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.
(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.
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.
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 = 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.
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.
