3D geometry: distance, solids, angles with lines and planes – IB Maths exam questions with mark scheme

1StandardCalculator6 marks

The cuboid $ABCDEFGH$ has a horizontal rectangular base $ABCD$ with $AB=8$ cm and $BC=6$ cm. The height $CG=5$ cm.

(a) Find the length of $AC$. [2]

(b) Find the length of $AG$. [2]

Pythagoras in the base: $AC^2=8^2+6^2$M1
$AC=10$ cmA1
Pythagoras in triangle $ACG$: $AG^2=10^2+5^2$M1
$AG=\sqrt{125}=11.2$ cm (accept $5\sqrt5$)A1
Angle is $\angle GAC$: $\tan\theta=\dfrac{5}{10}$ (or $\sin\theta=\dfrac{5}{\sqrt{125}}$)M1
$\theta=26.6^\circ$A1
Answer: (a) $10$ cm (b) $11.2$ cm (c) $26.6^\circ$
Examiner tip: Name the right-angled triangle first: the angle between a line and a plane uses the line's projection on the plane ($AC$), not the edge $AB$.

2HardCalculator7 marks

$VABCD$ is a right pyramid. The base $ABCD$ is a horizontal square of side $10$ cm. The apex $V$ is vertically above the centre $M$ of the base, and $VM=12$ cm.

(a) Find the length of $AM$. [2]

(b) Find the angle between the edge $VA$ and the base. [2]

(d) Find the volume of the pyramid. [1]

$AM=\tfrac12\sqrt{10^2+10^2}$M1
$AM=5\sqrt2=7.07$ cmA1
$\tan\theta=\dfrac{12}{5\sqrt2}$M1
$59.5^\circ$A1
Uses midpoint $N$ of $AB$, $MN=5$: $\tan\phi=\dfrac{12}{5}$M1
$67.4^\circ$A1
$V=\tfrac13\times10^2\times12=400$ cm$^3$A1
Answer: (a) $7.07$ cm (b) $59.5^\circ$ (c) $67.4^\circ$ (d) $400$ cm$^3$
Examiner tip: The angle between a face and the base is measured along a line at right angles to the shared edge, so use the midpoint of $AB$, not the corner $A$.

3CoreNon-calculator7 marks

The points $A$ and $B$ have coordinates $A(2,\,-1,\,5)$ and $B(-4,\,3,\,1)$.

(a) Find the coordinates of the midpoint $M$ of $[AB]$. [2]

(b) Find the exact distance $AB$, giving your answer in the form $a\sqrt{b}$ where $a,b\in\mathbb{Z}^+$. [2]

Attempt to average each pair of coordinates, e.g. $\left(\dfrac{2+(-4)}{2},\dfrac{-1+3}{2},\dfrac{5+1}{2}\right)$ (at least two correct components)M1
$M(-1,\,1,\,3)$ (accept as a column vector)A1
Correct substitution into the 3D distance formula: $\sqrt{(-6)^2+4^2+(-4)^2}$ (ignore sign of differences)M1
$AB=\sqrt{68}=2\sqrt{17}$ (A0 for $\sqrt{68}$ left unsimplified)A1
Sets up distance from their $M$ equal to $3$: $\sqrt{(1-(-1))^2+(p-1)^2+(3-3)^2}=3$ (or squared form)M1
$(p-1)^2=5$ (or equivalent quadratic $p^2-2p-4=0$)A1
$p=1\pm\sqrt5$ (both values required)A1
Answer: (a) $M(-1,1,3)$ (b) $2\sqrt{17}$ (c) $p=1+\sqrt5$ or $p=1-\sqrt5$
Examiner tip: In (c) the square root gives two values: candidates who write only $p=1+\sqrt5$ lose the final A1.

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4CoreCalculator6 marks

A solid right circular cone has base radius $5$ cm and slant height $13$ cm.

