid with square base of side 12 cm, slant height 10 cm
2 marks
Math_Sec2NA_SA2_2023_Bartley_Sec 2023
Q6

Solve the inequality 4x ≥ −10 and represent the solution on the number line.

A solid cylinder with base diameter 18 cm and vertical height 14 cm. A hemisphere has been removed from the top of the cylinder, creating a container. The hemisphere has the same diameter as the cylinder (18 cm).
📊 Diagram: A solid cylinder with base diameter 18 cm and vertical height 14 cm. A hemisphere has been removed from the top of the cylinder, creating a container. The hemisphere has the same diameter as the cylinder (18 cm).
2 marks
Math_Sec2NA_SA2_2023_Broadrick_Sec 2023
Q7

Hence, write down the smallest integer x that satisfies 4x ≥ −10.

Same container as shown in part (a) - cylinder with hemisphere removed from top.
📊 Diagram: Same container as shown in part (a) - cylinder with hemisphere removed from top.
2 marks
Math_Sec2NA_SA2_2023_Broadrick_Sec 2023
Q8

Calculate the volume of the cone.

A cone with base radius 12 cm, height 11.5 cm, and slant height l cm. The diagram shows the cone with the height marked as 11.5 cm pointing vertically from the apex, the slant height l marked on the slant edge, and the base radius 12 cm marked on the base.
📊 Diagram: A cone with base radius 12 cm, height 11.5 cm, and slant height l cm. The diagram shows the cone with the height marked as 11.5 cm pointing vertically from the apex, the slant height l marked on the slant edge, and the base radius 12 cm marked on the base.
2 marks
Math_Sec2NA_SA2_2023_Canberra_Sec 2023
Q9

A rectangular pyramid has a base of 9 cm by 12 cm and a height of 15 cm.

A rectangular pyramid with apex A and rectangular base BCDE. The base has dimensions 9 cm and 12 cm. The height from apex A to the base is 15 cm, shown by a vertical dashed line. Point F appears to be the center of the base. The slant heights are indicated.
📊 Diagram: A rectangular pyramid with apex A and rectangular base BCDE. The base has dimensions 9 cm and 12 cm. The height from apex A to the base is 15 cm, shown by a vertical dashed line. Point F appears to be the center of the base. The slant heights are indicated.
Math_Sec2NA_SA2_2023_Canberra_Sec 2023
Q10

Find the volume of the pyramid.

Diagram for question 14a
2 marks
Math_Sec2NA_SA2_2023_Canberra_Sec 2023
Q11

Find the slant height AG.

Diagram for question 14b
2 marks
Math_Sec2NA_SA2_2023_Canberra_Sec 2023
Q12

Given AF = 17 cm, calculate the total surface area of the pyramid.

Diagram for question 14c
3 marks
Math_Sec2NA_SA2_2023_Canberra_Sec 2023
Q13

Show that the volume of the chocolate, correct to one decimal place is 14.1 cm³.

Diagram I shows a spherical chocolate with radius 1.5 cm marked from the center to the edge.
📊 Diagram: Diagram I shows a spherical chocolate with radius 1.5 cm marked from the center to the edge.
2 marks
Math_Sec2NA_SA2_2023_Canberra_Sec 2023
Q14

The chocolate is now cut into half in the form of a hemisphere and wrapped in gold foil as shown in Diagram II. Calculate the total area of gold foil needed for one hemisphere, assuming there is no overlap.

Diagram II shows a hemisphere with radius 1.5 cm, depicting the flat circular top surface and the curved hemispherical surface.
📊 Diagram: Diagram II shows a hemisphere with radius 1.5 cm, depicting the flat circular top surface and the curved hemispherical surface.
2 marks
Math_Sec2NA_SA2_2023_Canberra_Sec 2023
Q15

Given that the cost of chocolate is $0.10 per cm³ and the cost of gold foil is $0.20 per cm², determine the selling price of one hemispherical chocolate so that a profit can be made.

Diagram for question 9c
3 marks
Math_Sec2NA_SA2_2023_Canberra_Sec 2023
Q16

A solid hemisphere has a volume of 144π cm³. (a) Find the radius of the hemisphere.

