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 = 450 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 = 450
(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= 600 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 = 600
(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 = 1000 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 = 1000
(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.
