Geometry and Measurement Sec 2 E-Mathematics

Congruence and Similarity

Key Concepts

  • Congruence and similarity are ways to compare shapes.
  • These ideas are very useful in geometry, measurement, map reading, scale drawings, and solving problems involving lengths, areas, and volumes.

Congruent figures

  • Two figures are congruent if they have:
    • the same shape, and
    • the same size.
  • If one figure can be moved by:
    • translation (sliding),
    • rotation (turning), or
    • reflection (flipping), so that it fits exactly onto the other, then the figures are congruent.
  • For congruent figures:
    • all corresponding sides are equal,
    • all corresponding angles are equal.

Congruent triangles

  • For triangles, you do not need to check all 6 parts every time.
  • There are special conditions that prove two triangles are congruent.

1. SSS Congruence

  • SSS means Side-Side-Side.
  • If the three sides of one triangle are equal to the three corresponding sides of another triangle, then the triangles are congruent.

2. SAS Congruence

  • SAS means Side-Angle-Side.
  • If two sides and the included angle between them in one triangle are equal to the corresponding two sides and included angle in another triangle, then the triangles are congruent.
  • The angle must be the angle between the two known sides.

3. AAS Congruence

  • AAS means Angle-Angle-Side.
  • If two angles and one corresponding side of one triangle are equal to those of another triangle, then the triangles are congruent.
  • Since the angles in a triangle add up to 180°, knowing two angles also fixes the third angle.

4. RHS Congruence

  • RHS means Right angle-Hypotenuse-Side.
  • This condition applies only to right-angled triangles.
  • If two right-angled triangles have:
    • one right angle,
    • equal hypotenuse,
    • and one corresponding side equal, then the triangles are congruent.

Similar figures

  • Two figures are similar if they have:
    • the same shape,
    • but not necessarily the same size.
  • For similar figures:
    • all corresponding angles are equal,
    • all corresponding lengths are in the same ratio.

Similar triangles

  • Two triangles are similar if:
    • their corresponding angles are equal, and
    • their corresponding sides are proportional.
  • Similar triangles may be:
    • enlarged,
    • reduced,
    • or turned/flipped.

AA Similarity Test

The AA (Angle-Angle) Similarity Test is the most commonly used test in Sec 2 exams:

If two pairs of corresponding angles in two triangles are equal, the triangles are similar.

Why only two angles? Because if two angles are equal, the third must also be equal (angles in a triangle sum to 180°).

Worked example — full similarity proof (write exactly this in exams):

Given: Triangle ABC and Triangle PQR, where ∠BAC = ∠QPR and ∠ABC = ∠PQR. Prove that the triangles are similar.

Statement Reason
∠BAC = ∠QPR Given
∠ABC = ∠PQR Given
∴ Triangle ABC is similar to Triangle PQR AA Similarity Test

Consequence: Once similarity is established, corresponding sides are proportional: [\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}]

Exam tip: Always write “AA Similarity Test” (not just “AA”) and list the two angle pairs with their reasons before stating the conclusion. One mark is typically awarded for each correct angle pair and one for the conclusion.

Scale factor

  • The scale factor compares corresponding lengths in similar figures.
  • If a figure is enlarged by scale factor kk:
    • every length is multiplied by kk.

Length scale factor

  • If corresponding lengths are in the ratio a:ba:b, then the length scale factor is:

    ab \frac{a}{b}

Area scale factor

  • If the length scale factor is kk, then the area scale factor is:

    k2 k^2
  • So if lengths double, area becomes:

    22=4 2^2 = 4

    times as large.

Volume scale factor

Not in Sec 2 2026: The volume scale factor for similar solids (k³) is a Sec 3/4 topic. The 2026 Sec 2 syllabus only requires the area scale factor (k²) for similar figures. The notes below are kept for completeness and future reference.

  • If the length scale factor is kk, then the volume scale factor is:

    k3 k^3
  • So if lengths triple, volume becomes:

    33=27 3^3 = 27

    times as large.

