Calculating the surface area of triangular prisms and pyramids & cylinders
🎯 In this topic you will
- Calculate the surface area of triangular prisms
- Calculate the surface area of pyramids
- Calculate the surface area of cylinders
🧠 Key Words
- net
- surface area
Show Definitions
- net: A 2D pattern that can be folded to form the surfaces of a 3D solid.
- surface area: The total area of all the faces of a 3D shape added together.
📐 Using Nets to Find Surface Area
You already know how to draw the net of a cube or cuboid to help you work out the surface area of the shape. You can use the same method to help you work out the surface area of triangular prisms and pyramids.
❓ EXERCISES
1. Copy and complete the workings to find the surface area of each shape.
a.
👀 Show answer
$A=8\times 12=96\ \text{cm}^2$; $B=6\times 12=72\ \text{cm}^2$; $C=10\times 12=120\ \text{cm}^2$.
Triangular ends: $D=E=\tfrac12\times 6\times 8=24\ \text{cm}^2$.
Surface area $=96+72+120+2\times 24=336\ \text{cm}^2$.
b.
👀 Show answer
Each triangular face: $B=\tfrac12\times 10\times 14=70\ \text{cm}^2$ (and $C,D,E$ are the same).
Surface area $=100+4\times 70=380\ \text{cm}^2$.
❓ EXERCISES
2. For each of these solids
i sketch a net
ii work out the surface area.
a.
b.
c.
d.
👀 Show answer
Nets: any correct nets showing all faces once each are acceptable.
a. Isosceles triangular prism with triangle sides $13,13,24$ and height $5$, length $30$.
Triangle area $=\tfrac12\times 24\times 5= $ $60\ \text{cm}^2$. Perimeter $=13+13+24= $ $50\ \text{cm}$. Lateral area $=50\times 30= $ $1500\ \text{cm}^2$. Total $=1500+2\times 60= $ $1620\ \text{cm}^2$.
b. Right-triangular prism with legs $6,8$, hypotenuse $10$, length $9$.
Triangle area $=\tfrac12\times 6\times 8= $ $24\ \text{cm}^2$. Perimeter $=6+8+10= $ $24\ \text{cm}$. Lateral area $=24\times 9= $ $216\ \text{cm}^2$. Total $=216+2\times 24= $ $264\ \text{cm}^2$.
c. Square-based pyramid, base side $18$, slant height $12$.
Base area $=18\times 18= $ $324\ \text{cm}^2$. Lateral area $=4\left(\tfrac12\times 18\times 12\right)= $ $432\ \text{cm}^2$. Total $=324+432= $ $756\ \text{cm}^2$.
d. Triangular-based pyramid with all triangles equal (regular tetrahedron). Each face has base $15$ and height $13$ (since an equilateral triangle of side $15$ has height $\approx 13$).
One face area $=\tfrac12\times 15\times 13= $ $97.5\ \text{cm}^2$. With $4$ faces: $390\ \text{cm}^2$.
3. The diagram shows a triangular prism and a cube. Which shape has the greater surface area? Show your working.
👀 Show answer
Prism cross-section is isosceles with sides $5,5,6$ and height $4$; length $15$.
Triangle area $=\tfrac12\times 6\times 4= $ $12\ \text{cm}^2$; perimeter $=5+5+6= $ $16\ \text{cm}$.
Lateral area $=16\times 15= $ $240\ \text{cm}^2$; total prism area $=240+2\times 12= $ $264\ \text{cm}^2$.
Cube side $=7\ \text{cm}$, area $=6\times 7^2=6\times 49= $ $294\ \text{cm}^2$.
Greater surface area: the cube, since $294>264$.
❓ EXERCISES
4. The diagram shows a triangular-based pyramid and a cuboid.
In the triangular-based pyramid, all triangles are the same size.
Show that the surface area of the triangular-based pyramid is $8\ \text{m}^2$ more than the surface area of the cuboid.
👀 Show answer
Pyramid: Each triangular face has base $4\ \text{m}$ and height $3.5\ \text{m}$, so area of one face $=\tfrac12\times 4\times 3.5=7\ \text{m}^2$. There are $4$ congruent faces, so total surface area $=4\times 7=28\ \text{m}^2$.
