IEB practice · Paper II

IEB Euclidean geometry practice

12 newly generated questions across all 14 euclidean geometry topics the IEB examines in Paper II, 137 marks in all. Work each one, then open its answer underneath to check yourself. Nothing here is taken from a past paper.

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The angle at the centre

2 questions · 26 marks

115 marks

In the diagram, O is the centre of the circle. A $\hat{P}$ B = 46°.

(a)
Determine, with reasons, the size of A $\hat{O}$ B.
(2)
(b)
Determine, with reasons, the size of P $\hat{B}$ O.
(2)
(c)
Determine, with reasons, the size of P $\hat{O}$ B.
(2)
In the diagram below, PA and PB are tangents to the circle from P. A $\hat{P}$ B = 66°.
(d)
Determine, with reasons, the size of P $\hat{A}$ B.
(3)
(e)
Determine, with reasons, the size of P $\hat{B}$ A.
(2)
(f)
Determine, with reasons, the size of B $\hat{A}$ U.
(2)
(g)
Determine, with reasons, the size of V $\hat{B}$ A.
(2)

Show the answers

(a) Answer: A $\hat{O}$ B = 92°

A $\hat{O}$ B = 92° (∠ at centre = 2 × ∠ at circumference)

(b) Answer: P $\hat{B}$ O = 46°

OB = OP (radii)
P $\hat{B}$ O = 46° (∠s opp equal sides)

(c) Answer: P $\hat{O}$ B = 88°

P $\hat{O}$ B = 88° (sum ∠s of Δ)

(d) Answer: P $\hat{A}$ B = 57°

PA = PB (Tans from common pt)
P $\hat{A}$ B = P $\hat{B}$ A (∠s opp equal sides)
P $\hat{A}$ B = 57° (sum ∠s of Δ)

(e) Answer: P $\hat{B}$ A = 57°

P $\hat{B}$ A = 57° (∠s opp equal sides)

(f) Answer: B $\hat{A}$ U = 123°

B $\hat{A}$ U = 123° (∠s on a str line)

(g) Answer: V $\hat{B}$ A = 123°

V $\hat{B}$ A = 123° (∠s on a str line)

211 marks

In the diagram, O is the centre of the circle and AB is joined. A $\hat{P}$ B = 31°.

(a)
Determine, with reasons, the size of A $\hat{O}$ B.
(2)
(b)
Determine, with reasons, the size of O $\hat{A}$ B.
(3)
(c)
Determine, with reasons, the size of O $\hat{B}$ A.
(2)
In the diagram below, ABCD is a cyclic quadrilateral and BC is produced to E. B $\hat{A}$ D = 73°.
(d)
Determine, with reasons, the size of D $\hat{C}$ B.
(2)
(e)
Determine, with reasons, the size of E $\hat{C}$ D.
(2)

Show the answers

(a) Answer: A $\hat{O}$ B = 62°

A $\hat{O}$ B = 62° (∠ at centre = 2 × ∠ at circumference)

(b) Answer: O $\hat{A}$ B = 59°

OA = OB (radii)
O $\hat{A}$ B = O $\hat{B}$ A (∠s opp equal sides)
O $\hat{A}$ B = 59° (sum ∠s of Δ)

(c) Answer: O $\hat{B}$ A = 59°

O $\hat{B}$ A = 59° (∠s opp equal sides)

(d) Answer: D $\hat{C}$ B = 107°

D $\hat{C}$ B = 107° (opp ∠s of cyclic quad)

(e) Answer: E $\hat{C}$ D = 73°

E $\hat{C}$ D = 73° (ext ∠ of cyclic quad)

Proving the theorem itself, not using it

3 questions · 33 marks

38 marks

In the diagram, ABCD is a cyclic quadrilateral in a circle with centre O.

(a)
Use the diagram to prove the theorem which states that the opposite angles of a cyclic quadrilateral are supplementary.
(4)
In the diagram below, ABCD is a quadrilateral with the diagonal AC drawn. B $\hat{A}$ C = 51°; A $\hat{C}$ B = 52°; C $\hat{D}$ A = 103°.
(b)
Prove that ABCD is a cyclic quadrilateral.
(4)

Show the answers

(a) Answer: See the proof below.

