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Grade 10 · Measurement
Measurement: Surface Area, Volume & Scaling (Grade 10)
MARKING GUIDELINE
Marks
42
Duration
1 hour 5 minutes
Questions
4
Name:
Class:
Date:
Mark
/ 42
Instructions and Information
- Answer ALL the questions in this question paper.
- Answer QUESTION 1 by circling the letter (A–D) in the answer grid at the end of that section.
- Show ALL calculations clearly.
- Show all units where applicable.
- Number the answers correctly according to the numbering system used in this question paper.
- A non-programmable calculator may be used, unless stated otherwise.
- Write neatly and legibly.
Question 1
[10 MARKS]Four options are given as possible answers to the following questions. Choose the correct answer and circle the letter (A–D) in the grid at the end of this section. If you want to change your choice, put a cross through the wrong letter and circle your new choice.
- 1.1A cylinder has radius 5 cm and height 10 cm. Its VOLUME is …(1)A)314,16 cm³B)785,40 cm³✓C)1 570,80 cm³D)157,08 cm³Answer: B — V = πr²h = π × 25 × 10 = 785,40 cm³.
- A — used 2πrh, which is the CURVED SURFACE area
- C — doubled the correct volume
- D — used πrh instead of πr²h
- 1.2A sphere has radius 6 cm. Its VOLUME is …(1)A)904,78 cm³✓B)150,80 cm³C)678,58 cm³D)452,39 cm³Answer: A — V = 4/3πr³ = 4/3 × π × 216 = 904,78 cm³.
- B — squared the radius instead of cubing it
- C — left out the factor of 4/3
- D — that is 4πr², the SURFACE AREA
- 1.3A cone has radius 3 cm and perpendicular height 8 cm. Its VOLUME is …(1)A)75,40 cm³✓B)226,19 cm³C)150,80 cm³D)25,13 cm³Answer: A — V = ⅓πr²h = 1/3 × π × 9 × 8 = 75,40 cm³.
- B — used πr²h — that is the CYLINDER, three times too big
- C — used two thirds instead of one third
- D — used ⅓πrh, forgetting to square the radius
- 1.4A closed cylinder has radius 4 cm and height 10 cm. Its TOTAL SURFACE AREA is …(1)A)301,59 cm²B)100,53 cm²C)351,86 cm²✓D)251,33 cm²Answer: C — TSA = 2πr² + 2πrh = 100,53 + 251,33 = 351,86 cm².
- A — included only ONE circular end instead of two
- B — gave only the two ends, leaving out the curved surface
- D — gave only the CURVED surface, leaving out the two circular ends
- 1.5If every dimension of a solid is multiplied by 3, its VOLUME is multiplied by …(1)A)9B)6C)27✓D)3Answer: C — Volume scales by k³, and 3³ = 27.
- A — that is the factor for an AREA
- B — doubled the factor instead of cubing it
- D — that is the factor for a LENGTH
- 1.6If every dimension of a solid is HALVED, its SURFACE AREA is multiplied by …(1)A)4B)1/2C)1/8D)1/4✓Answer: D — Area scales by k², and (½)² = ¼.
- A — halving makes the area smaller, not four times as large
- B — that is the factor for a LENGTH
- C — that is the factor for a VOLUME
- 1.7A right pyramid has a square base of side 16 cm, a perpendicular height h = 6 cm and a slant height ℓ = 10 cm, as in the diagram. Its TOTAL SURFACE AREA is …(1)A)448 cm²B)576 cm²✓C)320 cm²D)896 cm²Answer: B — TSA = 162 + 4 × 1/2 × 16 × 10 = 256 + 320 = 576 cm² — each face is a triangle whose height is the SLANT height.
- A — used the perpendicular height 6 cm for the faces; each face leans, so its height is the slant height 10 cm
- C — gave only the four triangular faces and left out the square base
- D — left out the 1/2 in the area of each triangle
- 1.8A cube has volume 125 cm³. Its TOTAL SURFACE AREA is …(1)A)30 cm²B)750 cm²C)150 cm²✓D)25 cm²Answer: C — The side is ∛125 = 5 cm, and a cube has 6 faces of 25 cm² each.
- A — multiplied the SIDE by 6 instead of the face AREA by 6
- B — used the volume 125 in place of the face area 25
- D — gave the area of ONE face only
- 1.9A cylindrical tank has radius 0,5 m and height 2 m. Its capacity is …(1)A)157,08 litresB)1,57 litresC)15,71 litresD)1 570,80 litres✓Answer: D — V = πr²h = 1,57 m³, and 1 m³ = 1 000 litres.