(a) Find the vertical height of the cone. [2]

(b) Find the volume of the cone. [2]

Uses Pythagoras with the slant height as hypotenuse: $h^2+5^2=13^2$M1
$h=12$ cmA1
Substitutes their $h$ into $V=\tfrac13\pi r^2h$: $\tfrac13\pi(5^2)(12)$M1
$V=100\pi=314$ cm$^3$ (accept $314.159\ldots$)A1
Adds curved surface area $\pi rl$ to the base area $\pi r^2$: $\pi(5)(13)+\pi(5^2)$ (M0 if the vertical height is used in $\pi rl$)M1
$90\pi=283$ cm$^2$ (accept $282.7$)A1
Answer: (a) $12$ cm (b) $314$ cm$^3$ (c) $283$ cm$^2$
Examiner tip: The curved surface formula $\pi r l$ uses the slant height, while the volume uses the vertical height: mixing them up is the commonest loss here.

5StandardCalculator8 marks

A grain silo is modelled as a closed vertical cylinder of radius $3.2$ m and height $9$ m, with a hemispherical roof of the same radius fixed to the top, as shown in the diagram. The silo stands on flat ground.

(a) Find the volume of the silo. [3]

The outside of the silo, excluding the base which rests on the ground, is to be painted.

(b) Find the area to be painted. [3]

One litre of paint covers $7.5$ m$^2$. Paint is sold only in $5$-litre tins.

Volume of cylinder $\pi(3.2)^2(9)$ $(=289.53\ldots)$M1
Volume of hemisphere $\tfrac23\pi(3.2)^3$ $(=68.63\ldots)$ (M0 for the full sphere $\tfrac43\pi r^3$)M1
$V=358$ m$^3$ (accept $358.2$)A1
Curved surface of cylinder $2\pi(3.2)(9)$ $(=180.96\ldots)$M1
Curved surface of hemisphere $2\pi(3.2)^2$ $(=64.34\ldots)$; M0 if a base circle $\pi r^2$ is also addedM1
Area $=245$ m$^2$ (accept $245.3$)A1
Divides their area by $7.5$ and then by $5$: $245.3\div7.5=32.7$ litres, $32.7\div5=6.54$M1
$7$ tins (must round up; A0 for $6$ or $6.54$)A1
Answer: (a) $358$ m$^3$ (b) $245$ m$^2$ (c) $7$ tins
Examiner tip: A hemisphere has curved surface area $2\pi r^2$ (half of $4\pi r^2$), and the circle where it joins the cylinder is not on the outside, so it must not be added.

6StandardCalculator9 marks

A solid metal sphere of radius $6$ cm is melted down and all of the metal is recast into solid right circular cones, each of base radius $2$ cm and height $4.5$ cm.

(a) Show that exactly $48$ cones are made. [3]

(b) Find the slant height of one cone. [2]

(c) Find the ratio of the total curved surface area of the $48$ cones to the surface area of the original sphere, giving your answer in the form $k:1$. [3]

(d) With reference to your answer to (c), comment on why manufacturers might prefer the metal to be cast as many small objects when it needs to cool quickly. [1]

Substitutes into both volume formulas: $\tfrac43\pi(6^3)$ and $\tfrac13\pi(2^2)(4.5)$M1
Volume of sphere $=288\pi$ and volume of one cone $=6\pi$ (or $904.8\ldots$ and $18.85\ldots$)A1
Divides: $\dfrac{288\pi}{6\pi}=48$ (AG; must see both volumes, exact or to at least 3 s.f., and the division)A1
$l=\sqrt{2^2+4.5^2}$M1
$l=4.92$ cm (accept $\sqrt{24.25}$, $\tfrac{\sqrt{97}}{2}$)A1
Total curved area $48\times\pi(2)(4.92\ldots)$ $(=1485\ldots)$M1
Sphere surface area $4\pi(6^2)=144\pi$ $(=452.4\ldots)$ and forms the ratioM1
$3.28:1$ (accept $3.28:1$ from $l=4.92$)A1
The cones have over three times the surface area of the sphere (for the same volume), so heat is lost much faster; accept any comment linking larger surface area to faster coolingR1
Answer: (a) $288\pi\div6\pi=48$ (b) $4.92$ cm (c) $3.28:1$ (d) much larger surface area for same volume so faster cooling
Examiner tip: In (c) use the unrounded slant height: rounding $l$ to $4.9$ early gives $3.27$, which is outside the accuracy allowed.