Proportional relationships in similar triangles

  • In similar triangles, corresponding sides are in equal ratios.

  • Example:

    • if triangle ABCABC is similar to triangle DEFDEF,
    • and ABDEAB \leftrightarrow DE, BCEFBC \leftrightarrow EF, ACDFAC \leftrightarrow DF, then
    ABDE=BCEF=ACDF \frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}
  • You can use these equal ratios to:

    • find unknown lengths,
    • compare heights and distances,
    • solve scale drawing problems.

Writing corresponding vertices correctly

  • The order of letters matters.

  • If

    ABCPQR \triangle ABC \cong \triangle PQR

    then:

    • APA \leftrightarrow P
    • BQB \leftrightarrow Q
    • CRC \leftrightarrow R
  • Therefore:

    • AB=PQAB = PQ
    • BC=QRBC = QR
    • AC=PRAC = PR
    • A=P\angle A = \angle P, etc.

Exterior Angle of a Triangle

Note: This theorem belongs to the Angles topic but is often tested alongside similarity questions, so it is included here for reference.

  • An exterior angle of a triangle is formed by extending one side of the triangle.
  • Exterior angle theorem: An exterior angle of a triangle is equal to the sum of the two non-adjacent interior angles (also called remote interior angles).
Exterior angle=Interior angle1+Interior angle2 \text{Exterior angle} = \text{Interior angle}_1 + \text{Interior angle}_2
  • Example: In triangle ABCABC, side BCBC is extended to point DD.

    • The exterior angle ACD=BAC+ABC\angle ACD = \angle BAC + \angle ABC
    • If BAC=45\angle BAC = 45^\circ and ABC=65\angle ABC = 65^\circ, then ACD=45+65=110\angle ACD = 45^\circ + 65^\circ = 110^\circ
    • Check: the three interior angles must add to 180180^\circ, so BCA=180110=70\angle BCA = 180^\circ - 110^\circ = 70^\circ, and 45+65+70=18045^\circ + 65^\circ + 70^\circ = 180^\circ. ✓
  • This theorem is useful when you know two angles of a triangle and need to find an exterior angle — or when you know an exterior angle and one interior angle.

Problem solving with congruent and similar figures

  • In geometry questions, you may need to:
    • identify equal sides or angles,
    • state the correct test for congruence,
    • state that triangles are similar,
    • use corresponding side ratios,
    • use scale factors to compare area or volume.
  • Always:
    1. identify corresponding parts carefully,
    2. write the correct ratio,
    3. substitute values,
    4. solve clearly.

Important Definitions

  • Congruent figures: figures that have the same shape and the same size.
  • Similar figures: figures that have the same shape but not necessarily the same size.
  • Corresponding parts: matching sides or angles in two figures that are in the same relative positions.
  • Triangle congruence: when two triangles are exactly equal in shape and size.
  • SSS congruence: a test for congruence where all three corresponding sides are equal.
  • SAS congruence: a test for congruence where two corresponding sides and the included angle are equal.
  • AAS congruence: a test for congruence where two corresponding angles and one corresponding side are equal.
  • RHS congruence: a test for congruence for right-angled triangles, using a right angle, hypotenuse, and one side.
  • Included angle: the angle between two given sides.
  • Right-angled triangle: a triangle with one angle equal to 9090^\circ.
  • Hypotenuse: the longest side in a right-angled triangle, opposite the right angle.
  • Scale factor: the ratio of corresponding lengths in two similar figures.
  • Proportional: having the same ratio.
  • Length scale factor: the ratio of corresponding lengths of similar figures.
  • Area scale factor: the ratio of corresponding areas of similar figures.
  • Volume scale factor: the ratio of corresponding volumes of similar solids.
  • Enlargement: a transformation that increases the size of a figure by a scale factor greater than 1.
  • Reduction: a transformation that decreases the size of a figure by a scale factor between 0 and 1.

