**Chapter 4 Constructions of Triangles Practice Set 4.2**

Chapter 4 Constructions of Triangles Practice Set 4.2

**Question 1.Construct ∆XYZ, such that YZ = 7.4 cm, ∠XYZ = 45° and XY – XZ = 2.7 cm.Solution:**

Here, XY – XZ = 2.7 cm

∴ XY > XZ

As shown in the rough figure draw seg YZ = 7.4 cm

Draw a ray YP making an angle of 45° with YZ

Take a point W on ray YP, such that

YW = 2.7 cm.

Now, XY – XW = YW [Y-W-X]

∴ XY – XW = 2.7 cm ….(i)

Also, XY – XZ = 2.7 cm ….(ii) [Given]

∴ XY – XW = XY – XZ [From (i) and (ii)]

∴ XW = XZ

∴ Point X is on the perpendicular bisector of seg ZW

∴ Point X is the intersection of ray YP and the perpendicular bisector seg ZW

Steps of construction:

i. Draw seg YZ of length 7.4 cm.

ii. Draw ray YP, such that ∠ZYP = 45°.

iii. Mark point W on ray YP such that l(YW) = 2.7 cm.

iv. Join points W and Z.

v. Join the points X and Z.

Hence, ∆XYZ is the required triangle.

**Question 2.Construct ∆PQR, such that QR = 6.5 cm, ∠PQR = 60° and PQ – PR = 2.5 cm.Solution:**

Here, PQ – PR = 2.5 cm

∴ PQ > PR

As shown in the rough figure draw seg QR = 6.5 cm

Draw a ray QT making on angle of 60° with QR

Take a point S on ray QT, such that QS = 2.5 cm.

Now, PQ – PS = QS [Q-S-T]

∴ PQ – PS = 2.5 cm ……(i) [Given]

Also, PQ – PR = 2.5 cm …..(ii) [From (i) and (ii)]

∴ PQ – PS = PQ – PR

∴ PS = PR

∴ Point P is on the perpendicular bisector of seg RS

∴ Point P is the intersection of ray QT and the perpendicular bisector of seg RS

Steps of construction:

i. Draw seg QR of length 6.5 cm.

ii. Draw ray QT, such that ∠RQT = 600.

iii. Mark point S on ray QT such that l(QS) = 2.5 cm.

iv. Join points S and R.

v. Draw perpendicular bisector of seg SR intersecting ray QT. Name the point as P.

vi. Join the points P and R.

Hence, ∆PQR is the required triangle.

**Question 3.Construct ∆ABC, such that BC = 6 cm, ∠ABC = 100° and AC – AB = 2.5 cm.Solution:**

Here, AC – AB = 2.5 cm

∴ AC > AB

As shown in the rough figure draw seg BC = 6 cm

Draw a ray BT making an angle of 100° with BC.

Take a point D on opposite ray of BT, :

such that BD 2.5 cm.

Now, AD – AB = BD [A-B-D]

∴ AD – AB = 2.5cm …..(i)

Also, AC – AB = 2.5 cm …..(ii) [Given]

∴ AD – AB = AC – AB [From (i) and (ii)]

∴ AD = AC

∴ Point A is on the perpendicular bisector of seg DC

∴ Point A is the intersection of ray BT and the perpendicular bisector of seg DC

Steps of construction:

i. Draw seg BC of length 6 cm.

ii. Draw ray BT, such that ∠CBT = 100°.

iii. Take point D on opposite ray of BT such that l(BD) = 2.5 cm.

iv. Join the points D and C.

v. Draw the perpendicular bisector of seg DC intersecting ray BT. Name the point as A.

vi. Join the points A and C.

Hence, ∆ABC is the required triangle.