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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 👈
