Page 190 - ICSE Math 7

Solution: (a) AC = PR (each 5 cm)
CB = PQ (each 4 cm)

BA ≠ QR
∴ ∆ ACB ∆ RPQ
(b) AB = AC (each 3.7 cm)

BD = CD (each 2.6 cm)
AD = AD (common)
∴ ∆ ABD ≅ ∆ ACD (by SSS congruence criterion)
Example 5: In the given figure, AB = AC and D is the midpoint of BC. A
(a) State the three pairs of equal parts in ∆ ADB and ∆ ADC.

(b) Is ∆ ADB ≅ ADC? Give reasons.
(c) Is ∠BAD = ∠CAD? Why?

Solution: (a) AB = AC (given)
B D C
BD = CD (  D is the midpoint of BC)
AD = AD (common)
(b) Yes, ∆ ADB  ∆ ADC (by SSS congruence criterion)

(c) Yes, ∠BAD = ∠CAD because ∆ ADB  ∆ ADC, clearly ∠BAD corresponds to
∠CAD.
Case II: Side-angle-side (SAS) condition
If the two sides and the included angle of one triangle are equal to the corresponding two sides and
the included angle of the other triangle, then two triangles are congruent.
Draw a ∆ ABC with AB = 6 cm, BC = 5 cm and ∠B = 45º. Draw another ∆ PQR with PQ = 6 cm,
QR = 5 cm and ∠Q = 45º.

Now put trace copy of ∆ ABC on ∆ PQR in such a way that AB falls on PQ, ∠B falls on ∠Q and
BC coincides with QR. In this process we shall observe that two triangles cover each other exactly.
Therefore, ∆ ABC  ∆ PQR.

Example 6: In the given figures, measures of some parts of the triangles are indicated. By applying
SAS congruence criterion, state the pairs of congruent triangles, if any, in each case.
In case of congruent triangles, write them in symbolic form.

D P P 3.8 cm Q
40°

(a) 4 cm 4 cm (b)

40° 40° 40°
E 3 cm F Q 3 cm R S 3.8 cm R
Solution: (a) In ∆ DFE and ∆ PQR
DF = PQ (each 4 cm)
∠ F = ∠Q (each 40°)

FE = QR (each 3 cm)
∴ ∆ DFE  ∆ PQR (by SAS congruence criterion)

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