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Congruence and Similarity Tests

Secondary 3 practice

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Congruence and Similarity Tests practice questions

Secondary 3 topics from the O-Level Mathematics 4052 syllabus. Every question below has a full worked solution.

When a student gets a question wrong in Math Amigo, it does not show the answer straight away. It reads the working, identifies the step that went wrong, and asks a question that points at the correction. The worked solution is there once a genuine attempt has been made.

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What this topic covers

Worked examples

Example 1

AOC and BOD are straight lines. OA = OC and OB = OD. △AOB ≡ △COD. ∠OAB = 42°42° and ∠AOB = 32°32°. Find ∠OCD.
42°32°AOB?CODNot drawn to scale
Show the worked solution
  • △AOB≡△CODgiven\triangle AOB \equiv \triangle COD \quad \scriptsize\textit{given}
  • ∠OCD corresponds to ∠OABA↔C, O↔O, B↔D in congruence statement\angle\text{OCD corresponds to }\angle\text{OAB} \quad \scriptsize\textit{A}{\leftrightarrow}\textit{C, O}{\leftrightarrow}\textit{O, B}{\leftrightarrow}\textit{D in congruence statement}
  • ∠OCD=∠OAB=42°\angle OCD = \angle OAB = 42°

Answer: 4242

Example 2

In △ABC\triangle ABC, DEDE is parallel to BCBC where DD is on ABAB and EE is on ACAC. AD=5AD = 5 cm, DB=10DB = 10 cm and DE=6DE = 6 cm. Find BCBC.
?DE6 cmABCNot drawn to scale
Show the worked solution
  • DEDE is parallel to BCBC, so △ADE\triangle ADE is similar to △ABC\triangle ABC (AA similarity).
  • AB=AD+DB=5+10=15AB = AD + DB = 5 + 10 = 15 cm
  • BCDE=ABADcorresponding sides in proportion\frac{BC}{DE} = \frac{AB}{AD} \quad \scriptsize\textit{corresponding sides in proportion}
  • BC6=155\frac{BC}{6} = \frac{15}{5}
  • BC=6×155BC = 6 \times \frac{15}{5}
  • =6×3= 6 \times 3
  • =18= 18 cm

Answer: 1818

Example 3

AOC and BOD are straight lines. OA = OC = 55 cm and OB = OD = 1212 cm. △AOB ≡ △COD (SAS) and AB = 1111 cm. Find the perimeter of △COD.
5 cm12 cm11 cmAOB5 cm12 cmCODNot drawn to scale
Show the worked solution
  • △AOB≡△CODgiven\triangle AOB \equiv \triangle COD \quad \scriptsize\textit{given}
  • OC corresponds to OA, OD corresponds to OB, CD corresponds to AB\text{OC corresponds to OA, OD corresponds to OB, CD corresponds to AB}
  • OC=OA=5\text{OC} = \text{OA} = 5 cm
  • OD=OB=12\text{OD} = \text{OB} = 12 cm
  • CD=AB=11\text{CD} = \text{AB} = 11 cm
  • Perimeter of △COD=OC+OD+CD\text{Perimeter of }\triangle COD = \text{OC} + \text{OD} + \text{CD}
  • =5+12+11= 5 + 12 + 11
  • =28= 28 cm

Answer: 2828

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Which syllabus does this follow?

O-Level Mathematics (4052), for Singapore secondary students. Math Amigo covers Secondary 1 to 4, both O-Level Mathematics and Additional Mathematics.

Will it just show my child the answer?

No. On a wrong answer it looks at the working, points to the step that went wrong, and asks a question that helps the student find the correction. The full worked solution is available after a genuine attempt.

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