Worked Examples

Example 1: Proving triangles are congruent

Two triangles have:

  • AB=PQ=7 cmAB = PQ = 7\text{ cm}
  • AC=PR=5 cmAC = PR = 5\text{ cm}
  • A=P=40\angle A = \angle P = 40^\circ

Show that ABC\triangle ABC is congruent to PQR\triangle PQR.

Step 1: Identify the given equal parts

  • AB=PQAB = PQ
  • AC=PRAC = PR
  • A=P\angle A = \angle P

Step 2: Check the position of the angle

  • A\angle A is between sides ABAB and ACAC
  • P\angle P is between sides PQPQ and PRPR

So the equal angle is the included angle.

Step 3: Apply the congruence condition

  • Two sides and the included angle are equal.

Therefore,

ABCPQR by SAS \triangle ABC \cong \triangle PQR \text{ by SAS}

Final answer

  • ABCPQR\triangle ABC \cong \triangle PQR by SAS.

Example 2: Finding an unknown length in similar triangles

ABCDEF\triangle ABC \sim \triangle DEF

Given:

  • AB=6 cmAB = 6\text{ cm}
  • DE=9 cmDE = 9\text{ cm}
  • BC=8 cmBC = 8\text{ cm}
  • Find EFEF

Step 1: Write corresponding sides

Since ABCDEF\triangle ABC \sim \triangle DEF,

  • ABDEAB \leftrightarrow DE
  • BCEFBC \leftrightarrow EF

So,

ABDE=BCEF \frac{AB}{DE}=\frac{BC}{EF}

Step 2: Substitute values

69=8EF \frac{6}{9}=\frac{8}{EF}

Step 3: Simplify ratio

23=8EF \frac{2}{3}=\frac{8}{EF}

Step 4: Cross multiply

2(EF)=3(8) 2(EF)=3(8)
2EF=24 2EF=24
EF=12 EF=12

Final answer

EF=12 cm EF = 12\text{ cm}

Example 3: Using scale factors for area and volume

Two similar solids have length scale factor 2:52:5.

The smaller solid has:

  • surface area 40 cm240\text{ cm}^2
  • volume 24 cm324\text{ cm}^3

Find the surface area and volume of the larger solid.

Step 1: Find the area scale factor

Length scale factor:

2:5 2:5

Area scale factor:

22:52=4:25 2^2:5^2 = 4:25

Step 2: Find the larger surface area

small arealarge area=425 \frac{\text{small area}}{\text{large area}}=\frac{4}{25}

So,

large area=40×254 \text{large area}=40\times \frac{25}{4}
=10×25 =10\times 25
=250 cm2 =250\text{ cm}^2

Step 3: Find the volume scale factor

Volume scale factor:

23:53=8:125 2^3:5^3 = 8:125

Step 4: Find the larger volume

small volumelarge volume=8125 \frac{\text{small volume}}{\text{large volume}}=\frac{8}{125}

So,

large volume=24×1258 \text{large volume}=24\times \frac{125}{8}
=3×125 =3\times 125
=375 cm3 =375\text{ cm}^3

Final answers

  • Larger surface area =250 cm2= 250\text{ cm}^2
  • Larger volume =375 cm3= 375\text{ cm}^3

Common Mistakes to Avoid

  • Mixing up congruent and similar.
    • Congruent = same shape and same size.
    • Similar = same shape only.
  • Using the wrong order of vertices.
    • If ABCDEF\triangle ABC \sim \triangle DEF, do not match ABAB with DFDF unless the order shows that.
  • Using SAS when the angle given is not the included angle.
  • Forgetting that RHS works only for right-angled triangles.
  • Assuming triangles are congruent just because two angles are equal.
    • Equal angles alone show same shape, not necessarily same size.
  • Using length scale factor directly for area or volume.
    • Area uses k2k^2
    • Volume uses k3k^3
  • Writing side ratios upside down halfway through the solution.
    • Stay consistent from start to end.
  • Forgetting units:
    • length in cm,
    • area in cm²,
    • volume in cm³.
  • Not marking corresponding sides and angles clearly in a diagram.
  • Cross-multiplying wrongly when solving proportions.
  • Assuming figures are similar without checking corresponding angles or side ratios.

