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Joy of Mathematics Class 8 Solutions Chapter 17 Circles
Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of Joy of Mathematics Class 8 Book, Chapter 17 Circles. Here students can easily find step by step solutions of all the problems for Circles. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Here students will find solutions for Exercise 17
Circles Exercise 17 Solution :
Question no – (1)
Solution :
The part of the circle along which the insect is crawling diameter.
Question no – (2)
Solution :
Here,
AC = 12 cm
AB = 5 cm
and, let the longest part of the right angled triangle is x
Now,
x2 = (AC)2 + (AB)2
=> x2 = (12)2 + (5)2
=> x = √144 + 25
= √169
= 13
∴ Thus the longest part of the triangle is 13.
Question no – (3)
Solution :
From the given figure,
AP = 3 cm
OA = 8/2
= 4 cm
∴ OP2 = (OA)2 + (AP)2
= 42 + 32
=> OP = √16 + 9
=> Op = √25
=> OP = 5
∴ Thus the required distance between O and P is 5 cm.
Question no – (4)
Solution :
Here,
the radius of the circle = 25cm
(a) OA is 14cm then A will lie in the interior of the circle.
(b) OB is 28 cm then B will lie in the exterior of the circle.
(c) OC is 25 cm, then C will lie in the circle.
Question no – (5)
Solution :
(a)
Here,
<ACB = 90°
=> y = 90°
and, <z = 60° (∵ <DPB = <CPA)
From △ADB
<ADB = 90°
and <DBA = 90/2
=> <x = 45°
Therefore the required angles are
x = 45°, y = 90°, z = 60°
(b)
As O is the center of the circle,
<ACB = 90°
From the △ACB
<ACB + <ABC + <BAC = 180
=> 90 + x + 39 = 180
=> x = 180 – 129
=> x = 51
Therefore the required angle is x = 51°
(c)
From the △ABC
<ABC = 90°
<BAC = 35°
So, <BCA = 180 – (90 + 35)
=> x = 180 – 125
=> x = 55
and, the △ADC, <ADC = 90°, <DAC = 55
<ACD = 180 – (90 + 55)
=> y = 180 – 145
=> y = 35
∴ Thus the required value of
x = 55°
y = 35°
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