Given: ABCD is a cyclic quadrilateral in a circle with centre O.
To prove: $\hat{A}$ + $\hat{C}$ = 180°
Construction: Join OB and OD.
Let ${\hat{O}}_{1}$ be the angle at the centre on C's side of BD, and ${\hat{O}}_{2}$ the other one.
${\hat{O}}_{1}$ = 2 × $\hat{A}$ (∠ at centre = 2 × ∠ at circumference)
${\hat{O}}_{2}$ = 2 × $\hat{C}$ (∠ at centre = 2 × ∠ at circumference)
${\hat{O}}_{1}$ + ${\hat{O}}_{2}$ = 360° (∠s round a pt)
∴ 2 $\hat{A}$ + 2 $\hat{C}$ = 360°
∴ $\hat{A}$ + $\hat{C}$ = 180°

(b) Answer: See the proof below.

A $\hat{B}$ C = 180° − 51° − 52° = 77° (sum ∠s of Δ)
A $\hat{B}$ C + A $\hat{D}$ C = 77° + 103° = 180°
∴ ABCD is a cyclic quadrilateral (opp ∠s of quad supp)

413 marks

In the diagram, ΔABC with D on AB and E on AC, and DE ∥ BC.

(a)
Use the diagram to prove the theorem which states that a line parallel to one side of a triangle divides the other two sides proportionally.
(4)
In the diagram below, ΔABC is drawn. D is the mid-point of AB. E lies on AC with DE ∥ BC. H lies on BC with BH : HC = 2 : 3, and AH cuts DE at K. BC = 10 units.
(b)
Prove that E is the mid-point of AC.
(2)
(c)
Calculate the length of DE.
(2)
(d)
Calculate the length of DK.
(3)
(e)
Determine DK : KE.
(2)

Show the answers

(a) Answer: See the proof below.

Given: ΔABC with D on AB and E on AC, and DE ∥ BC.
To prove: AD/DB = AE/EC
Construction: Join BE and CD. Draw EX ⊥ AB and DY ⊥ AC.
area ΔADE / area ΔBDE = AD/DB (same height)
area ΔADE / area ΔCDE = AE/EC (same height)
area ΔBDE = area ΔCDE (same base; same height)
∴ AD/DB = AE/EC

(b) Answer: See the proof below.

AD = DB (given)
DE ∥ BC (given)
∴ AE = EC (line from midpoint ∥ to one side)

(c) Answer: DE = 5 units

DE = ½BC (Midpt Theorem)
DE = ½ × 10 = 5

(d) Answer: DK = 2 units

BH = 2/5 × 10 = 4
In ΔABH: AD = DB and DK ∥ BH (given)
∴ AK = KH (line from midpoint ∥ to one side)
DK = ½BH = 2 (Midpt Theorem)

(e) Answer: DK : KE = 2 : 3

In ΔAHC: KE = ½HC (Midpt Theorem)
∴ DK : KE = ½BH : ½HC = 2 : 3

512 marks

In the diagram, circle with centre O. AB is a chord. OM ⊥ AB.

(a)
Use the diagram to prove the theorem which states that the line from the centre perpendicular to a chord bisects the chord.
(6)
In the diagram below, O is the centre of the circle and OM ⊥ AB, with M on AB. O $\hat{A}$ B = 34°.
(b)
Determine, with reasons, the size of O $\hat{M}$ A.
(2)
(c)
Determine, with reasons, the size of O $\hat{M}$ B.
(2)
(d)
Determine, with reasons, the size of A $\hat{O}$ M.
(2)

Show the answers

(a) Answer: See the proof below.

