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Geometry of 3D Shapes

Master IB Math AI Geometry of 3D Shapes with notes created by examiners and strictly aligned with the syllabus.

IB Syllabus Requirements for Geometry of 3D Shapes

3.1

Geometry and measurement in three dimensions

3.1

GEOMETRY AND MEASUREMENT IN THREE DIMENSIONS

Working in three dimensions

A three-dimensional solid is a geometric object that has length, width and height. Dimension makes a difference. A two-dimensional figure has area but no volume; a three-dimensional solid has both surface area and volume. Since a sketch on paper is still two-dimensional, don’t assume hidden edges, right angles or perpendicular heights from its appearance. Label them.

Keep to one length unit throughout a calculation. When measurements use different units, convert them before substituting. Surface area uses squared units, while volume uses cubed units.

Distance and midpoint in three-dimensional space

An ordered triple locates a point in three-dimensional Cartesian space. Let P(x1,y1,z1)P(x_1,y_1,z_1) and Q(x2,y2,z2)Q(x_2,y_2,z_2) be two points, where x1x_1, y1y_1 and z1z_1 are the coordinates of PP (m), and x2x_2, y2y_2 and z2z_2 are the coordinates of QQ (m).

The differences between corresponding coordinates give three mutually perpendicular displacements. Apply Pythagoras' theorem across the three directions:

d=(x2x1)2+(y2y1)2+(z2z1)2d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}

Because each difference is squared, changing the order of subtraction doesn’t change the answer.

You can picture three-dimensional distance as Pythagoras in two stages. First calculate a diagonal in a horizontal coordinate plane, then combine it with the vertical displacement.

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A midpoint is a point that divides a line segment into two equal lengths. Find it by averaging each pair of corresponding coordinates:

M=(x1+x22,y1+y22,z1+z22)M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right)

Match coordinate with coordinate; don’t average all six numbers together.

The right-angled triangle hidden inside a solid

Most three-dimensional problems involving lengths or angles become easier once you isolate the correct cross-section. Sketch the solid and mark the required points. Then redraw just the plane containing the relevant triangle. Check from the geometry that the triangle really is right-angled—a perspective drawing doesn’t prove it.

For a right-angled triangle,

a2+b2=c2a^2+b^2=c^2

and, relative to an acute angle θ\theta,

sinθ=ac,cosθ=bc,tanθ=ab\sin\theta=\frac{a}{c},\qquad \cos\theta=\frac{b}{c},\qquad \tan\theta=\frac{a}{b}

When finding an angle, choose the appropriate inverse trigonometric function and make sure the calculator is in degree mode.

Useful triangles often involve a face diagonal and two edges, or a space diagonal together with a face diagonal and another edge. Another common case is a vertical cross-section through the centre or apex of a solid. Questions about three-dimensional shapes in this part of the syllabus use right-angled trigonometry.

Volume and surface area

Volume is a measure of the three-dimensional space enclosed by a solid. Surface area is the total area of the exposed faces or curved surfaces bounding a solid. Volume formulas give cubic units because they combine dimensions. Surface-area formulas give square units.

For a right prism or cylinder, use the general volume relationship

V=BhV=Bh

Perpendicular is the key word here: a sloping edge isn’t automatically the height.

For a right pyramid,

V=13BhV=\frac{1}{3}Bh

A right pyramid is a pyramid whose apex lies directly above the centre of its base. To find its surface area, add the base area to the areas of all the triangular faces. For a regular right pyramid, this becomes

A=B+12ppA=B+\frac{1}{2}p\ell_p

Don’t mix up the slant height with the vertical height in the volume formula.

A right cone is a cone whose apex lies directly above the centre of its circular base. Its volume and total surface area are

V=13πr2hV=\frac{1}{3}\pi r^2h

and

A=πr2+πrA=\pi r^2+\pi r\ell

where rr is the radius of the circular base (m), \ell is the cone's slant height (m), and π\pi is the dimensionless circle constant. Here, πr2\pi r^2 gives the base and πr\pi r\ell gives the curved surface. In a central cross-section, rr, hh and \ell form a right triangle.

A sphere is a solid whose surface points are all the same distance from its centre. Its volume and surface area are

V=43πr3V=\frac{4}{3}\pi r^3

and

A=4πr2A=4\pi r^2

A hemisphere is one of the two congruent solids formed when a sphere is cut through its centre. Its volume is

V=23πr3V=\frac{2}{3}\pi r^3

The curved surface area is 2πr22\pi r^2. When the circular base is included, the total surface area is 3πr23\pi r^2. Read the context carefully. An open hemispherical bowl has no circular base in its exposed area, whereas a closed hemispherical end does.

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Break a composite solid into familiar solids. Add the volumes of joined pieces; subtract the volumes of holes or removed sections. Surface area needs different care: count only the surfaces exposed on the outside. For example, the circular face where a cone joins a hemisphere is internal, so it must not be counted. A right-angled triangle in a central cross-section often supplies unknown radii, heights or slant heights.

Angles between lines and planes

Two intersecting lines form supplementary pairs of angles. The angle between two intersecting lines is the smaller angle through which one line can be rotated to match the direction of the other, unless a larger angle is explicitly requested. For a solid, mark points on both lines and find a triangle in the plane determined by them. If it is right-angled, use the trigonometric ratios above. Subtract the acute result from 180180^\circ to get the supplementary angle.

The orthogonal projection of a line onto a plane is the line in that plane obtained by dropping perpendiculars from the original line to the plane. The angle between a line and a plane is the acute angle between the line and its orthogonal projection onto that plane. In general, it isn’t the angle between the line and any arbitrary edge drawn in the plane.

Image

To build the required triangle, select a point on the line and drop a perpendicular from it to the plane. Join the foot of that perpendicular to the point where the line meets the plane. The original line forms the hypotenuse, its projection lies in the plane, and the perpendicular is the height above the plane. If an answer gives the angle between the line and the plane's normal instead, that angle is complementary to the required line-plane angle.

Applications, technology and mathematical foundations

These methods appear in architecture and design. Coordinates locate structural points, while right triangles give inaccessible lengths. Surface area and volume can then estimate material requirements and capacity. Design technology relies on the same distinction between external surfaces, internal joins and enclosed volume. There is also a link to physics through spherical models: the sphere formula gives a star's approximate volume, while radiation spreading across spherical surfaces helps explain why intensity follows an inverse-square relationship.

Dynamic geometry software or a graphing package can rotate a model, making the relevant cross-section easier to see. Technology is particularly helpful when checking a space diagonal or line-plane angle. Even so, the calculation still requires a labelled triangle showing why the chosen lengths and angle belong together.

An axiomatic system is a deductive mathematical framework that begins with accepted statements and derives further results using stated rules of logic. An axiom is a starting statement accepted without proof within such a framework. These foundations support results involving perpendicular lines, Euclidean distance and angle measurement. Axioms aren’t necessarily self-evident to everyone; they make the model’s assumptions explicit. Different choices of axioms can lead to different geometries, so a convincing diagram can’t replace a valid deduction.

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