Chapter 4 - Constructions Of Triangles Mathematics Part II Solutions for Class 9 Math Practice Set 4.3

Chapter 4 - Constructions Of Triangles

Mathematics Part II Solutions for Class 9 Math

Practice Set 4.3

Question 1:

Construct ∆ PQR,  in which ∠Q = 70⁰,  ∠R = 80⁰ and PQ + QR + PR = 9.5 cm.

Answer: 

Steps of construction:

(1) Draw seg AB of 9.5 cm length.

(2) Draw a ray making angle of 35⁰at point A.

(3) Draw another ray making an angle of 40⁰at 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. 

Question 2:

Construct ∆ XYZ, in which ∠Y = 58⁰, ∠X = 46⁰ 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 29⁰at point P.

(3) Draw another ray making an angle of 23⁰at 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.

(6) Draw seg XZ and seg YZ.

Therefore, ∆ is the required triangle. 

Question 3:

Construct ∆LMN, in which ∠M = 60⁰, ∠N = 80⁰  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 30⁰at point A.

(3) Draw another ray making an angle of 40⁰at 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.

(6) Draw seg LM and seg LN.

Therefore, ∆ LMN is the required triangle.