# Chapter 4 - Constructions Of Triangles

# Mathematics Part II Solutions for Class 9 Math

Practice Set 4.3

**Question 1:**

**Construct ∆ PQR, in which**

**∠**

**Q = 70**

^{0},**∠**

**R = 80**

^{0}and PQ + QR + PR = 9.5 cm.**Answer:**

Steps of construction:

(1) Draw seg AB of 9.5 cm length.

(1) Draw seg AB of 9.5 cm length.

(2) Draw a ray making angle of 35

^{0}at point A.
(3) Draw another ray making an angle of 40

^{0}at point B.
(4) Name the point of intersection of the two rays as P.

(5) Draw the perpendicular bisector of seg AP and seg BP.

Name the points as Q and R respectively where the perpendicular bisectors intersect line AB.

(6) Draw seg PQ and seg PR.

Therefore, ∆ PQR is the required triangle.

Therefore, ∆ PQR is the required triangle.

**Question 2:**

**Construct ∆ XYZ, in which**

**∠**

**Y = 58**

^{0},**∠**

**X = 46**

^{0}and perimeter of triangle is 10.5 cm.

**Answer:**

Steps of construction:

(1) Draw seg PQ of 10.5 cm length.

(2) Draw a ray making angle of 29

^{0}at point P.
(3) Draw another ray making an angle of 23

^{0}at point Q.
(4) Name the point of intersection of the two rays as Z.

(5) Draw the perpendicular bisector of seg PZ and seg QZ.

Name the points as Y and X respectively where the perpendicular bisectors intersect line PQ.

Name the points as Y and X respectively where the perpendicular bisectors intersect line PQ.

(6) Draw seg XZ and seg YZ.

Therefore, ∆ is the required triangle.

Therefore, ∆ is the required triangle.

**Question 3:**

**Construct ∆LMN, in which**

**∠**

**M = 60**

^{0},**∠**

**N = 80**

^{0}and LM + MN + NL = 11 cm.

**Answer:**

Steps of construction:

(1) Draw seg AB of 11 cm length.

(2) Draw a ray making angle of 30

^{0}at point A.
(3) Draw another ray making an angle of 40

^{0}at point B.
(4) Name the point of intersection of the two rays as L.

(5) Draw the perpendicular bisector of seg AL and seg BL.

Name the points as M and N respectively where the perpendicular bisectors intersect line AB.

Name the points as M and N respectively where the perpendicular bisectors intersect line AB.

(6) Draw seg LM and seg LN.

Therefore, ∆ LMN is the required triangle.

Therefore, ∆ LMN is the required triangle.