Exam Tips

  • When proving triangles congruent, write the full statement clearly:

    • ABCDEF\triangle ABC \cong \triangle DEF by SSS”
    • PQRXYZ\triangle PQR \cong \triangle XYZ by RHS”
  • Use the correct symbol:

    • congruent: \cong
    • similar: \sim
  • Always state the reason:

    • “since AB=DEAB = DE, BC=EFBC = EF, and AC=DFAC = DF
    • “therefore the triangles are congruent by SSS”
  • For similarity questions, write the ratio in matching order:

    ABDE=BCEF=ACDF \frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}
  • In scale factor questions:

    • identify whether the question is about length, area, or volume
    • then use kk, k2k^2, or k3k^3 correctly
  • If there are parallel lines, look for equal angles:

    • alternate angles,
    • corresponding angles,
    • common angles.
  • In written explanations, useful mark-earning phrases include:

    • “corresponding sides are equal”
    • “included angle”
    • “right angle”
    • “hypotenuse”
    • “corresponding angles are equal”
    • “corresponding sides are proportional”
  • If the answer is a scale factor, state the direction clearly:

    • “scale factor from small to large is 3”
    • or “from large to small is 13\frac{1}{3}
  • Draw or annotate your own marks on the diagram if allowed. This helps avoid matching the wrong sides.


Quick Summary

  • Congruent figures have the same shape and same size.
  • Similar figures have the same shape but may have different sizes.
  • For congruent triangles, use:
    • SSS
    • SAS
    • AAS
    • RHS
  • SAS needs the included angle between the two known sides.
  • RHS applies only to right-angled triangles.
  • In similar figures, corresponding angles are equal and corresponding sides are proportional.
  • If ABCDEF\triangle ABC \sim \triangle DEF, keep the vertex order correct when matching sides.
  • If the length scale factor is kk, then:
    • area scale factor =k2= k^2
    • volume scale factor =k3= k^3
  • Use proportions to find unknown lengths in similar triangles.
  • Always check whether the question is about length, area, or volume before calculating.
  • Write complete mathematical statements such as:
    • ABCDEF\triangle ABC \cong \triangle DEF by SSS
    • PQRXYZ\triangle PQR \sim \triangle XYZ
  • Include correct units: cm, cm², cm³.
✏️ 28 practice questions available

30 questions from school exam papers

Q1

Construct the triangle PQR where PQ = QR = 7 cm and PR = 5 cm.

Empty space provided for construction of triangle PQR
📊 Diagram: Empty space provided for construction of triangle PQR
Math_Sec2NA_SA2_2023_ACS_Barker 2023
Q2

Measure and write down the size of angle QPR.

Diagram for question 8b
2 marks
Math_Sec2NA_SA2_2023_ACS_Barker 2023
Q3

Calculate the percentage of the children who read more than 20 books in Play Hub.

Diagram for question 8c
1 mark
Math_Sec2NA_SA2_2023_ACS_Barker 2023
Q4

Triangles ABC and XYZ are similar. Find angle XYZ.

Two triangles shown: Triangle ABC with angle A = 40°, side AB = 6 cm, side BC = 4 cm. Triangle XYZ with angle Z = 75°, side ZX = 15 cm, side ZY = 16 cm.
📊 Diagram: Two triangles shown: Triangle ABC with angle A = 40°, side AB = 6 cm, side BC = 4 cm. Triangle XYZ with angle Z = 75°, side ZX = 15 cm, side ZY = 16 cm.
2 marks
Math_Sec2NA_SA2_2023_ACS_Barker 2023
Q5

Triangles ABC and XYZ are similar. Find the length of XY.

Two triangles shown: Triangle ABC with angle A = 40°, side AB = 6 cm, side BC = 4 cm. Triangle XYZ with angle Z = 75°, side ZX = 15 cm, side ZY = 16 cm.
📊 Diagram: Two triangles shown: Triangle ABC with angle A = 40°, side AB = 6 cm, side BC = 4 cm. Triangle XYZ with angle Z = 75°, side ZX = 15 cm, side ZY = 16 cm.
2 marks
Math_Sec2NA_SA2_2023_ACS_Barker 2023
Q6

Construct a triangle ABC such that BC = 8 cm and AC = 6.8 cm. AB has been drawn for you.

