Perimeter, Area and Volume
Perimeter
Back to Top ↑Perimeter
Perimeter is the total distance around the edge of a 2D shape. Add the lengths of all sides.
- Rectangle: \( P = 2(l + w) \)
- Square: \( P = 4s \)
- Triangle: \( P = a + b + c \)
- Circle (circumference): \( C = 2\pi r \) or \( C = \pi d \)
For composite shapes, label any missing sides and add all edges on the outside of the shape.
Perimeter – Quick Quiz
Area
Back to Top ↑Area
Area is the amount of space inside a 2D shape, measured in square units.
- Rectangle: \( A = lw \)
- Square: \( A = s^{2} \)
- Triangle: \( A = \tfrac{1}{2}bh \)
- Parallelogram: \( A = bh \)
- Circle: \( A = \pi r^{2} \)
For composite areas, split the figure into rectangles, triangles, or circles; find each area and then add or subtract as needed. Use units such as \( \text{cm}^{2} \) or \( \text{m}^{2} \).
Area & Perimeter of Rectangles (Interactive)
Use this GeoGebra model to explore how the length and width of a rectangle affect its perimeter \( P = 2(l + w) \) and area \( A = lw \).
Try:
- Keeping the perimeter the same and finding rectangles with different areas.
- Keeping the area the same and comparing the perimeters.
- Predicting the formulas from the patterns you notice on the screen.
Area – Quick Quiz
Circumference of a Circle — π Diameters (Interactive)
This GeoGebra model provides a visual explanation of why the circumference of a circle is given by \( C = \pi d = 2\pi r \).
Area of a Circle — Dissection & Rearrangement (Interactive)
This GeoGebra model explores why the area of a circle (disc) is \( A = \pi r^2 \) by cutting the disc into equal sectors and rearranging them to form an approximate rectangle.
As the number of sectors increases, the rearranged shape becomes closer to a true rectangle.
Observe:
- The length of the rectangle approaches the radius \( r \).
- The width approaches half the circumference, \( \pi r \).
- The area therefore approaches \( (\pi r)\times r = \pi r^2 \).
Nets & Surface Area
Back to Top ↑Nets & Surface Area
A net is a 2D pattern that folds to make a 3D solid. Surface area is the total area of all faces of the solid. At Higher Level this includes spheres and related shapes.
- Rectangular prism (cuboid): \( SA = 2(lw + lh + wh) \)
- Cube: \( SA = 6s^{2} \)
- Cylinder: total surface area \( = 2\pi r^{2} + 2\pi rh \); curved surface area \( = 2\pi rh \)
- Triangular prism: \( SA = \text{(area of both triangles)} + \text{(areas of all rectangles)} \)
- Sphere (HL): \( SA = 4\pi r^{2} \)
- Hemisphere (HL): curved surface area \( = 2\pi r^{2} \); curved surface area + circular base \( = 3\pi r^{2} \)
- Combinations: break the solid into familiar pieces (prisms, cylinders, hemispheres), find each surface area, then add.
Volume of a Cuboid (Interactive)
Explore the relationships between the dimensions and volume of a cuboid.
Surface Area of a Cuboid (Interactive)
Explore how the dimensions of a cuboid affect its total surface area.
Cylinder: Nets & Volume (Interactive)
Use this GeoGebra model to investigate how the height, radius and the net of a cylinder are connected to its volume and surface area.
Nets & Surface Area – Quick Quiz
Volume
Back to Top ↑Volume
Volume is the amount of 3D space inside a solid, measured in cubic units.
- Rectangular prism: \( V = lwh \)
- Cylinder: \( V = \pi r^{2}h \)
- Triangular prism: \( V = \left(\tfrac{1}{2}bh\right)L \)
- Sphere (HL): \( V = \tfrac{4}{3}\pi r^{3} \)
Check units: for example, \( 1\text{ m}^{3} = 1{,}000{,}000\text{ cm}^{3} \). Always convert lengths to the same unit before finding volume.
Surface Area & Volume of Spheres (Interactive)
Investigate how the radius affects both the surface area \( SA = 4\pi r^{2} \) and the volume \( V = \tfrac{4}{3}\pi r^{3} \) of a sphere.
Try:
- Dragging the slider and comparing how quickly volume grows compared to surface area.
- Predicting when the values will double.
- Comparing a sphere and hemisphere of the same radius.
Volume – Quick Quiz
OL Questions
Back to Top ↑OL 2025 – Question 4 (The Flag)
Suggested time: 5 minutesA rectangular flag has a length of \(100\text{ cm}\) and a width of \(60\text{ cm}\).
(a) Work out the area of the rectangular flag, in \(\text{cm}^2\).
(b) A circle is inside the rectangle. The radius of the circle is \(21\text{ cm}\). Work out the area, correct to the nearest \(\text{cm}^2\).
