2D Plane Figures & Composite Area Calculations
Composite shapes are created by combining or subtracting standard geometric plane figures such as polygons, circles, and sectors.
- Triangle: $A = \frac{1}{2}bh = \frac{1}{2}ab\sin C$
- Parallelogram: $A = bh$
- Trapezium: $A = \frac{1}{2}(a + b)h$, where $a$ and $b$ are parallel bases and $h$ is perpendicular height
- Rhombus & Kite: $A = \frac{1}{2}d_1 d_2$, where $d_1, d_2$ are perpendicular diagonals
- Circle: $A = \pi r^2$, Circumference $C = 2\pi r = \pi d$
- Circular Sector: $A = \frac{\theta}{360^\circ} \pi r^2$ (degrees) or $A = \frac{1}{2}r^2\theta$ (radians)
Decomposition Principle: When solving composite figures:
- Partition the shape into non-overlapping standard sub-regions, or enclose it in a bounding rectangle and subtract void areas.
- For perimeter, sum only the exposed exterior boundaries. Never include internal shared partition lines!
Surface Area & Volume of Prisms & Cylinders
A prism is a 3D solid with uniform cross-sectional area along its entire length. A cylinder is a circular prism.
- Volume: $V = \pi r^2 h$
- Curved Surface Area (CSA): $CSA = 2\pi rh$ (the unrolled lateral rectangle)
- Total Surface Area (Closed): $TSA = 2\pi r^2 + 2\pi rh = 2\pi r(r + h)$
- Open Cylinder (one end open): $SA = \pi r^2 + 2\pi rh$
Cylinder geometry: The curved surface area unrolls into a rectangle of length $2\pi r$ and height $h$.
Pyramids, Cones, Spheres & Frustums
Tapering solids come to an apex point, reducing their volume to one-third of the enclosing prism or cylinder.
- Pyramid Volume: $V = \frac{1}{3} A_{\text{base}} h$
- Right Cone Volume: $V = \frac{1}{3} \pi r^2 h$
- Cone Slant Height ($l$): $l = \sqrt{r^2 + h^2}$ (via Pythagoras)
- Cone Curved Surface Area (CSA): $CSA = \pi r l$
- Cone Total Surface Area: $TSA = \pi r^2 + \pi r l = \pi r(r + l)$
- Sphere Volume: $V = \frac{4}{3}\pi r^3$
- Sphere Surface Area: $A = 4\pi r^2$
- Solid Hemisphere Volume: $V = \frac{2}{3}\pi r^3$
- Solid Hemisphere Total Surface Area: $TSA = 2\pi r^2 + \pi r^2 = 3\pi r^2$ (curved bowl + flat circular lid)
A frustum is formed by slicing off the top apex parallel to the base: $$V_{\text{frustum}} = V_{\text{original solid}} - V_{\text{removed top solid}}$$ For a conical frustum with bottom radius $R$, top radius $r$, and height $h$: $$V = \frac{1}{3}\pi h (R^2 + Rr + r^2)$$ $$\text{Lateral Area } = \pi (R + r)l_{\text{slant}}$$
Geometric Similarity & Dimensional Scaling ($k, k^2, k^3$)
When two objects are geometrically similar, all corresponding linear dimensions are magnified by a linear scale factor $k = \frac{L_2}{L_1}$.
Mass and Capacity Connections: If two similar objects are made of the same uniform material (constant density $\rho = \frac{m}{V}$), their masses and capacities scale directly with volume: $$\frac{m_2}{m_1} = \frac{V_2}{V_1} = k^3$$
Worked Example 1: Composite Solid Surface Area & Volume
A storage silo consists of a vertical cylinder of radius $3\text{ m}$ and height $8\text{ m}$, topped with a right conical roof of height $4\text{ m}$.
a) Find the total internal capacity (volume) of the silo in terms of $\pi$ and to 3 significant figures.
b) Calculate the total exterior surface area (excluding the floor base) that needs painting.
