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**Joy of Mathematics Class 6 Solutions Chapter 16 Constructions**

Welcome to NCTB Solution. Here with this post we are going to help 6th class students for the Solutions of Joy of Mathematics Class 6 Book, Chapter 16 Constructions. Here students can easily find step by step solutions of all the problems for Constructions. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Students will get chapter 16 solutions. Here students will find exercise wise solutions for chapter 16 Exercise 16.1, 16.2, 16.3 and 16.4

**Constructions Exercise 16.2 Solution :**

**Question no – (1)**

**Solution :**

**(a) 30°**

**∴** ∠DBC = 30°

**(e)** 150°

**∴** ∠AOB = 150°

**(f)** 22 1/2°

**Question no – (2)**

**Solution :**

**(a)** 82°

**Question no – (3)**

**Solution :**

**∴** ∠AOB = 102° and, ∠COB = 51°

**Question no – (4)**

**Solution :**

120°

**Question no – (5)**

**Solution :**

90°

**∴** ∠AOB = 90° & ∠COB = 45°

**Question no – (6)**

**Solution :**

135°

**Question no – (7)**

**Solution :**

**Question no – (8)**

**Solution :**

**∴** ∠AOB = 88°

∠COB = 44°

**Question no – (9)**

**Solution :**

∠QPB = 120°

**Question no – (10)**

**Solution :**

**∴ **∠AXY = 90°

and, ∠ZXY = 45°

**Constructions Exercise 16.3 Solution :**

**Question no – (1)**

**Solution :**

**∴** ABC is required triangle.

**Question no – (2)**

**Solution :**

**∴** PQR is required triangle.

Whose, PQ = 4.4cm

QR = 5.3cm

PR = 3.5cm

**Question no – (3)**

**Solution :**

**∴** BAT is a required triangle, whose,

∠A = 45°, AT = 4.7cm, BA = 6.5cm.

**Question no – (4)**

**Solution :**

**∴** XYZ is required triangle.

**Question no – (5)**

**Solution :**

**Question no – (6)**

**Solution :**

**∴** ABC is required triangle.

**Question no – (7)**

**Solution :**

SAB is required triangle.

**Question no – (8)**

**Solution :**

**∴** LMN is required triangle.

**∴** ∠L = 90°, LN = 7.5cm, LM = 5.2cm.

**Question no – (9)**

**Solution :**

**∴** XYZ is a required triangle.

**Question no – (10)**

**Solution :**

**∴** PQR is a required triangle.

**Constructions Exercise 16.4 Solution :**

**Question no – (1)**

**Solution :**

Here, PQ = 3.7 cm

QL tangent = 4cm

**Question no – (2)**

**Solution :**

Diameter = 5cm

Radius = 5/2

= 2.5cm

**Question no – (3)**

**Solution :**

XY = 8.2 cm

OY = 4.1cm (Radius)

and, MN = perpendicular bisector.

**Question no – (4)**

**Solution :**

Diameter = 7.2cm

Radius = 7.2/2

= 3.6cm

AB = 7.2cm

OB = 3.6cm

ML perpendicular bisector.

**Question no – (5)**

**Solution :**

Here, ∠P = ∠Q = 60°

PQ = 3cm

**Question no – (6)**

**Solution :**

ABC is an equilateral triangle whose side AB = BC = AC = 5cm

**Question no – (7) **

**Solution :**

**(a) Circumcircle : **

This is the circumcircle of triangle XYZ

**(b) Incircle : **

**∴ **This is the incircle of triangle XYZ

**Next Chapter Solution : **

👉 Chapter 17 👈