Revision: geometry in space

Grade 11 geometry in space still appears in the graduation exam. In Part I (multiple choice) there are usually one or two questions on perpendicularity, angles or volume. In Part II or Part III there is usually one question on a distance, a dihedral angle or a volume in a real-life setting. Altogether this topic is usually worth about $0.5$ to $1$ mark.

This revision lesson covers four areas: perpendicularity, angles, distances and volumes. Every example is written exactly like an exam question. Most of them use the same familiar figure: a pyramid with one lateral edge perpendicular to the base.

The pyramid

S.ABCD

with

SA\perp(ABCD)

and a square base with centre

O

Parallel and perpendicular lines and planes

For a pyramid with $SA\perp(ABCD)$ and a square base, the following two "perpendicular pairs" always hold. Remember how to derive them rather than learning the results by heart:

$BC\perp AB,\ BC\perp SA\ \Rightarrow\ BC\perp(SAB);\qquad BD\perp AC,\ BD\perp SA\ \Rightarrow\ BD\perp(SAC).$

Multiple-choice questions often give four relations and ask which is correct. The reliable method is to find two intersecting lines in the plane that the line is perpendicular to.

Example Multiple choice: choose the correct perpendicular relation

The pyramid $S.ABCD$ has a square base $ABCD$ and $SA\perp(ABCD)$ . Which statement is correct?

A. $BD\perp(SAB)$ .

B. $BC\perp(SAB)$ .

C. $AC\perp(SBD)$ .

D. $CD\perp(SBC)$ .

Solution. We check whether each line is perpendicular to two intersecting lines of the plane.

$BC\perp AB\ (\text{square base}),\quad BC\perp SA\ (SA\perp(ABCD))\ \Rightarrow\ BC\perp(SAB).$

Option A is wrong because $BD$ is not perpendicular to $AB$ (the angle between them is $45^\circ$ ). Option C is wrong because $AC$ is not perpendicular to $SO$ . Option D is wrong because $CD\parallel AB$ and $AB$ is not perpendicular to $SB$ (triangle $SAB$ has its right angle at $A$ ), so $CD$ is not perpendicular to $SB$ .

So the answer is B.

Angles in space

To calculate an angle we reduce it to a right-angled triangle and use $\tan$ , $\sin$ or $\cos$ . In the exam, a true/false question on angles often tests both identifying the angle and calculating it.

Example True/false: the angle with a plane and a dihedral angle

The pyramid $S.ABCD$ has a square base with side $a$ and centre $O$ , $SA\perp(ABCD)$ and $SA=a\sqrt2$ .

a) The angle between $SC$ and the plane $(ABCD)$ is $\angle SCA$ .

b) The angle between $SC$ and the plane $(ABCD)$ is $45^\circ$ .

c) The dihedral angle $[S, BD, A]$ is $60^\circ$ .

d) The sine of the angle between $SB$ and the plane $(SAD)$ is $\dfrac{\sqrt3}{3}$ .

Solution. We judge each claim; each one reduces to a right-angled triangle.

a) True. $SA\perp(ABCD)$ , so $AC$ is the projection of $SC$ on the base.

b) True. The diagonal of the square is $AC=a\sqrt2$ , so in triangle $SAC$ with its right angle at $A$ :

$\tan\angle SCA=\frac{SA}{AC}=\frac{a\sqrt2}{a\sqrt2}=1\ \Rightarrow\ \angle SCA=45^\circ.$

c) False. Since $BD\perp(SAC)$ , we have $BD\perp SO$ and $BD\perp AO$ ; the dihedral angle is $\angle SOA$ . With $AO=\dfrac{a\sqrt2}{2}$ :

$\tan\angle SOA=\frac{SA}{AO}=\frac{a\sqrt2}{\frac{a\sqrt2}{2}}=2\ \Rightarrow\ \angle SOA\approx63.4^\circ.$

d) True. $AB\perp AD$ and $AB\perp SA$ , so $AB\perp(SAD)$ ; the projection of $SB$ on $(SAD)$ is $SA$ and the angle is $\angle BSA$ . We have $SB=\sqrt{a^2+2a^2}=a\sqrt3$ :

$\sin\angle BSA=\frac{AB}{SB}=\frac{a}{a\sqrt3}=\frac{\sqrt3}{3}.$

So the answers are a) True, b) True, c) False, d) True.

