Chapter 4 - Constructions Of Triangles Mathematics Part II Solutions for Class 9 Math Practice Set 4.2

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.

(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.

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.

(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.


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.

(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.

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