(a) Solution
Area \(= 100 \times 60 = 6000\text{ cm}^2\)
- Full Credit (5): \(100 \times 60\)
- Partial (2): writes correct formula
(b) Solution
\(A = \pi r^2 = \pi (21)^2 = 441\pi\)
\(441\pi \approx 1385\text{ cm}^2\)
- Partial (2): substitutes 21, finds 441
- Full Credit −1: incorrect rounding or gives \(441\pi\)
OL 2025 – Question 11 (Cabbage Patch)
Suggested time: 10 minutesTanya draws a scaled diagram of her garden, as shown below. The diagram includes a shed and a flower bed. (*Each small square has a width of 1 cm.)
(a) The diagram shows the distance \(x\) between the shed and the flower bed.
(i) Write down the value of \(x\) on the diagram. Give your answer in centimetres.
(ii) Hence, find the actual distance from the shed to the flower bed, in metres.
(iii) Tanya wants to put a rectangular vegetable patch in the garden. The vegetable patch will have a length of \(3\text{ m}\) and a width of \(1.5\text{ m}\). Draw this vegetable patch on the diagram so that it does not overlap with the shed or the flower bed.
(b) Tanya decides to plant cabbage in the vegetable patch. Each cabbage needs a square space of \(0.3\text{ m}\) by \(0.3\text{ m}\). What is the maximum number of cabbages that Tanya should plant in this vegetable patch?
(a)(i) & (ii) Solution
\((i)\) On the diagram, the distance marked \(x\) is \(3\text{ cm}\).
\((ii)\) Actual distance
\[
= 3 \times 0.5 = 1.5\text{ m}.
\]
The width of the garden on the diagram is \(10\text{ cm}\), so the actual width is
\[
10 \times 0.5 = 5\text{ m}.
\]
- Accept correct answers without units, or with incorrect units, in (i).
- Low Partial (3): work of merit, e.g. in (i) one value correct, or in (ii) one side length found in cm or m.
- High Partial (7): part (i) or (ii) correct, or work of merit in both parts.
- Full Credit (10) −1: apply * for missing or incorrect unit in (ii).
(a)(iii) Solution
Using the scale \(1\text{ cm} = 0.5\text{ m}\): \[ 3\text{ m} \rightarrow 6\text{ cm}, \qquad 1.5\text{ m} \rightarrow 3\text{ cm}. \] A rectangle measuring \(6\text{ cm}\) by \(3\text{ cm}\) is drawn on the diagram so that it does not overlap with the shed or the flower bed and stays inside the garden.
- Low Partial (2): work of merit, e.g. draws a line segment of \(6\text{ cm}\) or \(3\text{ cm}\), or does one correct conversion.
- High Partial (3):
- finds both \(6\text{ cm}\) and \(3\text{ cm}\), or
- draws rectangle of correct size but overlapping shed/flower bed or going outside garden, or
- draws rectangle in appropriate place with sides not correct lengths but in ratio \(1:2\).
- Full Credit (5): correct rectangle, correct position.
(b) Solution
Length direction: \(\dfrac{3}{0.3} = 10\) cabbages.
Width direction: \(\dfrac{1.5}{0.3} = 5\) cabbages.
Total number of cabbages
\[
10 \times 5 = 50.
\]
Alternatively,
\[
\dfrac{3 \times 1.5}{0.3 \times 0.3}
= \dfrac{4.5}{0.09} = 50.
\]
- Low Partial (2): work of merit, e.g. indicates two numbers with a correct operation such as \(3 \div 0.3\) or \(3 \times 1.5\); mentions \(5\) or \(10\).
- High Partial (3): correct answer \(50\) without supporting work.
- Full Credit (5): correct answer with supporting work.
OL 2025 – Question 12 (Pizza Box)
Suggested time: 10 minutes(b) The pizza is in a closed rectangular box with dimensions:
Length = \(28\text{ cm}\)
Width = \(25\text{ cm}\)
Height = \(4\text{ cm}\)
(i) Work out the volume of the pizza box.
(ii) Work out the surface area of the pizza box. Give your answer in \(\text{cm}^2\).
(b)(i) Solution
\(V = 28 \times 25 \times 4 = 2800\text{ cm}^3\)
- Accept correct answer without unit.
- Low Partial (4): shows relevant operation such as \(28 \times 25\).
- Mid Partial (6): (i) correct.
- High Partial (8): (i) correct and some face-area work for (ii).
- Full Credit (10) −1: incorrect/no unit in (i).
(b)(ii) Solution
\(A = 2[(25 \times 28) + (25 \times 4) + (28 \times 4)]\)
\(= 2[700 + 100 + 112]\)
\(= 2 \times 912 = 1824\text{ cm}^2\)
- Low Partial (4): indicates two numbers with operation.