Distances

With a square or rectangular base, the foot $K$ is $B$ itself, because $AB\perp BC$ . Changing the point lets us turn the distance from an "awkward" point ( $D$ , $C$ , a midpoint...) into the distance from $A$ .

Example Short answer: the distance to a lateral face

The pyramid $S.ABCD$ has a square base with side $3$ , $SA\perp(ABCD)$ and $SA=4$ . Find the distance from $D$ to the plane $(SBC)$ .

Solution. Since $AD\parallel BC$ , we have $AD\parallel(SBC)$ , so $d(D, (SBC))=d(A, (SBC))$ . Draw $AH\perp SB$ ; since $BC\perp(SAB)$ , $BC\perp AH$ , so $AH\perp(SBC)$ :

$\frac{1}{AH^2}=\frac{1}{4^2}+\frac{1}{3^2}=\frac{9+16}{144}=\frac{25}{144}\ \Rightarrow\ AH=\frac{12}{5}=2.4.$

So $d(D, (SBC))=2.4$ . On the answer sheet we write $2.4$ .

Look back. For the same reason, the distance between the skew lines $AD$ and $SB$ is also $2.4$ .

Volumes of pyramids and prisms

In a regular pyramid, the foot of the height is the centre of the base. So the height is usually found with Pythagoras' theorem in the triangle formed by a lateral edge, the height and half a diagonal of the base.

Example Short answer: the volume of a tent

A camping tent has the shape of a regular square pyramid with a square base of side $6$ m, and each lateral edge is $5$ m long. Find the volume of air inside the tent (in m³, to 1 decimal place).

Solution. Let $O$ be the centre of the base; $SO$ is the height. Half the diagonal of the base is $OA=\dfrac{6\sqrt2}{2}=3\sqrt2$ , so:

$SO=\sqrt{SA^2-OA^2}=\sqrt{25-18}=\sqrt7,\qquad V=\frac13\cdot 6^2\cdot\sqrt7=12\sqrt7\approx31.749.$

So the volume of air in the tent is about $31.7$ m³ (write $31.7$ ).

Look back. Using half the side, $3$ , instead of half the diagonal gives a height of $4$ and a volume of $48$ m³, a wrong answer that looks very "nice".

Key terms

English	Tiếng Việt
line perpendicular to a plane	đường thẳng vuông góc với mặt phẳng
perpendicular planes	hai mặt phẳng vuông góc
theorem of three perpendiculars	định lí ba đường vuông góc
orthogonal projection	hình chiếu vuông góc
dihedral angle	góc nhị diện
distance	khoảng cách
skew lines	đường thẳng chéo nhau
regular pyramid	hình chóp đều
right prism	hình lăng trụ đứng
volume	thể tích

Exercises

The exercises go from easy to hard, and each one states its exam format.

Exercise 1 Multiple choice

The right prism $ABC.A'B'C'$ has a base that is a right-angled triangle at $A$ , with $AB=3$ , $AC=4$ and $AA'=5$ . The volume of the prism is

A. $30$ . B. $10$ . C. $60$ . D. $20$ .

What you need and how to start

What you need: $V_{\text{prism}}=B\cdot h$ ; in a right prism the height equals the lateral edge.

Steps:

Find the area of the right-angled triangle $ABC$ .
Multiply by $AA'$ .

Show solution

The base area is $\dfrac12\cdot3\cdot4=6$ , so $V=6\cdot5=30$ . Option B multiplies by an extra $\dfrac13$ ; option C forgets the $\dfrac12$ .

Answer. A.

Exercise 2 Multiple choice

In the tetrahedron $OABC$ , the edges $OA$ , $OB$ , $OC$ are mutually perpendicular, $OA=2$ , $OB=3$ , $OC=4$ . The volume of the tetrahedron is

A. $24$ . B. $12$ . C. $8$ . D. $4$ .

What you need and how to start

What you need: $V=\dfrac16\,OA\cdot OB\cdot OC$ .

Steps:

Take $OAB$ (a right-angled triangle) as the base and $OC$ as the height.
Compute $\dfrac13\cdot S_{OAB}\cdot OC$ .

