The DAV Maths Class 7 Solutions and **DAV Class 7 Maths Chapter 9 Worksheet 3** Solutions of Construction of Triangles offer comprehensive answers to textbook questions.

## DAV Class 7 Maths Ch 9 WS 3 Solutions

Question 1.

In the given figure, ∠P = ∠Y = 40° and ∠Q = ∠Z = 60°. Find the third pair of corresponding parts which make ∆PQR ≅ ∆YZX, by ASA congruence condition.

Answer:

∠P = ∠Y = 40° (given)

∠Q = ∠Z = 60° (given)

And Third pair of corresponding part is

PQ = ZY

∆PQR ≅ ∆YZX (By SAS)

Question 2.

Which pairs of triangles are congruent by ASA congruence condition in the given figure? If congruent, write the congruence of the two triangles in symbolic form.

Answer:

In ∆ABC and ∆ ZXY,

∠C = ∠Y = 60° (given)

∠A = ∠Z = 40° (given)

AC = YZ = 3.5 cm (given)

∆ABC ≅ ∆ZXY (By ASA)

Answer:

In ∆ABO and ∆CDO,

∠ABO = ∠CDO = 80°

BO = OD = 4 cm

∠AOB = ∠COD

∆ABO ≅ ∆CDO

Answer:

In ∆ABC and ∆DCB,

∠ABC = ∠DCB

= 40° + 20°

= 60°

∠ACB = ∠DBC = 40°

BC = BC

∆ABC ≅ ∆ DCB

Answer:

In ∆ABC and ∆AED,

∠B = ∠E = 180° – (40° + 60°)

= 80°

∠C = ∠D = 60°

BC = DE = 4 cm

∆ ABC ≅ ∆ AED

Answer:

In ∆ABC and ∆ XZY,

No side of ∆ABC is given equal to the corresponding

Hence they are not congruent by ASA

Question 3.

In the given figure, QX bisects ∠PQR as well as ∠PSR.

State the three facts needed to ensure that ∆QRS ≅ ∆QPS.

Give reasons for each statement.

Answer:

In ∆QRS and ∆QPS,

∠PQS = ∠RQS (given)

∠PSQ = ∠RSQ (given)

QS = QS

∆QRS ≅ ∆QPS

Question 4.

In the given figure, PS bisects ∠P and PS ± QR.

(i) Is ∆PSQ ≅ ∆PSR?

Answer:

In ∆PQS and ∆PRS,

∠QPS = ∠RPS (given)

∠PSQ = ∠PSR = 90° (given)

PS = PS (Common)

∆PSQ ≅ ∆PSR (ASA)

(ii) State the three pairs of matching parts you have used in (i)

Answer:

∠QPS and ∠RPS, ∠PSQ and ∠PSR, PS and PS

(iii) Is it true to say that QS = SR? Why?

Answer:

Yes, QS = SR [Corresponding parts of congruent triangles]

Question 5.

In the given figure, AO = BO and ∠A = ∠B

(i) Is ∠AOC = ∠BOD ? Why?

(ii) Is ∆AOC ≅ ∆BOD by ASA congruence condition ?

(iii) State the three facts you have used to answer (ii)

(iv) Is ∠ACO = ∠BDO? Why?

Answer:

In ∆AOC and ∆BOD

AO = BO (given)

∠A= ∠B (given)

(i) ∠AOC = ∠BOD

(ii) ∴ ∆ AOC ≅ ∆ BOD

(iii) AO and BO, ∠A and ∠B, ∠AOC and ∠BOD

(iv) ∠ACO = ∠BDO (Corresponding parts of congruent triangles)