#### NCERT Solutions for Class 9th: Ch 11 Constructions Maths

**Exercise 11.1**

1. Construct an angle of 90Â° at the initial point of a given ray and justify the construction.

**Answer**

Step 1: A ray YZ is drawn.

Step 2: With Y as a centre and any radius, an arc ABC is drawn cutting YZ at C.

Step 3: With C as a centre and the same radius, mark a point B on the arc ABC.

Step 4: With B as a centre and the same radius, mark a point A on the arc ABC.

Step 5: With A and B as centre, draw two arcs intersecting each other with the same radius at X.

Step 6: X and Y are joined and a ray XY making an angle 90Â° with YZ is formed.

Justification for construction:

We constructed âˆ BYZ = 60Â° and also âˆ AYB = 60Â°.

Thus, âˆ AYZ = 120Â°.

Also, bisector of âˆ AYB is constructed such that:

âˆ AYB = âˆ XYA

â‡’ âˆ XYB = 1/2âˆ AYB

â‡’ âˆ XYB = 1/2Ã—60Â°

â‡’ âˆ XYB = 30Â°

Now,

âˆ XYZ = âˆ BYZ + âˆ XYB = 60Â° + 30Â° = 90Â°

2. Construct an angle of 45Â° at the initial point of a given ray and justify the construction.

**Answer**

Step 1: A ray OY is drawn.

Step 2: With O as a centre and any radius, an arc ABC is drawn cutting OY at A.

Step 3: With A as a centre and the same radius, mark a point B on the arc ABC.

Step 4: With B as a centre and the same radius, mark a point C on the arc ABC.

Step 5: With A and B as centre, draw two arcs intersecting each other with the same radius at X.

Step 6: X and Y are joined and a ray making an angle 90Â° with YZ is formed.

Step 7: With A and E as centres, two arcs are marked intersecting each other at D and the bisector of âˆ XOY is drawn.

Justification for construction:

By construction,

âˆ XOY = 90Â°

We constructed the bisector of âˆ XOY as DOY.

Thus,

âˆ DOY = 1/2 âˆ XOY

âˆ DOY = 1/2Ã—90Â° = 45Â°

3. Construct the angles of the following measurements:

(i) 30Â° (ii) 22.5Â° (iii) 15Â°

**Answer**

(i) 30Â°

Step 1: A ray OY is drawn.

Step 2: With O as a centre and any radius, an arc AB is drawn cutting OY at A.

Step 3: With A and B as centres, two arcs are marked intersecting each other at X and the bisector of is drawn.

Thus, âˆ XOY is the required angle making 30Â° with OY.

(ii) 22.5Â°

Step 1: An angle âˆ XOY = 90Â° is drawn.

Step 2: Bisector of âˆ XOY is drawn such that âˆ BOY = 45Â° is constructed.

Step 3: Again, âˆ BOY is bisected such that âˆ AOY is formed.

Thus, âˆ AOY is the required angle making 22.5Â° with OY.

(iii) 15Â°

Step 1: An angle âˆ AOY = 60Â° is drawn.

Step 2: Bisector of âˆ AOY is drawn such that âˆ BOY = 30Â° is constructed.

Step 3: With C and D as centres, two arcs are marked intersecting each other at X and the bisector of âˆ BOY is drawn.

Thus, âˆ XOY is the required angle making 15Â° with OY.

4. Construct the following angles and verify by measuring them by a protractor:

(i) 75Â° (ii) 105Â° (iii) 135Â°

**Answer**

(i) 75Â°

Steps of constructions:

Step 1: A ray OY is drawn.Step 2: An arc BAE is drawn with O as a centre.

Step 3: With E as a centre, two arcs are A and C are made on the arc BAE.

Step 4: With A and B as centres, arcs are made to intersect at X and âˆ XOY = 90Â° is made.

Step 5: With A and C as centres, arcs are made to intersect at D

Step 6: OD is joined and and âˆ DOY = 75Â° is constructed.

Thus, âˆ DOY is the required angle making 75Â° with OY.

(ii) 105Â°

Steps of constructions:

Step 1: A ray OY is drawn.Step 2: An arc ABC is drawn with O as a centre.

Step 3: With A as a centre, two arcs are B and C are made on the arc ABC.

Step 4: With B and C as centres, arcs are made to intersect at E and âˆ EOY = 90Â° is made.

Step 5: With B and C as centres, arcs are made to intersect at X

Step 6: OX is joined and and âˆ XOY = 105Â° is constructed.

Thus, âˆ XOY is the required angle making 105Â° with OY.

(iii) 135Â°

Step 2: An arc ACD is drawn with O as a centre.

Step 3: With A as a centre, two arcs are B and C are made on the arc ACD.

Step 4: With B and C as centres, arcs are made to intersect at E and âˆ EOY = 90Â° is made.

Step 5: With F and D as centres, arcs are made to intersect at X or bisector of âˆ EOD is constructed.

Step 6: OX is joined and and âˆ XOY = 135Â° is constructed.

Thus, âˆ XOY is the required angle making 135Â° with DY.

5. Construct an equilateral triangle, given its side and justify the construction.

**Answer**

Step 1: A line segment AB=4 cm is drawn.

Step 2: With A and B as centres, two arcs are made.

Step 4: With D and E as centres, arcs are made to cut the previous arc respectively and forming angle of 60Â° each.

Step 5: Lines from A and B are extended to meet each other at C.

Thus, ABC is the required triangle formed.

Justification:

By construction,

AB = 4 cm, âˆ A = 60Â° and âˆ B = 60Â°

We know that,

âˆ A

â‡’ 60Â° + 60Â° + âˆ C = 180Â°

â‡’ 120Â° + âˆ C = 180Â°

â‡’ âˆ C = 60Â°

BC = CA = 4 cm (Sides opposite to equal angles are equal)

AB = BC = CA = 4 cm

âˆ A = âˆ B = âˆ C = 60Â°

Page No: 195

**Exercise 11.2**

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

**Answer**

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

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

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

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

Step 4: Let CY intersect BX at A.

Thus, Î”ABC is the required triangle.

2. Construct a triangle ABC in which BC = 8cm, âˆ B = 45Â° 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 45Â° i.e. âˆ 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, Î”ABC is the required triangle.

3. Construct a triangle PQR in which QR = 6cm, âˆ Q = 60Â° and PR â€“ PQ = 2cm.

**Answer**

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 60Âº 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, Î”PQR is the required triangle.

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

**Answer**

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

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

Step 3: âˆ RPQ and âˆ 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, Î”XYZ 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: âˆ XBY = 90Â° 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, Î”ABC is the required triangle.

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