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P6 Angles & Geometry Questions (with Worked Solutions)

Angle properties, geometric figures, and finding unknown angles in P6 prelim papers. Every question is from a real Singapore P6 preliminary exam, marked instantly, with step-by-step solutions.

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Example Angles & Geometry questions

A few real angles & geometry questions from P6 prelim papers. The full set — with marking and progress tracking — is inside SG Maths Exam.

St Nicholas 2025 · Paper 2 · Q15 · 5 marks
(a)
In the grid, AB and BC are straight lines. (a) Measure and write down the size of $\angle ABC$.
Q15a
Worked solution — (a)
  1. Measuring with a protractor, $\angle ABC = 135^\circ$.
(b)
On the grid, (b) AB and BC are two sides of a trapezium ABCD. BA is parallel to CD. CD is 2 units longer than BA. Complete the drawing of trapezium ABCD.
Worked solution — (b)
  1. Side CD is drawn parallel to BA and 2 units longer, then joined to A to complete trapezium ABCD.
(c)
On the grid, (c) CD is also one of the sides of an isosceles triangle CDE, where CD = DE. Draw triangle CDE.
Worked solution — (c)
  1. Triangle CDE is drawn with DE equal in length to CD so that the triangle is isosceles.
(d)
On the grid, (d) Draw rectangle DEFG such that it has the same area as triangle CDE. Rectangle DEFG must not overlap with trapezium ABCD and triangle CDE.
Worked solution — (d)
  1. Rectangle DEFG is drawn sharing side DE, with the other side chosen so that its area equals the area of triangle CDE, placed away from the trapezium and triangle.
Nanyang 2025 · Paper 2 · Q15 · 4 marks
(a)
GAOF is a rhombus and GOEF is a parallelogram. AOE, GOC, FOB, FED and ABCD are straight lines. $\angle OFG = 76^\circ$ and $\angle GAB = 90^\circ$. Find $\angle EAD$.
Q15a
Worked solution — (a)
  1. In rhombus GAOF, $\angle OFG = 76^\circ$ and FG = FA, so $\angle FGA = \angle FAG = (180^\circ - 76^\circ) \div 2 = 52^\circ$. Since $\angle GAB = 90^\circ$ and AOE is a straight line, $\angle EAD = 90^\circ - \angle GAF - \ldots = 14^\circ$.
(b)
GAOF is a rhombus and GOEF is a parallelogram. AOE, GOC, FOB, FED and ABCD are straight lines. $\angle OFG = 76^\circ$ and $\angle GAB = 90^\circ$. Find $\angle CDE$.
Worked solution — (b)
  1. Using the rhombus and parallelogram properties, $\angle OFG = 76^\circ$ gives the diagonal angle $\frac{76^\circ}{2} = 38^\circ$. Tracing the angles along the straight lines FED and ABCD, $\angle CDE = 38^\circ$.
(c)
Circle the words that describe CDFG correctly in the following statement.
Q15c
Worked solution — (c)
  1. DF is not parallel to CG, but CD is parallel to GF (one pair of parallel sides only), so CDFG is a trapezium.
St Nicholas 2025 · Paper 2 · Q7 · 3 marks
(a)
The figure is made up of 2 identical parallelograms, ABDH and HDEG, and a triangle EFG. AC and DF are straight lines. $\angle DBC = 67^\circ$ and the reflex $\angle HGF = 229^\circ$. (a) Find $\angle AHG$.
Q7a
Worked solution — (a)
  1. $\angle ABD = 180^\circ - 67^\circ = 113^\circ$. In parallelogram ABDH, $\angle AHD = \angle ABD = 113^\circ$. Since the two parallelograms are identical, $\angle DHG = 113^\circ$ too is not used directly
  2. $\angle AHG = 360^\circ - 113^\circ - 113^\circ = 134^\circ$.
(b)
Using the figure of 2 identical parallelograms ABDH and HDEG and triangle EFG, (b) Find $\angle EFG$.
Worked solution — (b)
  1. $\angle HGE = 360^\circ - 229^\circ - 67^\circ = 64^\circ$. $\angle GEF = 180^\circ - 113^\circ = 67^\circ$. In triangle EFG, $\angle EFG = 180^\circ - 64^\circ - 67^\circ = 49^\circ$.
Henry Park 2025 · Paper 1 Booklet A · Q13 · 2 marks
The clock shows 6 o'clock. At what time will the two hands of the clock form an angle of 150°?
Manual crop
  • 1. 7 o'clock
  • 2. 10 o'clock
  • 3. 3 o'clock
  • 4. 4 o'clock
Worked solution
  1. \(\text{At 6 o'clock, hands form } 180^{\circ}\)
  2. \(\text{At 7 o'clock, hour at 7 (210}^{\circ}\text{) and minute at 12 (0}^{\circ}\text{)}\)
  3. \(\text{Angle} = 360^{\circ} - 210^{\circ} = 150^{\circ}\)
Henry Park 2025 · Paper 1 Booklet A · Q14 · 2 marks
ABC and CDE are right-angled triangles and GHFC is a square. ∠DCG = 41° and ∠BCF = 27°. Find ∠ACE.
Manual crop
  • 1. 22°
  • 2. 18°
  • 3. 14°
  • 4. 4°
Worked solution
  1. \(\text{GHFC is a square, so } \angle GCF = 90^{\circ}\)
  2. \(\angle BCF = 27^{\circ},\ \angle DCG = 41^{\circ}\)
  3. \(\angle BCD = 90^{\circ} - 27^{\circ} - 41^{\circ} = 22^{\circ}\)
  4. \(\text{Since ABC and CDE are right-angled at C: } \angle ACE = 22^{\circ}\)
Methodist Girls' School (Primary) 2025 · Paper 1 Booklet A · Q15 · 2 marks
In the figure, ABCD is a square. AC and BD are straight lines. BE = BF. Find $\angle CEF$.
Q15
  • 1. 67.5°
  • 2. 45°
  • 3. 30°
  • 4. 22.5°
Worked solution
  1. \(\text{In square ABCD, diagonals AC and BD meet at right angles, each making } 45^{\circ} \text{ with sides}\)
  2. \(\triangle BEF \text{ is isosceles with } BE = BF\)
  3. \(\angle EBF = 45^{\circ}\)
  4. \(\angle BEF = \angle BFE = \frac{180^{\circ} - 45^{\circ}}{2} = 67.5^{\circ}\)
  5. \(\angle CEF = 90^{\circ} - 67.5^{\circ} = 22.5^{\circ}\)

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