# NCERT Solutions for Chapter 11 Constructions Class 9 Maths

 Chapter Name NCERT Solutions for Chapter 11 Constructions Class Class 9 Topics Covered Some Basic ConstructionsConstruction of bisector of a line segmentConstruction of bisector of a given angleConstruction of Equilateral trianglesConstructions of Triangles Related Study Materials NCERT Solutions for Class 9 MathsNCERT Solutions for Class 9Revision Notes for Chapter 11 Constructions Class 9 MathsImportant Questions for Chapter 11 Constructions Class 9 MathsMCQ for for Chapter 11 Constructions Class 9 Maths

## Short Revision for Ch 11 Constructions Class 9 Maths

1. The bisector of an angle passes between the two arms of the angle.
2. Each point on the bisector of an angle is equidistant from the arms of the angle.
3. In the process of constructing the perpendicular bisector of a line segment, the radius of each arc must be greater than the half of the line segment.
4. Each point on the perpendicular bisector of a line segment is equidistant from the end points of the line segment.
5. To construct a triangle when the sum of the three sides and the two base angles are given, first we draw the line segment consisting the sum of the three sides and the two base angles.

### Exercise 11.1

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

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

3. Construct the angles of the following measurements :
(i) 30°

(iii) 15
°

4. Construct the following angles and verify by measuring them by a protractor:
(i) 75°

(ii) 105°

(iii) 135°

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

### Exercise 11.2

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

Solution

Steps of construction :
(i) Draw a line segment BC of length 7 cm.
(ii) Draw an angle, say XBC, of measure 75° at the end B of BC.

(iii) Cut a line segment BD = 13 cm from the ray BX.
(iv) Join DC and draw the perpendicular bisector of it to intersect BD at A.
(v) Join AC.
Then, △ABC is the required triangle.

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

Solution

Steps of construction :
(i) Draw a line segment BC of length 8 cm.
(ii) Make an angle, say XBC, of measure 45° at the end B of BC.
(iii) Cut a line segment BD of length 3.5 cm from the ray BX.
(iv) Join CD and draw the perpendicular bisector of it to intersect BX at A.
(v) Join AC.
Then, △ABC is the required triangle.

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

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

Solution

Steps of construction :
(i) Draw a line segment PQ of length 11 cm.
(ii) Draw angles of 30° at P and 90° at Q on the same side of PQ.
(iii) Bisect these angles. Let the bisectors intersect each other at a point X.
(iv) Draw perpendicular bisectors of PX and QX. Let these bisectors intersect the line segment PQ and Y and Z respectively.
(v) Join XY and XZ.
Then, △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.

Solution

Steps of construction:
(i) Draw a line segment BC of length 12 cm.
(ii) Make an angle BCD of 90° at C.
(iii) Cut a line segment CE of length 18 cm.
(iv) Join BE and draw the perpendicular bisector of it. Let the bisector intersect CE at A.
(v) Join AB.
Then, △ABC is the required triangle.