A horizontal line segment with endpoints labeled A (left) and B (right), representing the pre-drawn side AB of the triangle.
📊 Diagram: A horizontal line segment with endpoints labeled A (left) and B (right), representing the pre-drawn side AB of the triangle.
2 marks
Math_Sec2NA_SA2_2023_Bartley_Sec 2023
Q7

The diagram shows a rhombus ABCD. DBE is a straight line, BC = BE and angle CDE = 46°. Stating your reasons clearly, find angle BAD.

A rhombus ABCD with point E below B. The vertices are labeled A (bottom left), B (bottom middle), C (top right), and D (top left). There is a diagonal from D to B. Point E is located such that B, D, and E are collinear (straight line DBE). The angle at D between DC and DB is marked as 46°. Arrows on the sides indicate: AB is parallel to DC (double arrows), and AD is parallel to BC (single arrows). There are also single arrows on BC and CE indicating BC = BE.
📊 Diagram: A rhombus ABCD with point E below B. The vertices are labeled A (bottom left), B (bottom middle), C (top right), and D (top left). There is a diagonal from D to B. Point E is located such that B, D, and E are collinear (straight line DBE). The angle at D between DC and DB is marked as 46°. Arrows on the sides indicate: AB is parallel to DC (double arrows), and AD is parallel to BC (single arrows). There are also single arrows on BC and CE indicating BC = BE.
2 marks
Math_Sec2NA_SA2_2023_Bartley_Sec 2023
Q8

Triangles ABC and DBE are similar. Find the length of BE.

Same diagrams showing triangle ABC with points D and E positioned such that triangle DBE can be identified as similar to ABC
📊 Diagram: Same diagrams showing triangle ABC with points D and E positioned such that triangle DBE can be identified as similar to ABC
2 marks
Math_Sec2NA_SA2_2023_Bedok_View_Sec 2023
Q9

In the diagram below, ∠AOC = 90°, AC = 13 cm, OA = 12 cm, OB = y cm, OC = x cm and OD = 6 cm. △BOD is similar to △COA. Find the value of x.

A coordinate system with origin O. Point A is on the positive y-axis at 12 cm from O. Point C is on the positive x-axis at distance x cm from O. Point B is on the negative y-axis at distance y cm from O. Point D is on a line segment from B, positioned 6 cm above O on a vertical line through O. Triangle AOC is formed by connecting A, O, and C, with AC = 13 cm. Triangle BOD is formed by connecting B, O, and D. The angle at O in triangle AOC is 90°.
📊 Diagram: A coordinate system with origin O. Point A is on the positive y-axis at 12 cm from O. Point C is on the positive x-axis at distance x cm from O. Point B is on the negative y-axis at distance y cm from O. Point D is on a line segment from B, positioned 6 cm above O on a vertical line through O. Triangle AOC is formed by connecting A, O, and C, with AC = 13 cm. Triangle BOD is formed by connecting B, O, and D. The angle at O in triangle AOC is 90°.
2 marks
Math_Sec2NA_SA2_2023_Bedok_View_Sec 2023
Q10

Write down the coordinates of point D such that ABCD forms a parallelogram.

Diagram for question 9c
1 mark
Math_Sec2NA_SA2_2023_Broadrick_Sec 2023
Q11

Explain why triangle ABC is similar to triangle APQ.

A diagram showing triangle ABC with point P on AB and point Q on AC. PQ is parallel to BC. The triangle has vertices labeled A (bottom left), B (top right), C (bottom right), with P on segment AB and Q on segment AC. Arrows indicate the direction along the sides.
📊 Diagram: A diagram showing triangle ABC with point P on AB and point Q on AC. PQ is parallel to BC. The triangle has vertices labeled A (bottom left), B (top right), C (bottom right), with P on segment AB and Q on segment AC. Arrows indicate the direction along the sides.
2 marks
Math_Sec2NA_SA2_2023_Broadrick_Sec 2023
Q12

Given that BC = 10 cm, PQ = 4 cm and QC = 3 cm, find the length of AC.

