NCERT solution class 9 chapter 11 Constructions exercise 11.2 mathematics

EXERCISE 11.2


Page No 195:

Question 1:

Construct a triangle ABC in which BC = 7 cm, ∠B = 75° and AB + AC = 13 cm.

Answer:

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. ΔABC is the required triangle.


Question 2:

Construct a triangle ABC in which BC = 8 cm, ∠B = 45° and AB − AC = 3.5 cm.

Answer:

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. ΔABC is the required triangle.


Question 3:

Construct a triangle PQR in which QR = 6 cm, ∠Q = 60° and PR − PQ = 2 cm

Answer:

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.

ΔPQR is the required triangle.


Question 4:

Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.

Answer:

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.

ΔXYZ 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.

Answer:

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.

ΔABC is the required triangle.


 

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