Show solution

$V=\dfrac13\cdot\dfrac12\cdot2\cdot3\cdot4=4$ . Option A forgets both $\dfrac13$ and $\dfrac12$ ; option B forgets $\dfrac13$ ; option C forgets $\dfrac12$ .

Answer. D.

Exercise 3 Multiple choice

The pyramid $S.ABC$ has an equilateral base with side $a$ , $SA\perp(ABC)$ , and the angle between the plane $(SBC)$ and the base is $60^\circ$ . The volume of the pyramid is

A. $\dfrac{a^3\sqrt3}{24}$ . B. $\dfrac{a^3\sqrt3}{8}$ . C. $\dfrac{3a^3\sqrt3}{8}$ . D. $\dfrac{a^3}{4}$ .

What you need and how to start

What you need: let $M$ be the midpoint of $BC$ ; since $BC\perp(SAM)$ , the angle between $(SBC)$ and the base is $\angle SMA$ .

Steps:

Find $AM=\dfrac{a\sqrt3}{2}$ , then $SA=AM\tan60^\circ$ .
Compute $V=\dfrac13\cdot\dfrac{a^2\sqrt3}{4}\cdot SA$ .

Show solution

$SA=\dfrac{a\sqrt3}{2}\cdot\sqrt3=\dfrac{3a}{2}$ , so $V=\dfrac13\cdot\dfrac{a^2\sqrt3}{4}\cdot\dfrac{3a}{2}=\dfrac{a^3\sqrt3}{8}$ . Option A uses $\tan30^\circ$ by mistake, option C forgets $\dfrac13$ , and option D takes the angle to be $\angle SBA$ .

Answer. B.

Exercise 4 True/false

$ABCD.A'B'C'D'$ is a cube with side $a$ , and $O$ is the centre of the square $ABCD$ .

a) $AC\perp(BDD'B')$ .

b) The angle between the lines $AB'$ and $BC'$ is $60^\circ$ .

c) The distance from $A$ to the plane $(BDD'B')$ is $\dfrac{a\sqrt2}{2}$ .

d) The dihedral angle $[C', BD, C]$ is $45^\circ$ .

What you need and how to start

What you need: the criterion for a line perpendicular to a plane; the angle between two lines equals the angle between two lines parallel to them.

Steps:

$AC\perp BD$ and $AC\perp BB'$ .
Replace $BC'$ by the parallel line $AD'$ and look at triangle $AB'D'$ .
For the dihedral angle, use $C'O\perp BD$ and $CO\perp BD$ .

Show solution

a) True. $AC\perp BD$ (diagonals of a square) and $AC\perp BB'$ , so $AC\perp(BDD'B')$ .

b) True. $BC'\parallel AD'$ ; triangle $AB'D'$ has all three sides equal to $a\sqrt2$ , so it is equilateral and the angle is $60^\circ$ .

c) True. By a), $AO\perp(BDD'B')$ , so the distance is $AO=\dfrac{a\sqrt2}{2}$ .

d) False. The dihedral angle is $\angle C'OC$ with $\tan\angle C'OC=\dfrac{CC'}{CO}=\dfrac{a}{\frac{a\sqrt2}{2}}=\sqrt2$ , so the angle is about $54.7^\circ$ .

Answer. a) True; b) True; c) True; d) False.

Exercise 5 Short answer

$ABCD.A'B'C'D'$ is a cube. Find the angle between the line $A'C$ and the plane $(ABCD)$ (in degrees, to the nearest whole number).

What you need and how to start

What you need: the projection of $A'C$ on $(ABCD)$ is $AC$ .

Steps:

Call the side $a$ and find $AC=a\sqrt2$ .
Compute $\tan\angle A'CA=\dfrac{AA'}{AC}$ .

Show solution

$\tan\angle A'CA=\dfrac{a}{a\sqrt2}=\dfrac{1}{\sqrt2}$ , so $\angle A'CA\approx35.26^\circ$ .

Answer. $35$ .

Exercise 6 Short answer

The pyramid $S.ABC$ has an equilateral base with side $2$ , $SA\perp(ABC)$ and $SA=1$ . Find the distance from $A$ to the plane $(SBC)$ (to 2 decimal places).