7StandardCalculator9 marks

The diagram shows a cuboid $ABCDEFGH$ with $AB=9$ cm, $BC=4$ cm and $CG=6$ cm. The space diagonals $[AG]$ and $[BH]$ intersect at $O$.

(a) Find the length of $AG$. [2]

(b) Find the acute angle between the lines $(AG)$ and $(BH)$. [4]

$AG=\sqrt{9^2+4^2+6^2}$ (or via $AC=\sqrt{97}$ then $\sqrt{97+36}$)M1
$AG=\sqrt{133}=11.5$ cmA1
Recognises $AO=BO=\tfrac12AG$ $(=5.766\ldots)$ and uses triangle $AOB$ with $AB=9$M1
Correct substitution into the cosine rule: $\cos A\hat{O}B=\dfrac{5.766^2+5.766^2-9^2}{2(5.766)(5.766)}$ $(=-0.218\ldots)$M1
$A\hat{O}B=102.6^\circ$A1
Acute angle between the lines $=180^\circ-102.6^\circ=77.4^\circ$ (A0 if $102.6^\circ$ given as final answer)A1
Identifies $B$ as the foot of the perpendicular from $A$ to face $BCGF$, so the required angle is $A\hat{G}B$ (may be implied by working in triangle $ABG$, right-angled at $B$)M1
$BG=\sqrt{4^2+6^2}=\sqrt{52}$ and $\tan A\hat{G}B=\dfrac{9}{\sqrt{52}}$ (or $\sin A\hat{G}B=\dfrac{9}{\sqrt{133}}$)M1
$51.3^\circ$A1
Answer: (a) $11.5$ cm (b) $77.4^\circ$ (c) $51.3^\circ$
Examiner tip: The angle between two lines is always quoted as the acute one, and for the angle with a vertical face you must first find where the perpendicular from $A$ meets that face (here at $B$).

8HardCalculator10 marks

The diagram shows a right pyramid $VABCD$ with a horizontal rectangular base $ABCD$, where $AB=8$ cm and $BC=6$ cm. The apex $V$ is vertically above $M$, the point where the diagonals of the base meet, and $VM=10$ cm.

(a) Find the angle between the edge $[VA]$ and the base. [3]

Let $N$ be the midpoint of $[AB]$.

(b) Find the angle between the line $(VN)$ and the base. [2]

(c) Let $K$ be the midpoint of $[BC]$. The angle between the line $(VK)$ and the base is $68.2^\circ$. Explain, without further calculation, why this is smaller than your answer to (b). [1]

(d) Find the total surface area of the pyramid. [4]

$AC=\sqrt{8^2+6^2}=10$ so $AM=5$M1
$\tan V\hat{A}M=\dfrac{10}{5}$ (or equivalent using $VA=\sqrt{125}$)M1
$63.4^\circ$A1
Uses $MN=\tfrac12BC=3$ in right-angled triangle $VMN$: $\tan\theta=\dfrac{10}{3}$ (M0 if $MN=4$ used)M1
$73.3^\circ$A1
The distance $MK=4$ cm is greater than $MN=3$ cm, while the height $VM$ is the same, so $(VK)$ is less steep than $(VN)$ (accept: $\tan$ of the angle is $\tfrac{10}{4}<\tfrac{10}{3}$)R1
Slant height of a face on $[AB]$: $\sqrt{10^2+3^2}=\sqrt{109}$ or slant height of a face on $[BC]$: $\sqrt{10^2+4^2}=\sqrt{116}$M1
Both slant heights correct (may be seen as $10.4$ and $10.8$)A1
Total $=8\times6+2\times\tfrac12(8)(\sqrt{109})+2\times\tfrac12(6)(\sqrt{116})$ (their slant heights; must include base and four triangles)M1
$196$ cm$^2$ (accept $196.1$)A1
Answer: (a) $63.4^\circ$ (b) $73.3^\circ$ (c) $MK=4$ cm $>MN=3$ cm with the same height, so $(VK)$ is less steep (d) $196$ cm$^2$
Examiner tip: For a rectangular base the four triangular faces are not congruent: the face on the longer edge has the shorter perpendicular distance to $M$, and each pair needs its own slant height.