Given: Circle with centre O. AB is a chord. OM ⊥ AB.
To prove: AM = MB
Construction: Join OA and OB.
In ΔOMA and ΔOMB:
OA = OB (radii)
OM = OM (common)
O $\hat{M}$ A = O $\hat{M}$ B = 90° (given)
∴ ΔOMA ≡ ΔOMB (RHS)
∴ AM = MB (≡ Δs)

(b) Answer: O $\hat{M}$ A = 90°

O $\hat{M}$ A = 90° (line from centre ⊥ to chord)

(c) Answer: O $\hat{M}$ B = 90°

O $\hat{M}$ B = 90° (line from centre ⊥ to chord)

(d) Answer: A $\hat{O}$ M = 56°

A $\hat{O}$ M = 56° (sum ∠s of Δ)

Tangents, and tangents from a point outside

1 questions · 14 marks

614 marks

In the diagram, SAT is a tangent at A to the circle with centre O, and B lies on the circle. B $\hat{A}$ T = 57°.

(a)
Determine, with reasons, the size of B $\hat{A}$ S.
(2)
(b)
Determine, with reasons, the size of O $\hat{A}$ S.
(2)
(c)
Determine, with reasons, the size of O $\hat{A}$ T.
(2)
(d)
Determine, with reasons, the size of O $\hat{B}$ A.
(2)
(e)
Determine, with reasons, the size of B $\hat{O}$ A.
(2)
In the diagram below, the chords AC and BD cut at P inside the circle. A $\hat{P}$ B = 68°; P $\hat{A}$ B = 87°.
(f)
Determine, with reasons, the size of P $\hat{B}$ A.
(2)
(g)
Determine, with reasons, the size of D $\hat{P}$ A.
(2)

Show the answers

(a) Answer: B $\hat{A}$ S = 123°

B $\hat{A}$ S = 123° (∠s on a str line)

(b) Answer: O $\hat{A}$ S = 90°

O $\hat{A}$ S = 90° (tan ⊥ radius)

(c) Answer: O $\hat{A}$ T = 90°

O $\hat{A}$ T = 90° (tan ⊥ radius)

(d) Answer: O $\hat{B}$ A = 33°

B $\hat{A}$ O = 33°
OA = OB (radii)
O $\hat{B}$ A = 33° (∠s opp equal sides)

(e) Answer: B $\hat{O}$ A = 114°

B $\hat{O}$ A = 114° (sum ∠s of Δ)

(f) Answer: P $\hat{B}$ A = 25°

P $\hat{B}$ A = 25° (sum ∠s of Δ)

(g) Answer: D $\hat{P}$ A = 112°

D $\hat{P}$ A = 112° (∠s on a str line)

Similar triangles

2 questions · 20 marks

710 marks

In the diagram, ΔABC is drawn.

D is a point on AB and E is a point on AC so that DE ∥ BC.
BE and DC intersect at T.
DE = 11 units, DT = 7 units, ET = 9 units and TC = 21 units.

(a)
Prove that ΔDET ||| ΔCBT.
(3)
(b)
Determine the length of BC.
(2)
(c)
Determine the length of BT.
(2)
(d)
Determine AD : DB.
(3)

Show the answers

(a) Answer: See the proof below.

In ΔDET and ΔCBT:
E $\hat{D}$ T = B $\hat{C}$ T (alt ∠s; DE ∥ BC)
D $\hat{E}$ T = C $\hat{B}$ T (alt ∠s; DE ∥ BC)
D $\hat{T}$ E = B $\hat{T}$ C (vert opp ∠s =)
∴ ΔDET ||| ΔCBT (∠∠∠)

(b) Answer: BC = 33 units

BC/DE = CT/DT (∥∥∥ Δs)
BC = 11 × 21/7 = 33

(c) Answer: BT = 27 units

BT/ET = CT/DT (∥∥∥ Δs)
BT = 9 × 21/7 = 27

(d) Answer: AD : DB = 1 : 2

In ΔADE and ΔABC:
$\hat{A}$ = $\hat{A}$ (common)
A $\hat{D}$ E = A $\hat{B}$ C (corresp ∠s equal; DE ∥ BC)
∴ ΔADE ||| ΔABC (∠∠∠)
AD/AB = DE/BC = 11/33 = 1/3 (∥∥∥ Δs)
∴ AD : DB = 1 : 2

810 marks

In the diagram, ΔABC is drawn.

