NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.2
Question 1:
Construct a triangle ABC in which BC = 7 cm, ∠B = 75° and AB + AC = 13 cm.
Solution 1:
The below given steps will be followed to construct the required triangle.
Step I: Draw a line segment BC of 7 cm. At point B, draw an angle of 75°, say ∠XBC.
Step II: Cut a line segment BD = 13 cm (that is equal to AB + AC) from the ray BX.
Step III: Join DC and make an angle DCY equal to ∠BDC.
Step IV: Let CY intersect BX at A.
Triangle ABC is the required triangle.
Question 2:
Construct a triangle ABC in which BC = 8 cm, ∠B = 45° and AB − AC = 3.5 cm.
Solution 2:
The below given steps will be followed to draw the required triangle.
Step I: Draw the line segment BC = 8 cm and at point B, make an angle of 45°, say ∠XBC.
Step II: Cut the line segment BD = 3.5 cm (equal to AB − AC) on ray BX.
Step III: Join DC and draw the perpendicular bisector PQ of DC.
Step IV: Let it intersect BX at point A. Join AC.
Triangle ABC is the required triangle.
Question 3:
Construct a triangle PQR in which QR = 6 cm, ∠Q = 60° and PR − PQ = 2 cm
Solution 3:
The below given steps will be followed to construct the required triangle.
Step I: Draw line segment QR of 6 cm. At point Q, draw an angle of 60°, say ∠XQR.
Step II: Cut a line segment QS of 2 cm from the line segment QT extended in the opposite side of line segment XQ. (As PR > PQ and PR − PQ = 2 cm). Join SR.
Step III: Draw perpendicular bisector AB of line segment SR. Let it intersect QX at point P. Join PQ, PR.
Triangle PQR is the required triangle.
Question 4:
Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.
Solution 4:
The below given steps will be followed to construct the required triangle.
Step I: Draw a line segment AB of 11 cm.
(As XY + YZ + ZX = 11 cm)
Step II: Construct an angle, ∠PAB, of 30° at point A and an angle, ∠QBA, of 90° at point B.
Step III: Bisect ∠PAB and ∠QBA. Let these bisectors intersect each other at point X.
Step IV: Draw perpendicular bisector ST of AX and UV of BX.
Step V: Let ST intersect AB at Y and UV intersect AB at Z.
Join XY, XZ.
Triangle XYZ is the required triangle.
Question 5:
Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm.
Solution 5:
The below given steps will be followed to construct the required triangle.
Step I: Draw line segment AB of 12 cm. Draw a ray AX making 90° with AB.
Step II: Cut a line segment AD of 18 cm (as the sum of the other two sides is 18) from ray AX.
Step III: Join DB and make an angle DBY equal to ADB.
Step IV: Let BY intersect AX at C. Join AC, BC.
Triangle ABC is the required triangle.