- Mid Partial (6): finds area of two faces.
- High Partial (8): finds area of 4 faces or three unique faces + work.
- Full Credit (10) −1: apply * for incorrect unit.
OL 2024 – Question 9 (The Baking Tin)
Suggested time: 10 minutesA baking tin is in the shape of an open rectangular box. Alex makes a scaled model of the tin. It is \(7\text{ cm}\) long, \(5\text{ cm}\) wide and \(2\text{ cm}\) high, as shown in the diagram.
(a) Use the lengths in the diagram above to draw a net of the open rectangular box. One side has already been done for you. Each small square in the grid has a side of length \(1\text{ cm}\).
(b) Find the surface area of Alex’s open rectangular box.
(c) Alex made his model using a scale of \(1:3\). Complete the table to show the lengths of the sides of the actual tin. One length has already been done for you.
(a) Net & (b) Surface Area — Solutions
The net shows the base and the four side faces (no top). For the surface area:
Base: \(7 \times 5 = 35\text{ cm}^2\)
Two faces \(7 \times 2: \; 7 \times 2 = 14,\; 14 \times 2 = 28\text{ cm}^2\)
Two faces \(5 \times 2: \; 5 \times 2 = 10,\; 10 \times 2 = 20\text{ cm}^2\)
Total surface area:
\[
35 + 28 + 20 = 83\text{ cm}^2.
\]
- Accept correct answer for (b) without work, i.e. \(83\text{ cm}^2\).
- Low Partial (2): correct area of one face in (b), or any horizontal/vertical line of length 2, 5 or 7 cm on the net.
- Mid Partial (3–4): one part correct or work of merit in both parts.
- Full Credit (5) −1: incorrect or missing units; net of a closed rectangular box.
(c) Scale Diagram — Solution
Scale \(1:3\). Scaled length \(7\text{ cm}\) corresponds to actual \(21\text{ cm}\) (given).
For width: \(5 \times 3 = 15\text{ cm}\).
For height: \(2 \times 3 = 6\text{ cm}\).
So the missing actual lengths are \(15\text{ cm}\) and \(6\text{ cm}\).
- Partial (6): one of \(15\text{ cm}\) or \(6\text{ cm}\) correct.
- Full Credit (10): both \(15\text{ cm}\) and \(6\text{ cm}\) correct.
OL 2023 – Question 4 (Hexagon Classroom)
Suggested time: 10 minutesThe hexagon below is a scaled diagram of a classroom in a school. All sides are equal in length.
The length of the side of the hexagon measures \(3\text{ cm}\).
(b) Find the perimeter of the hexagon. Give your answer in cm.
(c) The diagram is to a scale of \(1\text{ cm} = 2\text{ m}\). Find the actual perimeter of the classroom. Give your answer in metres.
(b) Solution
Perimeter \[ 6 \times 3 = 18\text{ cm}. \]
- Accept answer using given side length.
- Partial (2): work of merit (e.g. sums two sides; finds 6 × side).
(c) Solution
Actual perimeter \[ 18 \times 2 = 36\text{ m}. \]
- Partial (2): divides the perimeter by 2 instead of multiplying.
OL 2023 – Question 10 (Yoghurt Carton)
Suggested time: 5 minutesMartina buys a carton of yoghurt. The carton is roughly in the shape of a cylinder. It has the dimensions shown in the diagram below.
(a) Work out the volume of the carton in \(\text{cm}^3\). Give your answer in terms of \(\pi\).
(b) The carton contains fruit and yoghurt and weighs \(450\text{ g}\). The ratio of fruit to yoghurt is \(4:21\). Work out how many grams of fruit are in the carton.
(a) Solution
Volume of a cylinder: \(V = \pi r^2 h\).
Here \(r = 4\text{ cm}\), \(h = 11\text{ cm}\):
\[
V = \pi (4)^2(11) = \pi(16)(11) = 176\pi\text{ cm}^3.
\]
- Accept correct answer without unit.
- Low Partial (5): work of merit in one part.
- Mid Partial (10): full correct substitution in (a), i.e. \(\pi(4)^2(11)\).
- High Partial (15): (a) or (b) fully correct, or work of merit in both parts.
- Full Credit (20) −1: answer \(176\) (omits \(\pi\)).
(b) Solution
Total number of parts in the ratio: \[ 4 + 21 = 25. \] Fruit is \(\dfrac{4}{25}\) of the total mass: \[ \dfrac{4}{25} \times 450 = 72\text{ g}. \] So there are \(72\text{ g}\) of fruit in the carton.
- High Partial (15): (a) or (b) correct, or work of merit in both parts.
- Full Credit (20) −1: finds grams of yoghurt instead, i.e. \(378\text{ g}\).
