NCERT TEXTBOOK QUESTIONS SOLVED
Construction Class 9 EXERCISE 11.2
Q1. Construct a triangle ABC in which BC = 7 cm, ∠B = 75° and AB + AC = 13 cm.
Sol. Steps of construction:
I. Draw a ray BX.
II. From 
, cut off 
= 7 cm.
, cut off = 7 cm. III. At B, construct ∠CBY = 75°.
VI. From 
, cut off BD = 13 cm (= AB + BC)
, cut off BD = 13 cm (= AB + BC) V. Join D and C.
VI. Bisect DC such that the bisector of DC meets BD at A.
VII. Join AC.
Thus, ΔABC is the required triangle.
Q2. Construct a triangle ABC in which BC = 8 cm, ∠B = 45° and AB – AC = 3.5 cm.
Sol. Steps of construction:
I. Draw a ray BX.
II . From 
, cut off 
= 8 cm.
, cut off = 8 cm. III. Construct ∠CBY = 45°.
IV. From 
, cut off 
= 3.5 cm.
, cut off = 3.5 cm. V. Join D and C.
VI. Draw PQ, perpendicular bisector of DC, which intersects 
at A.
at A. VII. Join AC.
Thus, ABC is the required triangle.
Q3. Construct a triangle PQR in which QR = 6 cm, ∠Q = 60° and PR – PQ = 2cm.
Sol. Steps of construction:
I . Draw a ray 
.
. II . From 
, cut off QR = 6 cm.
, cut off QR = 6 cm. III. Construct a line YQY’ such that ∠RQY = 60°.
IV. Cut off QS = 2 cm (from QY’).
V. Join ‘S’ and ‘R’.
VI. Draw MN, perpendicular bisector of SR, which intersects QY at P.
VII.Join P and R.
Thus, PQR is the required triangle.
Q4. Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.
Sol. Steps of construction:
I. Draw a line segment AB = 11 cm = (XY + YZ + ZX)
II. Construct ∠BAP = 30° = ∠Y
III. Construct ∠ABQ = 90° = ∠Z
IV. Draw 
, the bisector of ∠BAP.
, the bisector of ∠BAP. V. Draw 
, the bisector of ∠ABQ, such that 
and 
intersect each other at X.
, the bisector of ∠ABQ, such that and intersect each other at X. VI. Draw perpendicular bisector of AX, which intersects AB at Y.
VII.Draw perpendicular bisector of XB, which intersects AB at Z.
VIII. Join XY and XZ.
Thus, XYZ is the required triangle.
Class 9 EXERCISE 11.2
Q5. Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm.
Sol. Steps of construction:
I. Draw 
= 12 cm.
= 12 cm. II. Construct ∠CBY = 90°.
III. From 
, cut off BX = 18 cm.
, cut off BX = 18 cm. VI. Join CX.
V. Draw PQ, the perpendicular bisector of CX, such that PQ meets BX at A.
VI. Join AC.
Thus, ABC is the required triangle.