A solid hemisphere shown in cross-section with a label pointing to the radius at the top flat surface
📊 Diagram: A solid hemisphere shown in cross-section with a label pointing to the radius at the top flat surface
2 marks
Math_Sec2NA_SA2_2023_Chung_Cheng_Yishun 2023
Q17

Find the total surface area of the hemisphere, correct your answer to 3 significant figures.

Diagram for question 14b
2 marks
Math_Sec2NA_SA2_2023_Chung_Cheng_Yishun 2023
Q18

Hence find the area of triangle ABC.

Diagram for question 5d
1 mark
Math_Sec2NA_SA2_2023_Chung_Cheng_Yishun 2023
Q19

The diagram shows a candle in the shape of a pyramid with a vertical height of 10 cm. The base of the pyramid is a square CDEF and the volume of the pyramid is 235.2 cm³. Show that the length of CD is 8.4 cm.

A pyramid with apex A and square base CDEF. The vertical height from A to the base is marked as 10 cm. Point O is marked at the center of the base. Vertices are labeled C, D, E, F for the base corners, and A for the apex. The height is shown as a dashed line from A perpendicular to O on the base.
📊 Diagram: A pyramid with apex A and square base CDEF. The vertical height from A to the base is marked as 10 cm. Point O is marked at the center of the base. Vertices are labeled C, D, E, F for the base corners, and A for the apex. The height is shown as a dashed line from A perpendicular to O on the base.
2 marks
Math_Sec2NA_SA2_2023_Chung_Cheng_Yishun 2023
Q20

Find the total surface area of the candle.

Diagram for question 6c
3 marks
Math_Sec2NA_SA2_2023_Chung_Cheng_Yishun 2023
Q21

A ladder BC is leaning against the wall AB and touching the top of the wall at B. The height of the wall is 6 m and the distance from the foot of the ladder to the foot of the wall, AC, is 9.6 m. Find the length of the ladder BC.

A cylindrical water tanker on a truck. The cylindrical tank has radius 120 cm and length 600 cm. The diagram shows the tank dimensions labeled.
📊 Diagram: A cylindrical water tanker on a truck. The cylindrical tank has radius 120 cm and length 600 cm. The diagram shows the tank dimensions labeled.
2 marks
Math_Sec2NA_SA2_2023_Chung_Cheng_Yishun 2023
Q22

Hence find the area of triangle ABC.

Triangle ABC with the working shown: Area = 1/2 × 5.7 × 13 = 37.05 cm²
📊 Diagram: Triangle ABC with the working shown: Area = 1/2 × 5.7 × 13 = 37.05 cm²
1 mark
Math_Sec2NA_SA2_2023_Chung_Cheng_Yishun 2023
Q23

The area of triangle ABC is 71.5 cm². Find BC.

Same triangle ABC as in part (a), with the given information that area of triangle ABC is 71.5 cm².
📊 Diagram: Same triangle ABC as in part (a), with the given information that area of triangle ABC is 71.5 cm².
2 marks
Math_Sec2NA_SA2_2023_Springfield_Sec 2023
Q24

The baked bean cans are packed vertically as shown in Fig. 1. By considering the dimensions of the medium box, show that x = 6.

Fig. 1 shows a cylindrical can with diameter 7 cm and height 3 cm. A table shows box dimensions and maximum number of cans per box: Large box (21 × 21 × 17 cm) holds 18 cans at $1.20 per box; Medium box (14 × 21 × 9 cm) holds x cans at $0.50 per box.
📊 Diagram: Fig. 1 shows a cylindrical can with diameter 7 cm and height 3 cm. A table shows box dimensions and maximum number of cans per box: Large box (21 × 21 × 17 cm) holds 18 cans at $1.20 per box; Medium box (14 × 21 × 9 cm) holds x cans at $0.50 per box.
1 mark
Math_Sec2NA_SA2_2023_Springfield_Sec 2023
Q25

An open cone has a circular top of diameter 12 cm and a slant height of 7 cm. Calculate its curved surface area, giving your answer correct to 2 significant figures.