OL 2022 – Question 3 (The Airbed)
Suggested time: 10 minutesJoshua estimates that an airbed is roughly in the shape of a rectangular solid. The dimensions of the airbed are \(180\text{ cm}\), \(80\text{ cm}\), and \(20\text{ cm}\).
(a) Use these dimensions to show that the volume of the airbed is \(288\,000\text{ cm}^3\).
(b) An electric pump blows air into the airbed at a rate of \(800\text{ cm}^3\) per second. Work out how many minutes it will take to fill the airbed.
(c) Joshua starts to blow up the airbed at \(9{:}35\text{ p.m.}\). It takes \(45\) minutes to fill completely. At what time will the airbed be full?
(a) Solution
\(180 \times 80 \times 20 = 288\,000\text{ cm}^3\)
- Full Credit (5): \(180 \times 80 \times 20\) (−1 slip allowed)
- Partial (2): uses 180, 80, 20, or computes \(180 \times 80\), or writes volume formula.
(b) Solution
In one minute: \(60 \times 800 = 48\,000\text{ cm}^3\)
\(\displaystyle \frac{288\,000}{48\,000} = 6\text{ minutes}\)
- Full Credit (10): correct time \(6\) minutes.
- High Partial (7): one correct calculation, e.g. \(60 \times 800\), or \(\frac{288000}{60}\), or \(\frac{288000}{800}\).
- Low Partial (4): work of merit (e.g. mentions 60, or uses 800 meaningfully).
(c) Solution
\(9{:}35 + 0{:}45 = 9{:}80 = 10{:}20\text{ p.m.}\)
- Full Credit (10): \(10{:}20\text{ p.m.}\)
- Partial (5): work of merit (e.g. indicates 60 minutes, or adds two times correctly).
HL Questions
Back to Top ↑HL 2025 – Question 7 (Golf Course)
Suggested time: 15 minutesOne section of a golf course, shown in the diagram below, is in the shape of a rectangle with a semicircle at one end. The shape has a width of \(50\text{ m}\) and a total length of \(300\text{ m}\).
(a) Rectangular field with semicircle
(i) Work out the distances marked \(x\) and \(y\) on the diagram, where \(x\) is the radius of the semicircle.
(ii) Hence, work out the total area of this section of the golf course. Give your answer correct to the nearest \(\text{m}^2\).
Solution
(i) The width of the rectangle is the diameter of the semicircle, so \[ x = \frac{50}{2} = 25\text{ m}. \] The total length is \(300\text{ m}\), so the straight part of the rectangle has length \[ y = 300 - x = 300 - 25 = 275\text{ m}. \]
(ii) Area of rectangle:
\[
A_{\text{rect}} = 50 \times 275 = 13\,750\text{ m}^2.
\]
Area of full circle of radius \(25\text{ m}\):
\[
A_{\text{circle}} = \pi (25)^2 = 625\pi \approx 1963.5\text{ m}^2.
\]
Area of semicircle:
\[
A_{\text{semi}} = \tfrac12 \times 625\pi \approx 981.7\text{ m}^2.
\]
Total area:
\[
A_{\text{total}} = 13\,750 + 981.7 \approx 14\,731.7
\approx 14\,732\text{ m}^2.
\]
(i) Scale 5B (0, 3, 5)
- Accept correct answers without work.
- Partial (3): \(x\) or \(y\) correct.
- Full (5) −1: apply * once for incorrect or missing units.
- Accept correct answer \(\approx 14\,732\text{ m}^2\) without unit.
- View solution as four steps:
- Finds area of rectangle.
- Finds area of circle.
- Finds area of semicircle.
- Adds curvilinear and rectilinear areas.
- Low partial (4): work of merit in any step.
- High partial (6): three steps correct.
- Full (10) −1: incorrect rounding, otherwise fully correct.
(b) Fertiliser for the whole course
Maciej is spreading fertiliser on the whole golf course. The packet of fertiliser has these instructions:
Use 6 litres of fertiliser per \(250\text{ m}^2\) of ground to be covered.
Maciej needs to cover \(809\,371\text{ m}^2\) with fertiliser. How many litres of fertiliser will Maciej need to use? Give your answer correct to the nearest whole number.
Solution
Each \(250\text{ m}^2\) requires \(6\) litres. Number of litres needed: \[ \text{Litres} = \frac{809\,371 \times 6}{250} = 19\,424.904\ldots \approx 19\,425\text{ litres}. \]
Alternative: \(1\) litre covers \(\dfrac{250}{6}\text{ m}^2\). Then \[ \text{Litres} = \frac{809\,371}{250/6} \approx 19\,425. \]
- Low partial (2): work of merit, e.g. two numbers with a relevant operation such as \(809\,371\times 6\) or \(\tfrac{6}{250}\).