Cuboid: Dimensions $2\ \text{m}\times 2\ \text{m}\times 1.5\ \text{m}$. Surface area $=2\big(2\times 2+2\times 1.5+2\times 1.5\big)=2(4+3+3)=20\ \text{m}^2$.
Difference $=28-20=8\ \text{m}^2$. Hence, the triangular-based pyramid has surface area $8\ \text{m}^2$ more than the cuboid.
🧠 Think like a Mathematician
5. This square-based pyramid has a base side length of $x$ cm. The perpendicular height of each triangular face is double the base side length.
a) Write an expression for the area of the base of the pyramid.
b) Write an expression for the area of one of the triangular faces of the pyramid.
c) Write a formula for the surface area of the pyramid.
d) In Pyramid A$x=5$. In Pyramid B$x=7$. Use your formula to work out the difference in surface area between the two pyramids.
e) Compare and discuss your answers to parts c and d.
👀 Show Answer
a) Base is a square of side $x$ ⇒ area $x^2\ \text{cm}^2$.
b) Each triangular face has base $x$ and perpendicular height $2x$ ⇒ area $\tfrac12 \cdot x \cdot 2x = x^2\ \text{cm}^2$.
c) Surface area = base + 4 triangular faces $= x^2 + 4(x^2) = 5x^2\ \text{cm}^2$.
d) For A: $x=5 \Rightarrow 5x^2 = 5(25)=125\ \text{cm}^2$.
For B: $x=7 \Rightarrow 5x^2 = 5(49)=245\ \text{cm}^2$.
Difference: $245-125=120\ \text{cm}^2$.
e) The surface area grows with $x^2$ (quadratically). Increasing the side from 5 to 7 increases area by $5(7^2-5^2)=5(24)=120$, matching part d.
❓ EXERCISES
6. The base of a triangular pyramid is an equilateral triangle with base length $6\ \text{cm}$ and perpendicular height $5.2\ \text{cm}$.
The sides of the triangular pyramid are isosceles triangles with base length $6\ \text{cm}$ and perpendicular height $8.7\ \text{cm}$.
Work out the surface area of the pyramid.
👀 Show answer
Each side face (isosceles triangle): area $=\tfrac12 \times 6 \times 8.7 = 26.1\ \text{cm}^2$.
There are $3$ such faces, so total for sides $= 78.3\ \text{cm}^2$.
Total surface area $=15.6 + 78.3 = \mathbf{93.9\ \text{cm}^2}$.
🧠 Reasoning Tip
Draw a diagram to help you.
7. This triangular prism has a volume of $180\ \text{cm}^3$.
The area of the triangular cross-section of the prism is $A$.
Use the information given to work out the surface area of the triangular prism.
Given: $l=5\ \text{cm},\ h=\sqrt{A},\ b=2h,\ x=2\tfrac15 \times l,\ y=1.4\times l$.
👀 Show answer
Then $h=\sqrt{36}=6\ \text{cm}$, so $b=2h=12\ \text{cm}$.
$x=2.2\times 5=11\ \text{cm}$, $y=1.4\times 5=7\ \text{cm}$.
Surface area $=$ 2 triangular cross-sections + 3 rectangles.
Two triangles $=2A=72\ \text{cm}^2$.
Rectangles: $b \times l=12\times 5=60$, $x\times l=11\times 5=55$, $y\times l=7\times 5=35$.
Total lateral area $=60+55+35=150\ \text{cm}^2$.
Total surface area $=72+150=\mathbf{222\ \text{cm}^2}$.
📐 From nets to cylinders
You already know how to draw the net of a triangular prism and of a pyramid. You also know how to use nets to work out the surface area of these shapes. You can use the same method to work out the surface area of a cylinder.
✅ Quick Fact
A cylinder’s net is a rectangle plus two circles. The rectangle’s width equals the circumference $2\pi r$ and its height equals the cylinder’s length (height). So the lateral (curved) area is $2\pi r \times h$, and the total surface area is $2\pi r h + 2\pi r^2$.
❓ EXERCISES
🧠 Reasoning Tip
Sketch the net of the cylinder to help with calculations.
1. Copy and complete the workings to find the surface area of this cylinder.
👀 Show answer
Circumference $=\pi d=\pi \times 10=31.42\ \text{cm}$ (2 d.p.)
Area of rectangle $=31.42 \times 12=377.04\ \text{cm}^2$ (2 d.p.)