- A — multiplied by 100 instead of by 1 000
- B — left the answer in cubic METRES instead of converting to litres
- C — multiplied by 10 instead of by 1 000
- 1.10A square pyramid has base side 6 cm and perpendicular height 10 cm. Its VOLUME is …(1)A)60 cm³B)360 cm³C)20 cm³D)120 cm³✓Answer: D — V = ⅓ × base area × height = 1/3 × 36 × 10 = 120 cm³.
- A — used 6 × 10, forgetting both the third and the square
- B — used base area × height — that is the PRISM, three times too big
- C — used ⅓ × 6 × 10, forgetting to square the base side
Answer grid — marking guideline
| 1.1 | A | B | C | D |
| 1.2 | A | B | C | D |
| 1.3 | A | B | C | D |
| 1.4 | A | B | C | D |
| 1.5 | A | B | C | D |
| 1.6 | A | B | C | D |
| 1.7 | A | B | C | D |
| 1.8 | A | B | C | D |
| 1.9 | A | B | C | D |
| 1.10 | A | B | C | D |
Question 2
[10 MARKS]A grain silo is made of a right cylinder with a hemispherical roof, as in the diagram below. The radius of the silo is 4 m and its total height, from the floor to the top of the roof, is 16 m. Round off your answers to TWO decimal places.
- 2.1Calculate the volume of the silo.(5)Height of the cylinder = 16 − 4 = 12 m (1)
V = πr2h + 2/3πr3 (1)
= π(4)2(12) + 2/3π(4)3 (1)
= 603,19 + 134,04 (1) = 737,23 m3 (1) - 2.2The walls and the roof are covered with metal sheeting; the floor is concrete. Calculate the area of sheeting needed, and explain why the circle where the roof meets the walls is not included.(5)Walls: 2πrh = 2π(4)(12) = 301,59 m2 (1)
Roof: 2πr2 = 2π(4)2 = 100,53 m2 (1)
No area is added for the floor (it is concrete) (1)
Sheeting = 301,59 + 100,53 = 402,12 m2 (1)
The joined circle is inside the silo, not on its outside, so no sheeting covers it (1)
Question 3
[13 MARKS]The diagram below shows a cone with a base radius of 3 cm and a perpendicular height of 4 cm, and a sphere with a radius of 6 cm. Use π ≈ 3,14, Vcone = 1/3πr2h, Vsphere = 4/3πr3 and SAsphere = 4πr2.
- 3.1Calculate the volume of the cone.(3)V = 1/3 × 3,14 × 9 × 4 (2) = 37,68 cm3 (1)
- 3.2Calculate the volume and the surface area of the sphere.(4)V = 4/3 × 3,14 × 216 = 904,32 cm3 (2)
SA = 4 × 3,14 × 36 = 452,16 cm2 (2) - 3.3ALL the dimensions of the cone are doubled. Determine by what factor its volume increases, and justify your answer.(4)Both r and h double (1)
V depends on r2h, so the factor is 22 × 2 (2)
The volume becomes 8 times as big (1) - 3.4Explain why doubling the dimensions multiplies the SURFACE AREA by only 4.(2)Surface area depends on TWO dimensions (1), so it grows by the square of the factor, 22 (1)
Question 4
[9 MARKS]The diagram below shows a right pyramid TABCD with a square base ABCD. O is the centre of the base and M is the midpoint of AB. AB = 10 cm and the perpendicular height TO = 12 cm.
- 4.1Show that the slant height TM = 13 cm.(3)OM = 1/2 × 10 = 5 cm (1)
TM2 = TO2 + OM2 = 122 + 52 = 169 (1)
TM = 13 cm (1) - 4.2Calculate the total surface area of the pyramid.(3)Base = 102 = 100 cm2 (1)
Four faces = 4 × 1/2 × 10 × 13 = 260 cm2 (1)
TSA = 360 cm2 (1) - 4.3Calculate the volume of the pyramid, and explain why TO, and not TM, is used.(3)V = 1/3 × 100 × 12 (1) = 400 cm3 (1)
Volume needs the PERPENDICULAR height, straight down from the apex to the base; TM leans along a face (1)
TOTAL: 42 marks
This question paper consists of 4 questions.