Triangles (Congruency) Class 9 ICSE Important Questions with Answers

Answer: (a) AAA.

AAA guarantees only similarity, not congruence, since the triangles may differ in size. SSS, SAS and RHS are all valid congruence conditions.

Answer: (a) SAS.

Two sides ($AB,BC$AB,BC) and the included angle ($\angle B$∠ B) equal the corresponding parts, so the triangles are congruent by SAS.

Answer: (a) $AC=PR$AC=PR.

The order of the letters gives the correspondence $A\leftrightarrow P$A↔ P, $B\leftrightarrow Q$B↔ Q, $C\leftrightarrow R$C↔ R, so $AC$AC corresponds to $PR$PR. The other options pair parts that do not correspond.

Answer: (a) RHS.

$\angle ADB=\angle ADC=90^\circ$∠ ADB=∠ ADC=90°, hypotenuse $AB=AC$AB=AC, and side $AD=AD$AD=AD is common. Hence $\triangle ABD\cong\triangle ACD$△ ABD≅ △ ACD by RHS.

Answer: (a) Both are true and R correctly explains A. By the angle sum property $\angle C=\angle F$∠ C=∠ F, so $BC$BC and $EF$EF become sides included between equal angles ($\angle B,\angle C$∠ B,∠ C and $\angle E,\angle F$∠ E,∠ F). That is why two angles and a non-included side (AAS) are enough for congruence.

Here two angles and the included side are equal, so by the ASA condition

$\triangle ABC\cong\triangle DEF\ \text{(ASA)}.$△ ABC≅ △ DEF (ASA).

The correspondence $A\leftrightarrow D,\ B\leftrightarrow E,\ C\leftrightarrow F$A↔ D, B↔ E, C↔ F is maintained.

The triangles are congruent by the RHS rule, since a right angle, equal hypotenuses and one equal side match:

$\triangle XYZ\cong\triangle LMN\ \text{(RHS)}.$△ XYZ≅ △ LMN (RHS).

By CPCTC, $\angle Y=\angle M$∠ Y=∠ M (a pair of corresponding equal angles).

In $\triangle ABD$△ ABD and $\triangle CDB$△ CDB:

$AB=CD\quad\text{(given)}$AB=CD (given)
$AD=CB\quad\text{(given)}$AD=CB (given)
$BD=DB\quad\text{(common side)}$BD=DB (common side)

All three pairs of sides are equal, so by the SSS congruence condition

$\triangle ABD\cong\triangle CDB.$△ ABD≅ △ CDB.

Hence proved.

In $\triangle ABD$△ ABD and $\triangle ACD$△ ACD:

$\angle BAD=\angle CAD\quad\text{(}AD\text{ bisects }\angle A\text{)}$∠ BAD=∠ CAD (AD bisects ∠ A)
$AD=AD\quad\text{(common side)}$AD=AD (common side)
$\angle ADB=\angle ADC=90^\circ\quad\text{(}AD\perp BC\text{)}$∠ ADB=∠ ADC=90° (AD⊥ BC)

By the ASA condition, $\triangle ABD\cong\triangle ACD$△ ABD≅ △ ACD.

Hence by CPCTC, $AB=AC$AB=AC. Proved.

In $\triangle AOB$△ AOB and $\triangle DOC$△ DOC:

$\angle OAB=\angle ODC\quad\text{(alternate angles, }AB\parallel CD\text{, transversal }AD\text{)}$∠ OAB=∠ ODC (alternate angles, AB∥ CD, transversal AD)
$\angle AOB=\angle DOC\quad\text{(vertically opposite angles)}$∠ AOB=∠ DOC (vertically opposite angles)
$OB=OC\quad\text{(given)}$OB=OC (given)

$OB$OB is not the side between the two equal angles, so by the AAS condition $\triangle AOB\cong\triangle DOC$△ AOB≅ △ DOC.

Hence by CPCTC, $AO=DO$AO=DO, so $O$O is the mid-point of $AD$AD. Proved.

Since $M$M is the midpoint of $BC$BC,

$BM=CM.\quad(1)$BM=CM. (1)

In right triangles $\triangle ABM$△ ABM and $\triangle DCM$△ DCM:

$\angle ABM=\angle DCM=90^\circ\quad\text{(given)}$∠ ABM=∠ DCM=90° (given)
$AM=DM\quad\text{(given, these are the hypotenuses)}$AM=DM (given, these are the hypotenuses)
$BM=CM\quad\text{[from (1)]}$BM=CM [from (1)]

A right angle, equal hypotenuses and one equal side match, so by the RHS condition

$\triangle ABM\cong\triangle DCM.$△ ABM≅ △ DCM.

Hence by CPCTC, $AB=DC$AB=DC. Proved.

(Also $\angle BAM=\angle CDM$∠ BAM=∠ CDM and $AM=DM$AM=DM as corresponding parts.)

In $\triangle ABC$△ ABC and $\triangle ADC$△ ADC:

$\angle BAC=\angle DAC\quad\text{(}AC\text{ bisects }\angle A\text{)}$∠ BAC=∠ DAC (AC bisects ∠ A)
$AC=AC\quad\text{(common side)}$AC=AC (common side)
$\angle BCA=\angle DCA\quad\text{(}AC\text{ bisects }\angle C\text{)}$∠ BCA=∠ DCA (AC bisects ∠ C)

Two angles and the included side $AC$AC are equal, so by the ASA condition

$\triangle ABC\cong\triangle ADC.$△ ABC≅ △ ADC.

Hence by CPCTC:

$AB=AD\qquad\text{and}\qquad CB=CD.$AB=AD and CB=CD.

Proved. (This also shows $ABCD$ABCD is a kite about diagonal $AC$AC.)

(i) He should compare $\triangle OPQ$△ OPQ and $\triangle ORS$△ ORS.

(ii) The SAS condition applies.

(iii) In $\triangle OPQ$△ OPQ and $\triangle ORS$△ ORS:

$OP=OR\quad\text{(constructed equal)}$OP=OR (constructed equal)
$\angle POQ=\angle ROS\quad\text{(vertically opposite angles)}$∠ POQ=∠ ROS (vertically opposite angles)
$OQ=OS\quad\text{(constructed equal)}$OQ=OS (constructed equal)

By SAS, $\triangle OPQ\cong\triangle ORS$△ OPQ≅ △ ORS.

(iv) By CPCTC, corresponding sides are equal, so $PQ=RS$PQ=RS. The surveyor therefore measures $RS$RS on dry land to obtain the pond width $PQ$PQ.