Step-by-step Solution:
Part a: Total Volume
Cylinder Volume: $V_{\text{cyl}} = \pi r^2 h_{\text{cyl}} = \pi (3^2)(8) = 72\pi\text{ m}^3$. [M1]
Cone Volume: $V_{\text{cone}} = \frac{1}{3}\pi r^2 h_{\text{cone}} = \frac{1}{3}\pi (3^2)(4) = \frac{1}{3}\pi (9)(4) = 12\pi\text{ m}^3$. [M1]
Total Volume: $V = 72\pi + 12\pi = 84\pi \approx 264\text{ m}^3$ (3 s.f.). [Criterion A: Knowledge & Understanding]
Part b: Exterior Surface Area (Excluding Floor)
The painted area consists of the cylinder's curved lateral wall plus the cone's curved roof (no floor, no internal dividing ceiling).
Cylinder CSA: $CSA_{\text{cyl}} = 2\pi r h_{\text{cyl}} = 2\pi (3)(8) = 48\pi\text{ m}^2$. [M1]
For the cone, slant height $l = \sqrt{r^2 + h_{\text{cone}}^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\text{ m}$. [M1]
Cone CSA: $CSA_{\text{cone}} = \pi r l = \pi (3)(5) = 15\pi\text{ m}^2$.
Total Area: $A = 48\pi + 15\pi = 63\pi \approx 198\text{ m}^2$ (3 s.f.). [Criterion C: Mathematical Communication]
Worked Example 2: Extended Frustum Volume & Mass Scaling
A solid metal trophy is made in the shape of a cone of base radius $10\text{ cm}$ and height $24\text{ cm}$. The top cone of height $12\text{ cm}$ is sliced off to produce a frustum.
a) Find the volume of the resulting frustum.
b) A larger similar trophy has a volume of $8$ times the volume of the original cone. What is the ratio of their surface areas?
Step-by-step Solution:
Part a: Frustum Volume
Original large cone: $r_1 = 10\text{ cm}, h_1 = 24\text{ cm}$.
$V_{\text{large}} = \frac{1}{3}\pi (10^2)(24) = 800\pi\text{ cm}^3$. [M1]
The removed top cone has height $h_2 = 12\text{ cm}$, which is $\frac{12}{24} = \frac{1}{2}$ the height of the original cone.
Linear scale factor $k = \frac{1}{2}$. By similarity, the top radius is $r_2 = \frac{1}{2}(10) = 5\text{ cm}$. [M1]
The volume scale factor is $k^3 = \left(\frac{1}{2}\right)^3 = \frac{1}{8}$.
Volume of top cone: $V_{\text{small}} = \frac{1}{8} \times 800\pi = 100\pi\text{ cm}^3$.
Frustum Volume: $V_{\text{frustum}} = 800\pi - 100\pi = 700\pi \approx 2199\text{ cm}^3 \approx 2.20\times 10^3\text{ cm}^3$. [A1]
Part b: Surface Area Scaling Ratio
Volume ratio $\frac{V_2}{V_1} = 8 \implies k^3 = 8 \implies k = \sqrt[3]{8} = 2$. [M1]
The surface area scale factor is $k^2 = 2^2 = 4$.
Therefore, the ratio of their surface areas is $4:1$. [Criterion A: Analysis & Conceptual Understanding]
Tiered Practice Exercises
10 problems advancing from standard 2D/3D area and volume calculations to complex composite frustums and real-world geometric modelling.
Find the area of a trapezium with parallel sides of length $14\text{ cm}$ and $22\text{ cm}$, and a perpendicular height of $9\text{ cm}$.
View Solution & Markscheme
Method & Marks:
Formula: $A = \frac{1}{2}(a + b)h$ [M1]
$A = \frac{1}{2}(14 + 22)(9) = \frac{1}{2}(36)(9) = 18 \times 9 = 162\text{ cm}^2$. [A1]
A solid rectangular prism has a length of $12\text{ cm}$, width of $7\text{ cm}$, and height of $5\text{ cm}$. Calculate its volume and total surface area.
