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LemonmathSelf-study sheet
Grade 12 · Lesson 7.5

Revision: geometry in space

Name: Date: Time spent: min

Goals

Exercises

Exercise 1 Multiple choice

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

A. 3030. B. 1010. C. 6060. D. 2020.

Exercise 2 Multiple choice

In the tetrahedron OABCOABC, the edges OAOA, OBOB, OCOC are mutually perpendicular, OA=2OA=2, OB=3OB=3, OC=4OC=4. The volume of the tetrahedron is

A. 2424. B. 1212. C. 88. D. 44.

Exercise 3 Multiple choice

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

A. a3324\dfrac{a^3\sqrt3}{24}. B. a338\dfrac{a^3\sqrt3}{8}. C. 3a338\dfrac{3a^3\sqrt3}{8}. D. a34\dfrac{a^3}{4}.

Exercise 4 True/false

ABCD.A′B′C′D′ABCD.A'B'C'D' is a cube with side aa, and OO is the centre of the square ABCDABCD.

a) AC⊥(BDD′B′)AC\perp(BDD'B').

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

c) The distance from AA to the plane (BDD′B′)(BDD'B') is a22\dfrac{a\sqrt2}{2}.

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

Exercise 5 Short answer

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

Exercise 6 Short answer

The pyramid S.ABCS.ABC has an equilateral base with side 22, SA⊥(ABC)SA\perp(ABC) and SA=1SA=1. Find the distance from AA to the plane (SBC)(SBC) (to 2 decimal places).

Exercise 7 Short answer

The cover of a skylight is a regular square pyramid with a square base of side 44 m and height 33 m. Find the angle between a sloping face of the cover and the base (in degrees, to the nearest whole number).

Exercise 8 Short answer

A water trough 22 m long is a right prism whose cross-section is an isosceles trapezium with a longer base (the top) of 0.60.6 m, a shorter base of 0.40.4 m and a depth of 0.30.3 m. At most how many litres of water can the trough hold?

Exercise 9 Written answer

The pyramid S.ABCDS.ABCD has a square base with side aa and centre OO, SA⊥(ABCD)SA\perp(ABCD) and SA=aSA=a. Find the distance between the skew lines SCSC and BDBD.

Exercise 10 Short answer (challenge)

The prism ABC.A′B′C′ABC.A'B'C' has volume 3030. Let MM be the midpoint of the edge CC′CC'. The pyramid M.ABCM.ABC is cut away from the prism. Find the volume of what remains.

Mini test

15 minutes, without looking back at the lesson. Part I: circle one letter; Part II: write T or F for each statement; Part III: write the answer in the box.

Part I. Multiple choice

Question 1. A pyramid has base area 1212 and height 55. Its volume is
A. 6060.B. 2020.C. 3030.D. 1010.
Question 2. The pyramid S.ABCDS.ABCD has SA⊥(ABCD)SA\perp(ABCD). The angle between the line SBSB and the plane (ABCD)(ABCD) is
A. ∠SBA\angle SBA.B. ∠BSA\angle BSA.C. ∠SBC\angle SBC.D. ∠SAB\angle SAB.
Question 3. ABCD.A′B′C′D′ABCD.A'B'C'D' is a cube with side 22. The distance between the lines AA′AA' and BDBD is
A. 22.B. 222\sqrt2.C. 2\sqrt2.D. 3\sqrt3.
Question 4. The pyramid S.ABCDS.ABCD has a rectangular base and SA⊥(ABCD)SA\perp(ABCD). Which plane is perpendicular to the plane (SAB)(SAB)?
A. (SCD)(SCD).B. (SBD)(SBD).C. (SAC)(SAC).D. (SBC)(SBC).

Part II. True or false

Question 1. S.ABCDS.ABCD is a regular square pyramid with all edges equal to 22, and OO is the centre of the base.
a) SO⊥(ABCD)SO\perp(ABCD).T / F
b) SO=2SO=\sqrt2.T / F
c) The angle between SASA and the plane (ABCD)(ABCD) is 60∘60^\circ.T / F
d) The volume of the pyramid is 423\dfrac{4\sqrt2}{3}. T / F

Part III. Short answer

Question 1. A right prism has an equilateral base with side 22 and lateral edges of length 33. Find its volume (to 1 decimal place).
Question 2. The pyramid S.ABCS.ABC has SA⊥(ABC)SA\perp(ABC), triangle ABCABC has its right angle at BB, and SA=AB=2SA=AB=2. Find the distance from AA to the plane (SBC)(SBC) (to 2 decimal places).

Score: ........ / 10

Hints (only when stuck)

Exercise 1.

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

Steps:

  1. Find the area of the right-angled triangle ABCABC.
  2. Multiply by AA′AA'.
Exercise 2.

What you need: V=16 OA⋅OB⋅OCV=\dfrac16\,OA\cdot OB\cdot OC.

Steps:

  1. Take OABOAB (a right-angled triangle) as the base and OCOC as the height.
  2. Compute 13⋅SOAB⋅OC\dfrac13\cdot S_{OAB}\cdot OC.
Exercise 3.

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

Steps:

  1. Find AM=a32AM=\dfrac{a\sqrt3}{2}, then SA=AMtan⁡60∘SA=AM\tan60^\circ.
  2. Compute V=13⋅a234⋅SAV=\dfrac13\cdot\dfrac{a^2\sqrt3}{4}\cdot SA.
Exercise 4.

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:

  1. AC⊥BDAC\perp BD and AC⊥BB′AC\perp BB'.
  2. Replace BC′BC' by the parallel line AD′AD' and look at triangle AB′D′AB'D'.
  3. For the dihedral angle, use C′O⊥BDC'O\perp BD and CO⊥BDCO\perp BD.
Exercise 5.

What you need: the projection of A′CA'C on (ABCD)(ABCD) is ACAC.

Steps:

  1. Call the side aa and find AC=a2AC=a\sqrt2.
  2. Compute tan⁡∠A′CA=AA′AC\tan\angle A'CA=\dfrac{AA'}{AC}.
Exercise 6.

What you need: with MM the midpoint of BCBC, BC⊥(SAM)BC\perp(SAM); 1AH2=1SA2+1AM2\dfrac{1}{AH^2}=\dfrac{1}{SA^2}+\dfrac{1}{AM^2}.

Steps:

  1. Find AM=3AM=\sqrt3.
  2. Substitute into the formula.
Exercise 7.

What you need: with OO the centre of the base and MM the midpoint of a base edge, the angle between the face and the base is ∠SMO\angle SMO.

Steps:

  1. OMOM is half the side of the base.
  2. Compute tan⁡∠SMO=SOOM\tan\angle SMO=\dfrac{SO}{OM}.
Exercise 8.

What you need: V=B⋅hV=B\cdot h with a trapezium as the base; 11 m³ =1000=1000 litres.

Steps:

  1. Find the area of the trapezium.
  2. Multiply by the length of the trough, then convert to litres.
Exercise 9.

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

Steps:

  1. Show that BD⊥OHBD\perp OH.
  2. Compute OH=OC⋅sin⁡∠SCAOH=OC\cdot\sin\angle SCA.
Exercise 10.

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

Steps:

  1. Write VM.ABC=13⋅B⋅h2V_{M.ABC}=\dfrac13\cdot B\cdot\dfrac h2 in terms of V=B⋅hV=B\cdot h.
  2. Subtract it from 3030.

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