D is a point on AB and E is a point on AC so that DE ∥ BC.
BE and DC intersect at T.
DE = 7 units, DT = 4 units, ET = 5 units and TC = 12 units.

(a)
Prove that ΔDET ||| ΔCBT.
(3)
(b)
Determine the length of BC.
(2)
(c)
Determine the length of BT.
(2)
(d)
Determine AD : DB.
(3)

Show the answers

(a) Answer: See the proof below.

In ΔDET and ΔCBT:
E $\hat{D}$ T = B $\hat{C}$ T (alt ∠s; DE ∥ BC)
D $\hat{E}$ T = C $\hat{B}$ T (alt ∠s; DE ∥ BC)
D $\hat{T}$ E = B $\hat{T}$ C (vert opp ∠s =)
∴ ΔDET ||| ΔCBT (∠∠∠)

(b) Answer: BC = 21 units

BC/DE = CT/DT (∥∥∥ Δs)
BC = 7 × 12/4 = 21

(c) Answer: BT = 15 units

BT/ET = CT/DT (∥∥∥ Δs)
BT = 5 × 12/4 = 15

(d) Answer: AD : DB = 1 : 2

In ΔADE and ΔABC:
$\hat{A}$ = $\hat{A}$ (common)
A $\hat{D}$ E = A $\hat{B}$ C (corresp ∠s equal; DE ∥ BC)
∴ ΔADE ||| ΔABC (∠∠∠)
AD/AB = DE/BC = 7/21 = 1/3 (∥∥∥ Δs)
∴ AD : DB = 1 : 2

The angle in a semi-circle

1 questions · 10 marks

910 marks

In the diagram, AB is a diameter of the circle with centre O, and P and Q lie on the circle. P $\hat{A}$ B = 30°.

(a)
Determine, with reasons, the size of A $\hat{P}$ B.
(2)
(b)
Determine, with reasons, the size of B $\hat{Q}$ A.
(2)
(c)
Determine, with reasons, the size of P $\hat{B}$ O.
(2)
In the diagram below, SAT is a tangent to the circle at A. S $\hat{A}$ B = 118°.
(d)
Determine, with reasons, the size of B $\hat{A}$ T.
(2)
(e)
Determine, with reasons, the size of A $\hat{C}$ B.
(2)

Show the answers

(a) Answer: A $\hat{P}$ B = 90°

A $\hat{P}$ B = 90° (∠s in semi-circle)

(b) Answer: B $\hat{Q}$ A = 90°

B $\hat{Q}$ A = 90° (∠s in semi-circle)

(c) Answer: P $\hat{B}$ O = 60°

P $\hat{B}$ O = 60° (sum ∠s of Δ)

(d) Answer: B $\hat{A}$ T = 62°

B $\hat{A}$ T = 62° (∠s on a str line)

(e) Answer: A $\hat{C}$ B = 62°

A $\hat{C}$ B = 62° (tan chord theorem)

Chords, and the line from the centre

1 questions · 12 marks

1012 marks

In the diagram, O is the centre of the circle and OM ⊥ AB, with M on AB. O $\hat{A}$ B = 34°.

(a)
Determine, with reasons, the size of O $\hat{M}$ A.
(2)
(b)
Determine, with reasons, the size of O $\hat{M}$ B.
(2)
(c)
Determine, with reasons, the size of A $\hat{O}$ M.
(2)
In the diagram below, AB is a diameter of the circle with centre O, and P and Q lie on the circle. P $\hat{A}$ B = 36°.
(d)
Determine, with reasons, the size of A $\hat{P}$ B.
(2)
(e)
Determine, with reasons, the size of B $\hat{Q}$ A.
(2)
(f)
Determine, with reasons, the size of P $\hat{B}$ O.
(2)

Show the answers

(a) Answer: O $\hat{M}$ A = 90°

O $\hat{M}$ A = 90° (line from centre ⊥ to chord)