View Solution & Markscheme
Method & Marks:
Volume: $V = lwh = 12 \times 7 \times 5 = 420\text{ cm}^3$. [A1]
Total Surface Area: $TSA = 2(lw + lh + wh) = 2(12 \times 7 + 12 \times 5 + 7 \times 5) = 2(84 + 60 + 35) = 2(179) = 358\text{ cm}^2$. [A1]
A closed cylinder has a radius of $4\text{ cm}$ and a height of $10\text{ cm}$. Find its volume and total surface area in terms of $\pi$.
View Solution & Markscheme
Method & Marks:
Volume: $V = \pi r^2 h = \pi (4^2)(10) = 160\pi\text{ cm}^3$. [A1]
Total Surface Area: $TSA = 2\pi r^2 + 2\pi rh = 2\pi (16) + 2\pi (4)(10) = 32\pi + 80\pi = 112\pi\text{ cm}^2$. [A1]
A solid right cone has base radius $5\text{ cm}$ and vertical height $12\text{ cm}$. Determine its slant height $l$, curved surface area, and total surface area (give answers in terms of $\pi$).
View Solution & Markscheme
Method & Marks:
Slant height via Pythagoras: $l = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\text{ cm}$. [M1]
Curved Surface Area: $CSA = \pi r l = \pi (5)(13) = 65\pi\text{ cm}^2$. [A1]
Total Surface Area: $TSA = \pi r^2 + CSA = \pi (5^2) + 65\pi = 25\pi + 65\pi = 90\pi\text{ cm}^2$. [A1]
A solid hemisphere has a radius of $6\text{ cm}$. Calculate its volume and total surface area (including the flat circular base) in terms of $\pi$.
View Solution & Markscheme
Method & Marks:
Volume: $V = \frac{2}{3}\pi r^3 = \frac{2}{3}\pi (6^3) = \frac{2}{3}\pi (216) = 144\pi\text{ cm}^3$. [A1]
Total Surface Area of solid hemisphere: $TSA = 2\pi r^2 + \pi r^2 = 3\pi r^2$. [M1]
$TSA = 3\pi (6^2) = 3\pi (36) = 108\pi\text{ cm}^2$. [A1]
A metal cylinder of radius $3\text{ cm}$ and height $16\text{ cm}$ is melted down and recast into spherical ball bearings, each of radius $1.5\text{ cm}$. Assuming no metal is lost, how many complete ball bearings can be made?
View Solution & Markscheme
Method & Marks:
Volume of cylinder: $V_{\text{cyl}} = \pi (3^2)(16) = 144\pi\text{ cm}^3$. [M1]
Volume of one sphere: $V_{\text{sph}} = \frac{4}{3}\pi (1.5)^3 = \frac{4}{3}\pi (3.375) = 4.5\pi\text{ cm}^3$. [M1]
Number of spheres: $N = \frac{144\pi}{4.5\pi} = \frac{144}{4.5} = 32$. [A1]
Two geometrically similar decorative statues have heights of $15\text{ cm}$ and $25\text{ cm}$. If the smaller statue has a surface area of $180\text{ cm}^2$ and a mass of $675\text{ g}$, find the surface area and mass of the larger statue.
View Solution & Markscheme
Method & Marks:
Linear scale factor $k = \frac{25}{15} = \frac{5}{3}$. [M1]
Surface area scales with $k^2$: $A_2 = A_1 \times k^2 = 180 \times \left(\frac{5}{3}\right)^2 = 180 \times \frac{25}{9} = 20 \times 25 = 500\text{ cm}^2$. [A1]
Mass scales with volume $k^3$: $m_2 = m_1 \times k^3 = 675 \times \left(\frac{5}{3}\right)^3 = 675 \times \frac{125}{27} = 25 \times 125 = 3125\text{ g} = 3.125\text{ kg}$. [A1]
A frustum of a right circular cone has a lower base radius of $12\text{ cm}$, an upper base radius of $6\text{ cm}$, and a height of $8\text{ cm}$. Calculate:
a) The volume of the frustum.
b) The total surface area of the closed frustum.
