Examsnet
Unconfined exams practice
Home
Exams
Banking Entrance Exams
CUET Exam Papers
Defence Exams
Engineering Exams
Finance Entrance Exams
GATE Exam Practice
Insurance Exams
International Exams
JEE Exams
LAW Entrance Exams
MBA Entrance Exams
MCA Entrance Exams
Medical Entrance Exams
Other Entrance Exams
Police Exams
Public Service Commission (PSC)
RRB Entrance Exams
SSC Exams
State Govt Exams
Subjectwise Practice
Teacher Exams
SET Exams(State Eligibility Test)
UPSC Entrance Exams
Aptitude
Algebra and Higher Mathematics
Arithmetic
Commercial Mathematics
Data Based Mathematics
Geometry and Mensuration
Number System and Numeracy
Problem Solving
Board Exams
Andhra
Bihar
CBSE
Gujarat
Haryana
ICSE
Jammu and Kashmir
Karnataka
Kerala
Madhya Pradesh
Maharashtra
Odisha
Tamil Nadu
Telangana
Uttar Pradesh
English
Competitive English
Certifications
Technical
Cloud Tech Certifications
Security Tech Certifications
Management
IT Infrastructure
More
About
Careers
Contact Us
Our Apps
Privacy
Test Index
CDS Model Paper 4 Mathematics
Show Para
Hide Para
Share question:
© examsnet.com
Question : 51
Total: 100
A wire is bent into the form of a circle, whose area is 154
cm
2
. If the same wire is bent into the form of an equilateral triangle, the approximate area of the equilateral triangle is
93.14
cm
2
90.14
cm
2
83.14
cm
2
39.14
cm
2
Validate
Solution:
Let
r
be the radius of circle.
π
r
2
=
154
cm
2
r
2
=
154
22
×
7
=
49
r
=
7
cm
length of wire
=
circumference of circle
=
2
×
22
7
×
7
=
44
cm
Now, Perimeter of equilateral triangle
=
44
cm
side
=
44
3
cm
Area of equilateral triangle
=
√
3
4
×
(
44
3
)
2
=
484
√
3
9
=
91.42
cm
2
Area of equilateral triangle is nearly equal to
90.14
cm
2
Hence, option (b) is correct.
© examsnet.com
Go to Question:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
Prev Question
Next Question