# Chapter 4 - Constructions Of Triangles

# Mathematics Part II Solutions for Class 9 Math

Practice Set 4.2

**Question 1:**

Construct ∆ XYZ, such that YZ = 7.4 cm, ∠XYZ = 45

^{0}and XY − XZ = 2.7 cm.**ANSWER:**

Steps of construction:

(1) Draw seg YZ of length 7.4 cm.

(1) Draw seg YZ of length 7.4 cm.

(2) Draw ray YP such that ∠ZYP = 45

^{0}
(3) Take point Q on ray YP such that YQ = 2.7 cm.

(4) Construct the perpendicular bisector of seg QZ.

(5) Name the point of intersection of ray YP and the perpendicular bisector of seg QZ as X.

(6) Draw seg XZ.

Therefore, ∆ XYZ is required triangle.

Therefore, ∆ XYZ is required triangle.

**Question 2:**

Construct ∆ PQR, such that QR = 7.4 cm, ∠PQR= 60

^{0}and PQ − PR = 2.5 cm.**ANSWER:**

Steps of construction:

(1) Draw seg QR of length 6.5 cm.

(1) Draw seg QR of length 6.5 cm.

(2) Draw ray QA such that ∠RQA = 60

^{0}
(3) Take point B on ray QA such that QB = 2.5 cm.

(4) Construct the perpendicular bisector of seg BR.

(5) Name the point of intersection of ray QA and the perpendicular bisector of seg BR as P.

(6) Draw seg PR.

Therefore, ∆ PQR is required triangle.

Therefore, ∆ PQR is required triangle.

**Question 3:**

Construct ∆ ABC, such that BC = 6 cm, ∠ABC = 100

^{0}and AC− AB = 2.5cm.**ANSWER:**

Steps of construction:

(1) Draw seg BC of length 6 cm.

(1) Draw seg BC of length 6 cm.

(2) Draw ray BE such that ∠CBE = 100

^{0}
(3) Take point D on the opposite of ray BE such that BD = 2.5 cm.

(4) Construct perpendicular bisector of seg DC.

(5) Name the point of intersection of ray BE and the perpendicular bisector of DC as A.

(6) Draw seg AC.

Therefore, ∆ ABC is required triangle.

Therefore, ∆ ABC is required triangle.