Diagram for question 10b
2 marks
Math_Sec2NA_SA2_2023_Broadrick_Sec 2023
Q13

Write down the coordinates of point D such that ABCD forms a parallelogram.

Diagram for question 9(c)
1 mark
Math_Sec2NA_SA2_2023_Broadrick_Sec 2023
Q14

Construct triangle PQR such that QR = 5 cm and PR = 6 cm. Line PQ has been provided for you.

A horizontal line segment with endpoints labeled P (on the left) and Q (on the right)
📊 Diagram: A horizontal line segment with endpoints labeled P (on the left) and Q (on the right)
1 mark
Math_Sec2NA_SA2_2023_Canberra_Sec 2023
Q15

In the figures below, △ABC is similar to △PQR. (a) Calculate length of CA.

Two triangles are shown. △ABC has: side AB = 7.5 cm, side BC = 6 cm, angle B = 18°. △PQR has: side PQ = 11.25 cm, side PR = 8.49 cm, angle P = 80°.
📊 Diagram: Two triangles are shown. △ABC has: side AB = 7.5 cm, side BC = 6 cm, angle B = 18°. △PQR has: side PQ = 11.25 cm, side PR = 8.49 cm, angle P = 80°.
2 marks
Math_Sec2NA_SA2_2023_Canberra_Sec 2023
Q16

Triangle ABC is congruent to triangle PQR. All the lengths are in centimetres.

Two triangles shown: Triangle ABC on the left with angle CAB = 101°, and vertices labeled C, A, B. Triangle PQR on the right with sides labeled: QR = 17 cm, RP = 11 cm, QP = 12 cm, and angle PQR = 35°.
📊 Diagram: Two triangles shown: Triangle ABC on the left with angle CAB = 101°, and vertices labeled C, A, B. Triangle PQR on the right with sides labeled: QR = 17 cm, RP = 11 cm, QP = 12 cm, and angle PQR = 35°.
Math_Sec2NA_SA2_2023_Canberra_Sec 2023
Q17

The diagram below shows two congruent quadrilaterals. It is given that angle SRU = 118°, angle STU = 78°, angle RST = 42°, angle ABC = 42°, AB = 10 cm, BC = 13 cm and RU = 3 cm.

Two congruent quadrilaterals are shown. The first quadrilateral ABCD has: AB = 10 cm (top side), angle ABC = 42° at vertex B, BC = 13 cm (right side), vertex D on the left with angle x° marked. The second quadrilateral RSTU has: angle STU = 78° at vertex T (top), RU = 3 cm (right side), angle SRU = 118° at vertex R (bottom right), angle RST = 42° marked at vertex S (bottom left). The quadrilaterals are positioned to show their corresponding parts.
📊 Diagram: Two congruent quadrilaterals are shown. The first quadrilateral ABCD has: AB = 10 cm (top side), angle ABC = 42° at vertex B, BC = 13 cm (right side), vertex D on the left with angle x° marked. The second quadrilateral RSTU has: angle STU = 78° at vertex T (top), RU = 3 cm (right side), angle SRU = 118° at vertex R (bottom right), angle RST = 42° marked at vertex S (bottom left). The quadrilaterals are positioned to show their corresponding parts.
Math_Sec2NA_SA2_2023_Chung_Cheng_Yishun 2023
Q18

State the figure that is congruent to ABCD.

Diagram for question 3a
1 mark
Math_Sec2NA_SA2_2023_Chung_Cheng_Yishun 2023
Q19

State the length of SR.

Diagram for question 3b
1 mark
Math_Sec2NA_SA2_2023_Chung_Cheng_Yishun 2023
Q20

In the diagram, triangle ABC is similar to triangle ADE. AC = 9 cm, CE = 10.3 cm and BC = 8.8 cm. Find the length of DE.