- High partial (3): indicates \(\dfrac{809\,371\times 6}{250}\) or equivalent expression.
- Full (5) −1: incorrect or missing rounding, otherwise correct.
(c) Ratio of ingredients
The fertiliser is made up of three ingredients, A, B, and C, in the following ratio.
Work out the number of millilitres of B in \(1\) litre of fertiliser.
Solution
Method 1 (fractions):
Total parts: \[ 1 + \frac{3}{2} + \frac{5}{3} = \frac{25}{6}. \] Fraction of the mixture that is \(B\): \[ \frac{3/2}{25/6} = \frac{3}{2}\times\frac{6}{25} = \frac{9}{25}. \] In \(1\) litre \(= 1000\text{ ml}\): \[ \text{Amount of B} = 1000\times\frac{9}{25} = 360\text{ ml}. \]
Method 2 (ratio):
Multiply all parts by \(6\): \[ 1 : \frac32 : \frac53 = 6 : 9 : 10. \] Total parts \(= 6 + 9 + 10 = 25\). \(B\) has \(9\) parts, so \[ \text{Amount of B} = 1000\times\frac{9}{25} = 360\text{ ml}. \]
- Accept correct answer \(360\text{ ml}\) without unit.
- Low partial (4): work of merit, e.g. correct conversion to ml, or indicates \(1+\tfrac32+\tfrac53\), or finds common denominator \(6\).
- High partial (6):
- In Method 1, finds \(\tfrac{25}{6}\) and does further relevant work, or
- In Method 2, finds \(6:9:10\) and total \(25\).
- Full (15) −1: answer in litres instead of ml, or finds ml for A or C only.
(d) Volume of fertiliser tank
The tank for spreading fertiliser is in the shape of a cylinder.
The radius of the tank is 90 centimetres and its length is 4 metres.
Work out the volume of the tank. Give your answer in \(\text{m}^3\), correct to 2 decimal places.
Solution
Convert radius to metres: \(r = 90\text{ cm} = 0.9\text{ m}\). Volume of a cylinder: \[ V = \pi r^2 h = \pi (0.9)^2(4) = \frac{81}{25}\pi \approx 10.178\ldots \approx 10.18\text{ m}^3. \]
Alternative: work in centimetres first, then convert \(\text{cm}^3\) to \(\text{m}^3\); both give \(10.18\text{ m}^3\).
- Accept correct answer \(10.18\text{ m}^3\) without unit.
- Low partial (2): work of merit, e.g. one correct conversion or some correct substitution into \(V=\pi r^2h\).
- High partial (3): correct formula fully substituted.
- Full (5) −1: incorrect rounding, otherwise correct.
(e) Cylindrical tank with given volume
The diagram on the right shows a different cylindrical tank with volume \(\dfrac{81\pi}{125}\text{ m}^3\). Its length is three times its radius.
Work out the radius of the tank, in metres.
Solution
Let the radius of the tank be \(r\) metres. Then the length is \(h = 3r\). Volume of a cylinder: \[ V = \pi r^2 h = \pi r^2 (3r) = 3\pi r^3. \] We are told \[ 3\pi r^3 = \frac{81\pi}{125}. \] Divide both sides by \(3\pi\): \[ r^3 = \frac{27}{125}. \] Hence \[ r = \sqrt[3]{\frac{27}{125}} = \frac{3}{5} = 0.6\text{ m}. \]
- Accept correct answer \(r = 0.6\text{ m}\) without unit.
- Consider solution as four steps:
- States \(h = 3r\).
- Sets up correct equation for volume.
- Isolates \(r^3\).
- Finds \(r\).
- Low partial (4): work of merit in one step.
- High partial (7): three steps correct.
- Full (20): all steps correct (−1 for missing/incorrect unit if penalised).
HL 2025 – Question 13 (Photo Frame)
Suggested time: 10 minutesIn this question, all lengths are in centimetres.
The diagram below shows a rectangular frame for a photo, where \(w,l,x \in \mathbb{R}\). The width of the frame is \(w\) and the length is \(l\). There is a rectangular hole in the middle for the photo, measuring \(12\) by \(18\). The internal width of the frame is \(x\), as shown.
(a) The length of the frame is \(l = 18 + 2x\). Write the width of the frame, \(w\), in terms of \(x\).
(b) Hence, show that \[ l \times w = 4x^2 + 60x + 216. \]
(a) & (b) Solution
From the diagram, vertically there is a strip of width \(x\) at the top, the inner height \(12\), and a strip of width \(x\) at the bottom. Therefore \[ w = x + 12 + x = 12 + 2x. \]
Using \(l = 18 + 2x\) and \(w = 12 + 2x\), \[ l \times w = (18 + 2x)(12 + 2x). \] Expand: \[ (18 + 2x)(12 + 2x) = 18\cdot12 + 18\cdot2x + 2x\cdot12 + 2x\cdot2x \] \[ = 216 + 36x + 24x + 4x^2 = 4x^2 + 60x + 216. \] Hence \(l \times w = 4x^2 + 60x + 216\), as required.