Total surface area $=2\times 78.54+377.04=534.12\ \text{cm}^2 \approx \mathbf{534\ \text{cm}^2}$ (3 s.f.)
2. Work out the surface area of each cylinder. Give your answers correct to one decimal place (1 d.p.).
👀 Show answer
a) $r=4,\ h=15$: $2\pi(4^2)+2\pi(4)(15)=100.53+376.99=477.5\ \text{cm}^2$
b) $r=2.5,\ h=18$: $2\pi(2.5^2)+2\pi(2.5)(18)=39.27+282.74=322.0\ \text{cm}^2$
c) Convert to cm: $r=2,\ h=1.4$. $2\pi(2^2)+2\pi(2)(1.4)=25.13+17.59=42.7\ \text{cm}^2$
Final answers: a) $\mathbf{477.5\ \text{cm}^2}$ b) $\mathbf{322.0\ \text{cm}^2}$ c) $\mathbf{42.7\ \text{cm}^2}$
❓ EXERCISES
3. The diagram shows a square-based pyramid and a cylinder. Which shape has the greater surface area? Show your working.
👀 Show answer
Pyramid (square base side $6\ \text{cm}$, triangular face perpendicular height $8\ \text{cm}$):
Base area $=6\times 6= \mathbf{36}\ \text{cm}^2$.
Each triangular face $=\tfrac12 \times 6 \times 8=\mathbf{24}\ \text{cm}^2$; there are $4$ faces $\Rightarrow 4\times 24=\mathbf{96}\ \text{cm}^2$.
Total pyramid area $=36+96=\mathbf{132\ \text{cm}^2}$.
Cylinder (radius $r=\mathbf{2}\ \text{cm}$, height $h=\mathbf{8}\ \text{cm}$):
Surface area $=2\pi r^2+2\pi r h=2\pi(2^2)+2\pi(2)(8)=8\pi+32\pi=40\pi \approx \mathbf{125.7\ \text{cm}^2}$.
Comparison: $\mathbf{132\ \text{cm}^2}>\mathbf{125.7\ \text{cm}^2}$, so the pyramid has the greater surface area.
🧠 Think like a Mathematician
4. Work with a partner to answer this question.
The diagram shows a cylinder.
The circular cross-section has radius $r$ cm.
The height of the cylinder is $h$ cm.
a Write a formula, using $r$ and $h$, for the surface area (SA) of the cylinder.
b Show that you can simplify the formula in part a to $SA = 2\pi r(r + h)$.
c When the height is twice the radius, show that the formula for the surface area of the cylinder can be simplified to $SA = 6\pi r^2$.
d Simplify the formula for the surface area of the cylinder when
i$h=3r$ ii$h=4r$ iii$h=5r$
e Can you spot a pattern in your answers to part d? Use this pattern to write the formula for the surface area of the cylinder when $h=19r$.
f Discuss your methods and answers to parts a to e with other pairs of learners in your class.
👀 Show Answer
a) Total surface area of a cylinder: two circles plus curved surface $SA = 2\pi r^2 + 2\pi r h$.
b) Factorising: $SA = 2\pi r(r+h)$.
c) If $h=2r$, then $SA=2\pi r(r+2r)=2\pi r(3r)=6\pi r^2$.
d) Using $SA=2\pi r(r+h)$:
i$h=3r$ ⇒ $SA=2\pi r(r+3r)=8\pi r^2$
ii$h=4r$ ⇒ $SA=2\pi r(r+4r)=10\pi r^2$
iii$h=5r$ ⇒ $SA=2\pi r(r+5r)=12\pi r^2$.
e) Pattern: when $h=kr$, $SA=2\pi r(r+kr)=2\pi(1+k)r^2$. So for $h=19r$, $SA=2\pi(1+19)r^2=40\pi r^2$.
f) Notes for discussion: the surface area grows linearly with $h$ for fixed $r$ (slope $2\pi r$) and quadratically with $r$ when $h$ is a multiple of $r$.
❓ EXERCISES
5. The circular cross-section of a cylinder has radius $3\ \text{cm}$. The height of the cylinder is three times the radius. Work out the surface area of the cylinder correct to three significant figures.
👀 Show answer
Given $r=3\ \text{cm},\ h=3\times 3=9\ \text{cm}$.