What you need and how to start

What you need: with $M$ the midpoint of $BC$ , $BC\perp(SAM)$ ; $\dfrac{1}{AH^2}=\dfrac{1}{SA^2}+\dfrac{1}{AM^2}$ .

Steps:

Find $AM=\sqrt3$ .
Substitute into the formula.

Show solution

$\dfrac{1}{AH^2}=1+\dfrac13=\dfrac43$ , so $AH=\dfrac{\sqrt3}{2}\approx0.866$ .

Answer. $0.87$ .

Exercise 7 Short answer

The cover of a skylight is a regular square pyramid with a square base of side $4$ m and height $3$ m. Find the angle between a sloping face of the cover and the base (in degrees, to the nearest whole number).

What you need and how to start

What you need: with $O$ the centre of the base and $M$ the midpoint of a base edge, the angle between the face and the base is $\angle SMO$ .

Steps:

$OM$ is half the side of the base.
Compute $\tan\angle SMO=\dfrac{SO}{OM}$ .

Show solution

$OM=2$ and $\tan\angle SMO=\dfrac32$ , so $\angle SMO\approx56.31^\circ$ .

Answer. $56$ .

Exercise 8 Short answer

A water trough $2$ m long is a right prism whose cross-section is an isosceles trapezium with a longer base (the top) of $0.6$ m, a shorter base of $0.4$ m and a depth of $0.3$ m. At most how many litres of water can the trough hold?

What you need and how to start

What you need: $V=B\cdot h$ with a trapezium as the base; $1$ m³ $=1000$ litres.

Steps:

Find the area of the trapezium.
Multiply by the length of the trough, then convert to litres.

Show solution

$B=\dfrac{(0.6+0.4)\cdot0.3}{2}=0.15$ m², so $V=0.15\cdot2=0.3$ m³ $=300$ litres.

Answer. $300$ .

Exercise 9 Written answer

The pyramid $S.ABCD$ has a square base with side $a$ and centre $O$ , $SA\perp(ABCD)$ and $SA=a$ . Find the distance between the skew lines $SC$ and $BD$ .

What you need and how to start

What you need: $BD\perp(SAC)$ ; if $OH\perp SC$ in the plane $(SAC)$ , then $OH$ is the common perpendicular.

Steps:

Show that $BD\perp OH$ .
Compute $OH=OC\cdot\sin\angle SCA$ .

Show solution

Since $BD\perp(SAC)$ and $OH\subset(SAC)$ , $OH\perp BD$ ; also $OH\perp SC$ , so $d(SC, BD)=OH$ . We have $OC=\dfrac{a\sqrt2}{2}$ , $SC=\sqrt{a^2+2a^2}=a\sqrt3$ and $\sin\angle SCA=\dfrac{SA}{SC}=\dfrac{1}{\sqrt3}$ :

$OH=\frac{a\sqrt2}{2}\cdot\frac{1}{\sqrt3}=\frac{a\sqrt6}{6}.$

Answer. $d(SC, BD)=\dfrac{a\sqrt6}{6}$ .

Exercise 10 Short answer (challenge)

The prism $ABC.A'B'C'$ has volume $30$ . Let $M$ be the midpoint of the edge $CC'$ . The pyramid $M.ABC$ is cut away from the prism. Find the volume of what remains.

What you need and how to start

What you need: the pyramid and the prism share the base; the height of the pyramid is half the height of the prism.

Steps:

Write $V_{M.ABC}=\dfrac13\cdot B\cdot\dfrac h2$ in terms of $V=B\cdot h$ .
Subtract it from $30$ .

Show solution

$d(M, (ABC))=\dfrac12 d(C', (ABC))=\dfrac h2$ , so $V_{M.ABC}=\dfrac13\cdot B\cdot\dfrac h2=\dfrac{V}{6}=5$ . What remains has volume $30-5=25$ .

Answer. $25$ .

Parallel and perpendicular lines and planes

Angles in space

Distances

Volumes of pyramids and prisms

Key terms

Exercises

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer

a) $SO\perp(ABCD)$ .	True	False
b) $SO=\sqrt2$ .	True	False
c) The angle between $SA$ and the plane $(ABCD)$ is $60^\circ$ .	True	False
d) The volume of the pyramid is $\dfrac{4\sqrt2}{3}$ .	True	False