- Low partial (2): work of merit in (a) (e.g. writes an expression involving \(x\) and \(12\)) or in (b) (e.g. correctly substitutes at least one expression into \(l\times w\)).
- Mid partial (3–4): work of merit in both parts; for example, three of the four terms in \(216+36x+24x+4x^2\) correct.
- High/full (5): (a) correct and significant progress in (b), or fully correct expansion and simplification to \(4x^2+60x+216\).
(c) It is given that \(l \times w = 648\text{ cm}^2\). Use this, and the information from part (b), to find the value of \(x\). Give your answer in centimetres, correct to \(1\) decimal place.
(c) Solution
From part (b), \(l \times w = 4x^2 + 60x + 216\). It is also given that \(l \times w = 648\), so \[ 4x^2 + 60x + 216 = 648. \] Bring all terms to one side: \[ 4x^2 + 60x + 216 - 648 = 0 \quad\Rightarrow\quad 4x^2 + 60x - 432 = 0. \] Divide throughout by \(4\): \[ x^2 + 15x - 108 = 0. \] Using the quadratic formula, \[ x = \frac{-15 \pm \sqrt{15^2 - 4(1)(-108)}}{2} = \frac{-15 \pm \sqrt{225 + 432}}{2} = \frac{-15 \pm \sqrt{657}}{2}. \] \(\sqrt{657} \approx 25.6\), so the positive root is \[ x \approx \frac{-15 + 25.6}{2} \approx \frac{10.6}{2} \approx 5.3. \] Therefore \[ x \approx 5.3\text{ cm} \] (correct to one decimal place).
- Partial credit: forms a correct equation such as \(4x^2+60x+216=648\) or \(x^2+15x-108=0\).
- High partial: uses a correct quadratic method with minor numerical or rounding slips.
- Full credit: correct solution \(x \approx 5.3\text{ cm}\) to one decimal place, with valid algebraic work.
HL 2024 – Question 4 (Cylindrical glass)
Suggested time: 3 minutes(c) Each glass is approximately the shape of a cylinder with diameter \(6\text{ cm}\) and height \(10\text{ cm}\).
Find the volume of a glass, correct to the nearest \(\text{cm}^3\).
Volume of a cylinder: \(V = \pi r^2 h\).
The diameter is \(6\text{ cm}\), so the radius is \(r = \dfrac{6}{2} = 3\text{ cm}\). The height is \(h = 10\text{ cm}\).
\(V = \pi(3)^2(10) = \pi(9)(10) = 90\pi \approx 282.7\text{ cm}^3.\)
Correct to the nearest \(\text{cm}^3\), \(\boxed{V \approx 283\text{ cm}^3}\).
- Accept correct answer without work or with missing / incorrect units.
- Low partial (5): correct formula with some correct substitution, or finds \(r=3\text{ cm}\).
- High partial (7): fully correct substitution into \(V=\pi r^2 h\).
- Full credit (10) −1: incorrect rounding, or answer left in terms of \(\pi\).
HL 2024 – Question 6 (Toblerone Net)
Suggested time: 5 minutes
A chocolate bar in the shape of a triangular prism is shown. The triangular faces are equilateral triangles of side \(3\text{ cm}\), and the prism is \(9\text{ cm}\) long.
Complete an accurate net of the prism on the grid. One triangular face is already drawn. Each small square on the grid has side \(1\text{ cm}\).
Solution
The net consists of:
- three rectangles of size \(9\text{ cm} \times 3\text{ cm}\)
- two equilateral triangles of side \(3\text{ cm}\)
The rectangles form a strip, and the triangles attach at the ends.
- Four faces must be added correctly.
- Correct lengths for all edges.
- Tolerance: apex may lie anywhere within the correct square.
- Work of merit.
- Three faces correct OR fully drawn but incorrectly attached.
- All faces correct and correctly attached.
HL 2024 – Question 7 (Cube and Cuboids)
Suggested time: 10 minutesTwo rectangular solids, a cube \(A\) and a cuboid \(B\), are shown in the diagram below (not to scale). The edges of cube \(A\) have length \(x\text{ cm}\).
(a) Volume of cube \(A\)
The volume of \(A\) is \(216\text{ cm}^3\). Find the value of \(x\).
For a cube: \(V=x^3\).
Given \(x^3=216\).
Therefore \(x=6\text{ cm}\).
- Correct answer accepted without units.
- Partial for work of merit.