A diagram of an open cone showing a circular top with diameter labeled as 12 cm and a slant height labeled as 7 cm extending from the edge of the circle to the apex of the cone.
📊 Diagram: A diagram of an open cone showing a circular top with diameter labeled as 12 cm and a slant height labeled as 7 cm extending from the edge of the circle to the apex of the cone.
2 marks
Math_Sec2NA_SA2_2023_Zhonghua_Sec 2023
Q26

Ryan is a local sculptor and his latest sculpture made from recycled iron was selected for an exhibit in Germany. The sculpture is made up of a solid sphere of radius 0.5 m and a solid pyramid of a height 1.2 m with a square base of sides 1 m. Calculate the volume of recycled iron used to make this sculpture.

A diagram showing a sculpture composed of two parts: (1) A solid sphere with radius 0.5 m positioned on top, with a horizontal dotted line showing the radius measurement across the diameter. (2) Below it, a square pyramid with height 1.2 m and a square base with sides of 1 m. The pyramid is shown in perspective with dotted lines indicating hidden edges and internal structure. The label 'x cm' appears on one edge of the pyramid.
📊 Diagram: A diagram showing a sculpture composed of two parts: (1) A solid sphere with radius 0.5 m positioned on top, with a horizontal dotted line showing the radius measurement across the diameter. (2) Below it, a square pyramid with height 1.2 m and a square base with sides of 1 m. The pyramid is shown in perspective with dotted lines indicating hidden edges and internal structure. The label 'x cm' appears on one edge of the pyramid.
3 marks
Math_Sec2NA_SA2_2023_Zhonghua_Sec 2023
Q27

An open cone has a circular top of diameter 12 cm and a slant height of 7 cm. Calculate its curved surface area, giving your answer correct to 2 significant figures.

A diagram of an open cone shown upside down (vertex at bottom, circular opening at top). The circular top is labeled with diameter 12 cm. The slant height is labeled as 7 cm along the slant edge.
📊 Diagram: A diagram of an open cone shown upside down (vertex at bottom, circular opening at top). The circular top is labeled with diameter 12 cm. The slant height is labeled as 7 cm along the slant edge.
2 marks
Math_Sec2NA_SA2_2023_Zhonghua_Sec 2023
Q28

Ryan is a local sculptor and his latest sculpture made from recycled iron was selected for an exhibit in Germany. The sculpture is made up of a solid sphere of radius 0.5 m and a solid pyramid of a height 1.2 m with a square base of sides 1 m. Calculate the volume of recycled iron used to make this sculpture.

A diagram showing a sculpture composed of two parts: a solid sphere sitting on top of a pyramid. The sphere has a radius of 0.5 m (indicated by a dotted line diameter). The pyramid below has a square base with sides of 1 m and a height of 1.2 m. The slant edge of the pyramid is marked as x cm.
📊 Diagram: A diagram showing a sculpture composed of two parts: a solid sphere sitting on top of a pyramid. The sphere has a radius of 0.5 m (indicated by a dotted line diameter). The pyramid below has a square base with sides of 1 m and a height of 1.2 m. The slant edge of the pyramid is marked as x cm.
3 marks
Math_Sec2NA_SA2_2023_Zhonghua_Sec 2023
Q29

The diagram shows a smaller cone removed from a larger cone of the same height of 12 cm. The smaller cone has a radius of 5 cm and slant height of 13 cm. The larger cone has a radius of 9 cm and slant height of 15 cm. Find an expression, in terms of π, for the volume of the remaining solid.

A diagram showing a larger cone with a smaller cone removed from its top. The larger cone has radius 9 cm at the base and slant height 15 cm. The smaller cone has radius 5 cm at the top and slant height 13 cm. Both cones share the same height of 12 cm. The diagram shows the cross-section with measurements labeled.
📊 Diagram: A diagram showing a larger cone with a smaller cone removed from its top. The larger cone has radius 9 cm at the base and slant height 15 cm. The smaller cone has radius 5 cm at the top and slant height 13 cm. Both cones share the same height of 12 cm. The diagram shows the cross-section with measurements labeled.
2 marks
Math_Sec2_SA2_2023_Bedok_South_Sec 2023

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