$SA = 2\pi(3^2) + 2\pi(3)(9)$
$= 18\pi + 54\pi = 72\pi$
$= 226.19\ \text{cm}^2 \approx \mathbf{226\ \text{cm}^2}$ (3 s.f.).
🧠 Think like a Mathematician
6. Work on your own to answer this question.
The diagram shows a triangular prism.
The triangular cross-section is a right-angled triangle.
a Before you work out the surface area of the prism, answer these questions:
b Compare your answers to part a with a model solution.
c Work out the surface area of the prism.
👀 Show Answer
a i) The missing length is the hypotenuse of the right-angled triangular cross-section.
a ii) Use Pythagoras on the legs $6\ \text{cm}$ and $8\ \text{cm}$: $c=\sqrt{6^2+8^2}=\sqrt{36+64}= \sqrt{100}=10\ \text{cm}$.
c) Surface area = areas of two triangles $+$ areas of three rectangles.
- Each triangle: $\tfrac12 \times 6 \times 8 = 24\ \text{cm}^2$; two triangles $= 48\ \text{cm}^2$.
- Rectangles (length $15\ \text{cm}$ by each triangle side $6,8,10$): $15\times 6=90$, $15\times 8=120$, $15\times 10=150\ \text{cm}^2$.
- Total lateral area $=90+120+150=360\ \text{cm}^2$.
So, total surface area $=48+360=\mathbf{408\ \text{cm}^2}$.
❓ EXERCISES
7. Work out the surface area of these triangular prisms. The triangular cross-section of each prism is a right-angled triangle. Give your answer to part c correct to one decimal place (1 d.p.).
a.
👀 Show answer
Two triangles: $2\times\frac12\cdot 5\cdot 12 = 60\ \text{cm}^2$.
Three rectangles (length $20\ \text{cm}$): $20(5+12+13)=20\cdot 30=600\ \text{cm}^2$.
Total $=60+600=\mathbf{660\ \text{cm}^2}$.
b.
👀 Show answer
Two triangles: $2\times\frac12\cdot 12\cdot 15 = 180\ \text{mm}^2$.
Three rectangles (length $30\ \text{mm}$): $30(12+15+19.235)\approx 1387.1\ \text{mm}^2$.
Total $\approx 180+1387.1=\mathbf{1567.1\ \text{mm}^2}$.
c.
👀 Show answer
Two triangles: $2\times\frac12\cdot 1.1\cdot 1.670 \approx 1.837\ \text{m}^2$.
Three rectangles (length $4.5$): $4.5(1.1+1.670+2)\approx 21.465\ \text{m}^2$.
Total $\approx 1.837+21.465= \mathbf{23.3\ \text{m}^2}$ (to $1$ d.p.).
❓ EXERCISES
8. The diagram shows a cardboard tube. Work out the area of cardboard needed to make the tube. Give your answer correct to three significant figures ($3$ s.f.).
👀 Show answer
Diameter $=6\ \text{cm}\Rightarrow$ circumference $=\pi d=\pi\times 6=18.8496\ldots\ \text{cm}$.
Length $=40\ \text{cm}$.
Area $=18.8496\ldots \times 40=753.98\ldots\ \text{cm}^2 \approx \mathbf{754\ \text{cm}^2}$ ($3$ s.f.).
9. The diagram shows a tin of beans. A label is cut to the exact size of the curved surface of the tin and glued on to the tin. Labels are cut from a rectangular piece of paper that measures $35\ \text{cm}$ by $120\ \text{cm}$. What is the maximum number of labels that can be cut from one piece of paper? Show all your working.
👀 Show answer
Try orientation $10$ along the $35$ side: $\left\lfloor \dfrac{35}{10} \right\rfloor = 3$ rows; along $120$: $\left\lfloor \dfrac{120}{23.5619\ldots} \right\rfloor = 5$ columns $\Rightarrow 3\times 5=15$ labels.
Rotated: width $23.5619\ldots$ along the $35$ side gives only $\left\lfloor \dfrac{35}{23.5619\ldots} \right\rfloor = 1$; along $120$ with height $10$ gives $\left\lfloor \dfrac{120}{10} \right\rfloor = 12$ $\Rightarrow 12$ labels.
Maximum $=\mathbf{15}$ labels.