(b) Surface area of cuboid \(B\)
(i) Show that the surface area is \((22y+60)\text{ cm}^2\).
(ii) Given the surface area is \(269\text{ cm}^2\), find \(y\).
(i)
Surface area: \(S=2(lw+lh+wh)\).
Here \(l=5\), \(w=y\), \(h=6\): \(S=2(5y+30+6y)=2(11y+30)=(22y+60)\).
(ii)
\(22y+60=269\Rightarrow y=9.5\text{ cm}.\)
- Low partial: work of merit.
- Mid partial: one part fully correct.
- High partial: both parts correct.
(c) Volume of a different cuboid
The areas of the three faces are:
- Front: \(lm=96\)
- Top: \(mp=48\)
- Side: \(lp=32\)
Work out the volume of the cuboid.
Multiply face areas: \((lm)(mp)(lp)=96\cdot48\cdot32\).
LHS = \(l^2m^2p^2=(lmp)^2\).
\(lmp=\sqrt{147456}=384\).
Volume = \(\boxed{384\text{ cm}^3}\).
- Accept correct answer without working.
- Low: work of merit.
- High: two dimensions correct.
HL 2023 – Question 3 (Wax Block & Candles)
Suggested time: 10 minutesThe students in a Transition Year mini-company make and sell candles.
They buy a solid rectangular block of wax measuring 35 cm × 45 cm × 16 cm.
(a) Volume of block of wax
Work out the volume of the block of wax.
\(V = 35 \times 45 \times 16 = 25\,200\text{ cm}^3.\)
- Accept correct answer without work.
- Partial: some correct substitution.
- Full −1: missing/incorrect units.
(b) Volume of each candle
Each candle is a cylinder of radius \(r\text{ cm}\) and height \(9\text{ cm}\).
Work out the volume of each candle, in terms of \(r\) and \(\pi\).
\(V = \pi r^2 h = 9\pi r^2\text{ cm}^3\).
- Correct structure merits MP.
- Full: \(9\pi r^2\).
(c) Find the radius of each candle
The students make 100 candles. When making the candles, 10% of the wax is wasted.
Use this information to work out the value of \(r\). Give your answer correct to 1 decimal place.
Wax used: \(25\,200 \times 90\% = 22\,680\text{ cm}^3\).
100 candles → volume per candle \(=9\pi r^2\): \[ 100(9\pi r^2)=22\,680 \] \[ 9\pi r^2 = \frac{22\,680}{100}. \]
\[ r^2 = \frac{22\,680}{900\pi} \Rightarrow r = \sqrt{\frac{22\,680}{900\pi}} \approx 2.83. \]
\(r \approx 2.8\text{ cm}\).
- Steps required: candle volume → wax used → equation → \(r²\) → \(r\)
- Low: 1 step
- Mid: 2 steps
- High: 3 steps + finish
- Full −1: rounding errors or missing units
HL 2022 – Question 5 (Balloon Pump)
Suggested time: 10 minutesA balloon pump is made from a cylinder with an internal diameter of \(6\text{ cm}\) and a height of \(20\text{ cm}\). Each time the pump is pumped, it passes one full cylinder of air into a balloon.
(a) Volume of air from one pump
Show that the volume of one full cylinder of air is \(180\pi\text{ cm}^3\).
Volume of cylinder: \(V=\pi r^2h\).
Radius \(r=3\), height \(h=20\).
\(V=\pi(3)^2(20)=180\pi\text{ cm}^3\).
- Correct substitution.
- Radius identified.
- Correct conclusion \(180\pi\text{ cm}^3\).
(b) Inflating Darragh’s balloon
Darragh is inflating a spherical balloon of radius \(15\text{ cm}\).
(i) Find its volume in terms of \(\pi\).
Sphere volume \(V=\frac{4}{3}\pi r^3\).
\(r=15\): \(V=\frac{4}{3}\pi(15)^3=4500\pi\text{ cm}^3\).
- Correct formula + substitution.
(ii) How many seconds to inflate the balloon?
One pump: \(180\pi\).
Total required: \(4500\pi\).
Time \(=\frac{4500\pi}{180\pi}=25\) seconds.
- Full marks for 25 seconds.
(c) Gustav’s balloon
Gustav inflates a balloon for 50 seconds.
Find the final radius correct to 1 decimal place.
Total volume = \(50\times180\pi=9000\pi\).
\(\frac{4}{3}\pi r^3=9000\pi\Rightarrow r^3=6750\).
\(r=\sqrt[3]{6750}\approx18.9\text{ cm}\).
- Correct set-up.
- Correct cube root and rounding.
HL 2021 Sample - Question 9 (Cans in a box)
Suggested time: 15 minutesA drinks can is in the shape of a right circular cylinder with radius \(3.6\text{ cm}\) and height \(10\text{ cm}\).
