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**Exercise 11.2****1. Construct a triangle ABC in which BC = 7cm, ∠B = 75° and AB + AC = 13 cm. Answer**

Steps of Construction:

Step 1: A line segment BC of 7 cm is drawn.

Step 2: At point B, an angle ∠XBC is constructed such that it is equal to 75°.

Step 3: A line segment BD = 13 cm is cut on BX (which is equal to AB+AC).

Step 3: DC is joined and ∠DCY = ∠BDC is made.

Step 4: Let CY intersect BX at A.

Thus, ΔABC is the required triangle. **2. Construct a triangle ABC in which BC = 8cm, ∠B = 45° and AB - AC = 3.5 cm. Answer**

Steps of Construction:

Step 1: A line segment BC = 8 cm is drawn and at point B, make an angle of 45° i.e. ∠XBC.

Step 2: Cut the line segment BD = 3.5 cm (equal to AB - AC) on ray BX.

Step 3: Join DC and draw the perpendicular bisector PQ of DC.

Step 4: Let it intersect BX at point A. Join AC.

Thus, ΔABC is the required triangle.**3. Construct a triangle PQR in which QR = 6cm, ∠Q = 60° and PR – PQ = 2cm. Answer**

Steps of Construction:

Step 1: A ray QX is drawn and cut off a line segment QR = 6 cm from it.

Step 2:. A ray QY is constructed making an angle of 60º with QR and YQ is produced to form a line YQY'

Step 3: Cut off a line segment QS = 2cm from QY'. RS is joined.

Step 5: Draw perpendicular bisector of RS intersecting QY at a point P. PR is joined.

Thus, ΔPQR is the required triangle.**4. Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm. Answer**

Steps of Construction:

Step 1: A line segment PQ = 11 cm is drawn. (XY + YZ + ZX = 11 cm)

Step 2: An angle, ∠RPQ = 30° is constructed at point A and an angle ∠SQP = 90° at point B.

Step 3: ∠RPQ and ∠SQP are bisected . The bisectors of these angles intersect each other at point X.

Step 4: Perpendicular bisectors TU of PX and WV of QX are constructed.

Step V: Let TU intersect PQ at Y and WV intersect PQ at Z. XY and XZ are joined.

Thus, ΔXYZ is the required triangle.**5. Construct a right triangle whose base is 12cm and sum of its hypotenuse and other side is 18 cm.****Answer**

Steps of Construction:

Step 1: A ray BX is drawn and a cut off a line segment BC = 12 cm is made on it.

Step 2: ∠XBY = 90° is constructed.

Step 3: Cut off a line segment BD = 18 cm is made on BY. CD is joined.

Step 4: Perpendicular bisector of CD is constructed intersecting BD at A. AC is joined.

Thus, ΔABC is the required triangle.

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