Chapter 6: Triangles

Solution: –

Two rectangles are congruent, if they have same and length and breadth.

Solution:-

Two squares are congruent, if they have same side.

Solution: –

If ∆PQR and ∆XYZ are congruent under the correspondence QPR ↔ XYZ, then

(i) ∠R = ∠Z (ii) QR = XZ (iii) ∠P = ∠Y

(iv) QP = XY (v) ∠Q = ∠X (vi) RP = ZY

Solution: –

In Fig. 6.22, ∆PQR ≅ ∆XZY

From the figure, PQ = XZ = 3.5 cm

QR = ZY = 5cm

∠PQR = ∠XZY = 45^o

From SAS criterion, ∆PQR ≅ ∆XZY

Solution: –

In Fig. 6.23, ∆PQR ≅ ∆RSP

From the figure, PS = RQ = 4.1 cm

PR = PR … [common side for both triangles]

∠PRQ = ∠RPS = 45^o

From SAS criterion, ∆PQR ≅ ∆RSP

Solution:-

In Fig. 6.24, ∆DRQ ≅ ∆ PQR

From the figure,

QR = QR … [common side for both triangles]

∠PRQ = ∠DQR = 40^o

∠PQR = ∠DRQ = 70^o

From ASA criterion, ∆DRQ ≅ ∆ PQR

Solution:-

In Fig. 6.25, ∆ ARO ≅ ∆PQO

From the figure,

AO = PO = 2.5 cm

∠ARO = ∠PQO = 55^o

∠AOR = ∠POQ [vertically opposite angles]

From ASA criterion, ∆ARO ≅ ∆ PQO

Solution: –

In Fig. 6.26, AB = AD and ∠BAC = ∠ DAC.

Then (i) ∆ADC ≅ ∆ ABC.

(ii) BC = DC.

(iii) ∠BCA = ∠DCA.

(iv) Line segment AC bisects ∠BAD and ∠BCD.

Solution: –

In Fig. 6.27,

(i) ∠ TPQ = ∠ PQR + ∠ PRQ … [from exterior angle property]

(ii) ∠ UQR = ∠ QRP + ∠ QPR … [from exterior angle property]

(iii) ∠ PRS = ∠ RPQ + ∠ RQP … [from exterior angle property]

In questions 70 to 80, state whether the statements are True or False.

Solution: –

False

In a right angled triangle, sum of squares of two sides is equal to the square of the third side.