Twenty–four of these cans are to be packed tightly into a closed rectangular box, as shown in the diagram below. The cans touch each other and the sides of the box.
(a) Volume of one can
Work out the volume of one can. Give your answer in \(\text{cm}^3\), correct to two decimal places.
Solution
Volume of a cylinder: \(V = \pi r^2 h\).
Here \(r = 3.6\) and \(h = 10\):
\(V = \pi(3.6)^2(10) = \pi(12.96)(10) = 129.6\pi \approx 407.15\ \text{cm}^3.\)
\(\boxed{V \approx 407.15\ \text{cm}^3}\).
- Uses \(V = \pi r^2 h\).
- Substitutes \(r = 3.6\), \(h = 10\).
- Obtains \(129.6\pi\) or correct decimal.
- Rounds correctly to two decimal places.
(b) Dimensions and volume of the box
In the arrangement shown, the cans are stacked two high, four along the length, and three across the width of the box. The length of the box is \(28.8\text{ cm}\).
(i) Write down the height and the width of this box.
Solution (b)(i)
Diameter of one can: \(2r = 2(3.6) = 7.2\text{ cm}\).
Two cans stacked vertically give height \(= 2 \times 10 = 20\text{ cm}\).
Three cans across the width give width \(= 3 \times 7.2 = 21.6\text{ cm}\).
\(\boxed{\text{height} = 20\text{ cm},\ \text{width} = 21.6\text{ cm}}\).
- Finds diameter \(7.2\text{ cm}\).
- Height \(= 20\text{ cm}\).
- Width \(= 21.6\text{ cm}\).
(ii) Work out the volume of this rectangular box.
Solution (b)(ii)
\(V_{\text{box}} = \text{length} \times \text{width} \times \text{height}\).
\(V_{\text{box}} = 28.8 \times 21.6 \times 20\).
\(28.8 \times 21.6 = 622.08,\; 622.08 \times 20 = 12\,441.6\ \text{cm}^3.\)
\(\boxed{V_{\text{box}} = 12\,441.6\ \text{cm}^3}\).
- Uses \(V = lwh\).
- Substitutes correct dimensions.
- Correct multiplication to \(12\,441.6\ \text{cm}^3\).
(iii) Work out the percentage of the volume of this box that is taken up by the 24 cans. Give your answer correct to one decimal place.
Solution (b)(iii)
Total volume of cans: \(V_{\text{cans}} = 24 \times 407.15 \approx 9\,771.6\ \text{cm}^3.\)
Percentage of the box filled: \[ \frac{V_{\text{cans}}}{V_{\text{box}}}\times100 = \frac{9\,771.6}{12\,441.6}\times100 \approx 78.5398\%. \]
Correct to one decimal place: \(\boxed{78.5\%}\).
- Finds total volume of cans.
- Divides by box volume.
- Multiplies by \(100\).
- Rounds correctly to one decimal place.
(c) A different arrangement of the 24 cans
There are different ways to arrange the 24 cans so that they fit in a rectangular box. Find the dimensions of a rectangular box required for a different arrangement of the 24 cans. The cans should still touch each other and the sides of the box. Show your working clearly.
Solution
We need a different arrangement from the \(2 \times 4 \times 3\) layout.
Choose \(24 = 3 \times 4 \times 2\).
Arrange the cans as: 3 high, 4 along the length, 2 across the width.
Height: \(3 \times 10 = 30\text{ cm}\).
Length: \(4 \times 7.2 = 28.8\text{ cm}\).
Width: \(2 \times 7.2 = 14.4\text{ cm}\).
One valid set of dimensions is
\(\boxed{\text{height} = 30\text{ cm},\
\text{length} = 28.8\text{ cm},\
\text{width} = 14.4\text{ cm}}\).
Any other correct arrangement using all 24 cans is acceptable.
- Chooses a factorisation of \(24\) different from \(2 \times 4 \times 3\).
- Links factors correctly to stack height / length / width.
- Uses diameter \(7.2\text{ cm}\) and height \(10\text{ cm}\).
- Gives a correct set of dimensions.
Area & Volume — Junior Cycle Syllabus
Back to Top ↑Students should be able to:
GT.1 calculate, interpret, and apply units of measure and time
GT.2 investigate 2D shapes and 3D solids so that they can:
- draw and interpret scaled diagrams
- draw and interpret nets of rectangular solids, prisms (polygonal bases), cylinders
- find the perimeter and area of plane figures made from combinations of discs, triangles, and rectangles, including relevant operations involving π
- find the volume of rectangular solids, cylinders, triangular-based prisms, spheres, and combinations of these, including relevant operations involving π
- find the surface area and curved surface area (as appropriate) of rectangular solids, cylinders, triangular-based prisms, spheres, and combinations of these
