Page 192 - ICSE Math 7

Solution: (a) In ∆ ABC and ∆ FED (b) In ∆ ABC and ∆ BAD
∠A = ∠F (each 80°) ∠CAB = ∠DBA (each 30°)

AB = FE (each 4.3 cm) AB = BA (common)
∠B = ∠E (each 40°) ∠ABC = ∠BAD (each 70°)
∴ ∆ ABC  ∆ FED ∴ ∆ ABC  ∆ BAD
(by ASA congruence criterion) (by ASA congruence criterion)

Example 9: A line AB is drawn through O, the point of S A R
intersection of the diagonals of a parallelogram
as shown in the given figure.
(a) Is ∆ ROA  ∆ POB? Give reasons. O
(b) Is OA = OB? Justify your answer. P B Q

Solution: (a) In ∆ ROA and ∆ POB
RO = PO (diagonals of a parallelogram bisect each other)
∠ARO = ∠BPO (alternate interior angles)
∠ROA = ∠POB (vertically opposite angles)

∆ ROA  ∆ POB (by ASA congruence criterion)
(b) Since the corresponding parts of congruent triangles are equal, therefore OA = OB.

Case IV: The RHS congruence criterion P
If the hypotenuse and one side of a right-angled A
triangle are equal to the corresponding hypotenuse
and one side of the other right-angled triangle, then 5 cm 5 cm
the triangles are congruent. 90° 90°
Draw ∆ ABC with ∠B = 90º, BC = 4 cm and B C Q R
AC = 5 cm. Draw another ∆ PQR with ∠Q = 90º, 4 cm 4 cm
QR = 4 cm and PR = 5 cm.
Now trace a copy of ∆ ABC and place it on ∆ PQR in such a way that BC falls on QR and ∠B falls
on ∠Q. In this process we shall observe that two triangles cover each other exactly. Therefore,
∆ ABC  ∆ PQR.
It is evident that two congruent figures have equal areas but conversely two figures having equal
area may or may not be congruent.

Example 10: In the given figures, measures of some parts of triangles are given. By applying RHS
congruence rule, state which pairs of triangles are congruent. In case of congruent
triangles, write the result in symbolic form.
B
C D
90° 4.1 cm

90° 90°
(a) 3 cm 3 cm (b) C A

A 5 cm B 90° 4.1 cm

D

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