A diagram showing two similar triangles. Triangle ABC is nested within triangle ADE. Point A is at the top, with AC = 9 cm marked along the right side to point C. From C, a line extends 10.3 cm to point E. BC = 8.8 cm is marked as the base of the smaller triangle ABC. Points B and D are on the left, with D below B. Triangle ADE is the larger triangle with vertices A (top), D (bottom left), and E (bottom right).
📊 Diagram: A diagram showing two similar triangles. Triangle ABC is nested within triangle ADE. Point A is at the top, with AC = 9 cm marked along the right side to point C. From C, a line extends 10.3 cm to point E. BC = 8.8 cm is marked as the base of the smaller triangle ABC. Points B and D are on the left, with D below B. Triangle ADE is the larger triangle with vertices A (top), D (bottom left), and E (bottom right).
2 marks
Math_Sec2NA_SA2_2023_Chung_Cheng_Yishun 2023
Q21

Find, giving your reasons clearly, the length of AD.

Diagram for question 1a
2 marks
Math_Sec2NA_SA2_2023_Chung_Cheng_Yishun 2023
Q22

Construct a triangle such that AB = 8.8 cm, BC = 8.6 cm and CA = 13 cm. The line AB is shown below.

A horizontal line segment labeled with A on the left and B on the right, with the length marked as 8.8 cm below the line.
📊 Diagram: A horizontal line segment labeled with A on the left and B on the right, with the length marked as 8.8 cm below the line.
2 marks
Math_Sec2NA_SA2_2023_Chung_Cheng_Yishun 2023
Q23

Construct a triangle such that AB = 8.8 cm, BC = 8.6 cm and CA = 13 cm. The line AB is shown below.

A line segment AB is shown horizontally, measuring 8.8 cm. Point A is on the left and point B is on the right.
📊 Diagram: A line segment AB is shown horizontally, measuring 8.8 cm. Point A is on the left and point B is on the right.
2 marks
Math_Sec2NA_SA2_2023_Chung_Cheng_Yishun 2023
Q24

Measure and write down the angle opposite the longest side of the triangle.

A completed triangle ABC with AB = 8.8 cm (base), BC = 8.6 cm (right side), and CA = 13 cm (left side, the longest side). The angle opposite to CA (the longest side) is angle B.
📊 Diagram: A completed triangle ABC with AB = 8.8 cm (base), BC = 8.6 cm (right side), and CA = 13 cm (left side, the longest side). The angle opposite to CA (the longest side) is angle B.
1 mark
Math_Sec2NA_SA2_2023_Chung_Cheng_Yishun 2023
Q25

The triangle EFG has FG = 8 cm and angle GEF = 116°. The line EF has been drawn for you below.

A horizontal line segment with points labeled E on the left and F on the right, representing side EF of triangle EFG.
📊 Diagram: A horizontal line segment with points labeled E on the left and F on the right, representing side EF of triangle EFG.
Math_Sec2NA_SA2_2023_Springfield_Sec 2023
Q26

Construct and label the triangle EFG.

Diagram for question 13a
2 marks
Math_Sec2NA_SA2_2023_Springfield_Sec 2023
Q27

The diagram shows a rhombus ABCD. Angle DAB is 36°.

A rhombus ABCD with vertices labeled. A is at the bottom left, B is at the top, C is at the top right, and D is at the bottom right. The angle DAB at vertex A is marked as 36°. A diagonal BD is drawn inside the rhombus.
📊 Diagram: A rhombus ABCD with vertices labeled. A is at the bottom left, B is at the top, C is at the top right, and D is at the bottom right. The angle DAB at vertex A is marked as 36°. A diagonal BD is drawn inside the rhombus.
Math_Sec2NA_SA2_2023_Springfield_Sec 2023
Q28

Stating your reasons clearly, find angle BCD.

Diagram for question 14(a)(i)
1 mark
Math_Sec2NA_SA2_2023_Springfield_Sec 2023

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