9StretchCalculator10 marks

A solid tetrahedron $OABC$ has vertices $O(0,0,0)$, $A(4,0,0)$, $B(0,6,0)$ and $C(0,0,3)$, as shown in the diagram. Lengths are in centimetres.

(a) Find the lengths $AB$, $BC$ and $CA$, giving exact answers. [3]

(b) Find $A\hat{C}B$. [2]

(d) Find the volume of the tetrahedron. [1]

(e) Hence find the shortest distance from $O$ to the face $ABC$. [2]

Uses the 3D distance formula for at least one length, e.g. $AB=\sqrt{4^2+6^2+0^2}$M1
Two of $AB=\sqrt{52}=2\sqrt{13}$, $BC=\sqrt{45}=3\sqrt5$, $CA=5$ correctA1
All three correct (exact forms; accept $\sqrt{52}$, $\sqrt{45}$)A1
Cosine rule with $AB$ opposite: $\cos A\hat{C}B=\dfrac{45+25-52}{2(\sqrt{45})(5)}$ $\left(=\dfrac{3\sqrt5}{25}=0.268\ldots\right)$M1
$A\hat{C}B=74.4^\circ$ (accept $1.30$ radians)A1
$\text{Area}=\tfrac12(\sqrt{45})(5)\sin(74.4^\circ)$ (their angle)M1
$16.2$ cm$^2$ (accept $3\sqrt{29}$, $16.155\ldots$)A1
$V=\tfrac13\times\left(\tfrac12\times4\times6\right)\times3=12$ cm$^3$ (using face $OAB$ as base and $OC$ as height)A1
Uses the same volume with base $ABC$: $\tfrac13(16.155)d=12$M1
$d=2.23$ cm (accept $\tfrac{12}{\sqrt{29}}$, $2.228\ldots$)A1
Answer: (a) $AB=2\sqrt{13}$, $BC=3\sqrt5$, $CA=5$ (b) $74.4^\circ$ (c) $16.2$ cm$^2$ (d) $12$ cm$^3$ (e) $2.23$ cm
Examiner tip: Part (e) is the key idea: the volume is the same whichever face you call the base, so $d=3V\div\text{area of }ABC$. Keep the unrounded area or the answer drifts to $2.22$.

10HardCalculator9 marks

A solid model of an ice-cream is made from a right circular cone of height $10$ cm and base radius $r$ cm, joined to a hemisphere of radius $r$ cm along their common circular face, as shown in the diagram. The vertex of the cone is $P$.

The total volume of the model is $200$ cm$^3$.

(a) Write down an equation in $r$ for the total volume. [2]

(b) Hence find the value of $r$. [1]

(d) Find the total surface area of the model. [3]

Adds a cone volume and a hemisphere volume: $\tfrac13\pi r^2(10)+\tfrac23\pi r^3$M1
$\tfrac{10}{3}\pi r^2+\tfrac23\pi r^3=200$ (or equivalent)A1
$r=3.38$ (accept $3.376$; from GDC)A1
Right-angled triangle with the axis: $\tan\alpha=\dfrac{3.376}{10}$ (or $\sin$/$\cos$ with slant height $10.55$)M1
Half-angle $\alpha=18.7^\circ$A1
Required angle $=2\alpha=37.3^\circ$ (FT from their $r$)A1
Slant height $l=\sqrt{10^2+3.376^2}=10.55\ldots$ and curved surface of cone $\pi r l$ $(=111.95\ldots)$M1
Adds curved surface of hemisphere $2\pi r^2$ $(=71.63\ldots)$; M0 if any flat circular face is includedM1
$184$ cm$^2$ (accept $183.6$)A1
Answer: (a) $\tfrac{10}{3}\pi r^2+\tfrac23\pi r^3=200$ (b) $r=3.38$ (c) $37.3^\circ$ (d) $184$ cm$^2$
Examiner tip: The joined circular faces are inside the solid, so the outside surface is only the two curved surfaces; adding $\pi r^2$ is a common way to lose the final mark.

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