View Solution & Markscheme
Method & Marks:
Part a: Volume
Using the frustum volume formula: $V = \frac{1}{3}\pi h (R^2 + Rr + r^2)$ [M1]
$V = \frac{1}{3}\pi (8) (12^2 + 12 \times 6 + 6^2) = \frac{8\pi}{3}(144 + 72 + 36) = \frac{8\pi}{3}(252) = 8\pi \times 84 = 672\pi \approx 2111\text{ cm}^3$. [A1]
Part b: Total Surface Area
Slant height $l = \sqrt{h^2 + (R - r)^2} = \sqrt{8^2 + (12 - 6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10\text{ cm}$. [M1]
Lateral curved area: $CSA = \pi (R + r)l = \pi (12 + 6)(10) = 180\pi\text{ cm}^2$. [M1]
Two flat circular bases: $A_{\text{bases}} = \pi R^2 + \pi r^2 = \pi (144 + 36) = 180\pi\text{ cm}^2$.
$TSA = 180\pi + 180\pi = 360\pi \approx 1131\text{ cm}^2$. [A1]
A regular square-based pyramid has a base side length of $2a$ and a vertical height of $h$. If the total surface area of the pyramid is equal to three times the area of its square base, express $h$ in terms of $a$.
View Solution & Markscheme
Method & Marks:
Base area: $A_{\text{base}} = (2a)^2 = 4a^2$. [M1]
Given $TSA = 3 \times A_{\text{base}} = 12a^2$.
$TSA = A_{\text{base}} + 4 \times A_{\text{triangular face}} = 4a^2 + 4\left(\frac{1}{2} \times 2a \times l_{\text{slant}}\right) = 4a^2 + 4a l_{\text{slant}}$. [M1]
$4a^2 + 4a l_{\text{slant}} = 12a^2 \implies 4a l_{\text{slant}} = 8a^2 \implies l_{\text{slant}} = 2a$. [A1]
In the right triangle linking height, half-base, and slant height: $l^2 = h^2 + a^2$. [M1]
$(2a)^2 = h^2 + a^2 \implies 4a^2 = h^2 + a^2 \implies h^2 = 3a^2 \implies h = a\sqrt{3}$. [A1][R1]
Criterion B & D Real-World Investigation: A municipal water reservoir is engineered as an inverted conical tank of radius $R = 8\text{ m}$ and total depth $H = 12\text{ m}$. Water is being pumped into the empty tank at a constant rate of $15\text{ m}^3/\text{min}$.
a) Using similar triangles, express the volume $V$ of water as a function of the water depth $h$.
b) Calculate the depth of the water after $20\text{ minutes}$ of pumping.
c) Evaluate how long it will take to fill the tank completely.
View Solution & Markscheme
Method & Marks:
Part a: In the vertical cross-section, the water cone is similar to the tank cone: $\frac{r}{h} = \frac{R}{H} = \frac{8}{12} = \frac{2}{3} \implies r = \frac{2}{3}h$. [M1]
Volume $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{2}{3}h\right)^2 h = \frac{1}{3}\pi \left(\frac{4}{9}h^2\right)h = \frac{4\pi}{27}h^3$. [A1]
Part b: After $20\text{ min}$, water volume $V = 15 \times 20 = 300\text{ m}^3$.