(b) Answer: O $\hat{M}$ B = 90°

O $\hat{M}$ B = 90° (line from centre ⊥ to chord)

(c) Answer: A $\hat{O}$ M = 56°

A $\hat{O}$ M = 56° (sum ∠s of Δ)

(d) Answer: A $\hat{P}$ B = 90°

A $\hat{P}$ B = 90° (∠s in semi-circle)

(e) Answer: B $\hat{Q}$ A = 90°

B $\hat{Q}$ A = 90° (∠s in semi-circle)

(f) Answer: P $\hat{B}$ O = 54°

P $\hat{B}$ O = 54° (sum ∠s of Δ)

The proportion theorem

1 questions · 8 marks

118 marks

In the diagram, ΔABC is drawn.

E and F are points on AC and AB respectively, with AE : EC = 2 : 1 and AF : FB = 1 : 1.
BC produced meets FE produced at D.
G is a point on FB so that FD ∥ GC.

(a)
Determine AF : FG.
(2)
(b)
Hence determine FG : GB.
(3)
(c)
Calculate BC : CD.
(3)

Show the answers

(a) Answer: AF : FG = 2 : 1

AF/FG = AE/EC = 2/1 (line ∥ one side of Δ; FE ∥ GC)
AF : FG = 2 : 1

(b) Answer: FG : GB = 1 : 1

Let AF = 2k and FB = 2k (given)
FG = 1/2 × 2k = 1k
GB = FB − FG = 2k − 1k = 1k
FG : GB = 1 : 1

(c) Answer: BC : CD = 1 : 1

BC/CD = BG/GF (line ∥ one side of Δ; GC ∥ FD)
BC/CD = 1k/1k
BC : CD = 1 : 1

Angles in the same segment

1 questions · 14 marks

1214 marks

In the diagram, the chords AC and BD cut at P inside the circle. A $\hat{P}$ B = 99°; P $\hat{A}$ B = 24°.

(a)
Determine, with reasons, the size of P $\hat{B}$ A.
(2)
(b)
Determine, with reasons, the size of D $\hat{P}$ A.
(2)
(c)
Determine, with reasons, the size of B $\hat{P}$ C.
(2)
(d)
Determine, with reasons, the size of P $\hat{D}$ C.
(2)
(e)
Determine, with reasons, the size of D $\hat{P}$ C.
(2)
In the diagram below, ABC is a triangle with BC produced to D. B $\hat{A}$ C = 100°; A $\hat{B}$ C = 37°.
(f)
Determine, with reasons, the size of A $\hat{C}$ B.
(2)
(g)
Determine, with reasons, the size of D $\hat{C}$ A.
(2)

Show the answers

(a) Answer: P $\hat{B}$ A = 57°

P $\hat{B}$ A = 57° (sum ∠s of Δ)

(b) Answer: D $\hat{P}$ A = 81°

D $\hat{P}$ A = 81° (∠s on a str line)

(c) Answer: B $\hat{P}$ C = 81°

B $\hat{P}$ C = 81° (∠s on a str line)

(d) Answer: P $\hat{D}$ C = 24°

P $\hat{D}$ C = 24° (∠s in the same seg)

(e) Answer: D $\hat{P}$ C = 99°

D $\hat{P}$ C = 99° (vert opp ∠s =)

(f) Answer: A $\hat{C}$ B = 43°

A $\hat{C}$ B = 43° (sum ∠s of Δ)

(g) Answer: D $\hat{C}$ A = 137°

D $\hat{C}$ A = 137° (∠s on a str line)

Download this booklet

The same questions as a booklet to print, with the memorandum as a separate file so you can work without the answers in front of you. In English and Afrikaans.

Questions (English) 22 pages, 0.3 MB
Memorandum (English) 13 pages, 0.2 MB
Vrae (Afrikaans) 22 pages, 0.3 MB
Memorandum (Afrikaans) 13 pages, 0.2 MB

Every question on this page was generated for this site by a computer engine, and its answer was confirmed by an independent method at the moment the question was made.

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