$\frac{4\pi}{27}h^3 = 300 \implies h^3 = \frac{300 \times 27}{4\pi} = \frac{8100}{4\pi} = \frac{2025}{\pi} \approx 644.577$. [M1]
$h = \sqrt[3]{644.577} \approx 8.64\text{ m}$. [A1]
Part c: Total capacity $V_{\text{total}} = \frac{1}{3}\pi (8^2)(12) = 256\pi \approx 804.25\text{ m}^3$. [M1]
Time to fill: $t = \frac{256\pi}{15} \approx \frac{804.25}{15} \approx 53.6\text{ minutes}$ (approx 53 min 37 sec). [A1][R1: Model evaluation]
Mensuration & 3D Solids Formula Sheet
Prisms
$$V = A_{\text{base}} \times h$$
Uniform cross-section.Cylinder
$$V = \pi r^2 h, \; CSA = 2\pi rh$$
$TSA = 2\pi r(r + h)$.Pyramids
$$V = \frac{1}{3} A_{\text{base}} h$$
One-third rule for pointed solids.Cone
$$V = \frac{1}{3}\pi r^2 h, \; CSA = \pi r l$$
$l = \sqrt{r^2 + h^2}$.Sphere
$$V = \frac{4}{3}\pi r^3, \; SA = 4\pi r^2$$
Uniform curved ball.Solid Hemisphere
$$V = \frac{2}{3}\pi r^3, \; TSA = 3\pi r^2$$
Includes the circular flat base.Conical Frustum
$$V = \frac{1}{3}\pi h(R^2 + Rr + r^2)$$
Truncated cone.Similarity Scaling
$$\frac{L_2}{L_1} = k, \; \frac{A_2}{A_1} = k^2, \; \frac{V_2}{V_1} = k^3$$
Linear, area, and volumetric scaling.Top 5 Common MYP Mensuration Pitfalls
- Confusing Cone Height ($h$) with Slant Height ($l$): Volume uses perpendicular vertical height $h$ ($V = \frac{1}{3}\pi r^2 h$), but Curved Surface Area uses slant height $l$ ($CSA = \pi r l$). Always check whether the question gives $h$ or $l$ and apply Pythagoras ($l = \sqrt{r^2 + h^2}$) when needed.
- Forgetting the Flat Base on Hemispheres & Cylinders: When computing total surface area for a solid hemisphere, students frequently write $2\pi r^2$ instead of $3\pi r^2$. Remember the flat bottom circle ($\pi r^2$) adds to the curved bowl ($2\pi r^2$).
- Including Internal Shared Faces in Composite Solids: When two solids are glued together (such as a cone on top of a cylinder), the shared interface circle is hidden inside. Do NOT include it in the exterior surface area.
- Squaring Instead of Cubing for Volume Scaling: If the linear dimensions of a solid are doubled ($k = 2$), its surface area increases by $k^2 = 4$, but its volume and mass increase by $k^3 = 8$. Never multiply volume by the linear scale factor!
- Unit Inconsistency: Always convert all dimensions to the same unit before computing (e.g. converting cm to m). Remember: $1\text{ m}^3 = 1{,}000{,}000\text{ cm}^3 = 1{,}000\text{ litres}$, and $1\text{ litre} = 1{,}000\text{ cm}^3$.
Self-Grading Mini Test
5 quiz questions on perimeter, area, volume, and similarity scaling.
A trapezium has parallel sides of $8\text{ cm}$ and $12\text{ cm}$, and perpendicular height $5\text{ cm}$. What is its area?
Explanation: $A = \frac{1}{2}(a + b)h = \frac{1}{2}(8 + 12)(5) = \frac{1}{2}(20)(5) = 50\text{ cm}^2$.
A cone has radius $r = 6\text{ cm}$ and vertical height $h = 8\text{ cm}$. What is its curved surface area in terms of $\pi$?
Explanation: Slant height $l = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10\text{ cm}$. Then $CSA = \pi r l = \pi (6)(10) = 60\pi\text{ cm}^2$.
What is the total surface area of a solid hemisphere of radius $r = 3\text{ cm}$?
Explanation: A solid hemisphere has curved area $2\pi r^2$ plus circular base $\pi r^2$, giving $TSA = 3\pi r^2 = 3\pi (3^2) = 27\pi\text{ cm}^2$.
Two similar cylinders have heights in the ratio $1 : 3$. If the smaller cylinder has a volume of $20\text{ cm}^3$, what is the volume of the larger cylinder?
Explanation: Linear ratio is $k = 3$. Volume scales as $k^3 = 3^3 = 27$. Volume of larger cylinder $= 20 \times 27 = 540\text{ cm}^3$.
A sphere has a diameter of $6\text{ cm}$. What is its volume in terms of $\pi$?
Explanation: Diameter $d = 6 \implies r = 3\text{ cm}$. $V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (3^3) = \frac{4}{3}\pi (27) = 36\pi\text{ cm}^3$.