Latest Elitmus Aptitude Question SOLUTION: if the angles of an n sided polygon are in A.P and a>=20 and d>=20 then wat is the maximum possible value of n? a)12 b)14 c)21 d)cant determine

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(#M40157507) ELITMUS Numerical Ability question geometry Keep an EYE

if the angles of an n sided polygon are in A.P and a>=20 and d>=20 then wat is the maximum possible value of n?
a)12
b)14
c)21
d)cant determine

Asked In Elitmus SHUBHAM (7 years ago)

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shrayashi dutta  (7 years ago)

Sum of Interior angles of a polygon = (n-2) × 180° where n = number of sides
There will be n angles which are in A.P. Therefore,
Since we need to find maximum value of n, put minimum value for a and d.i.e., take a=20 and d=20
Then sum of the angles
=n/2[2×20+(n−1)20]
Therefore,
10n(n+1)=(n−2)180
n=14.52
Since number of sides must be a positive integer, maximum value of n = 14

Is this solution Helpfull? Yes (7) | No (2)  

monika  (7 years ago)

Sum of Interior angles of a polygon = (n-2) × 180° where n = number of sides

There will be
n
n angles which are in A.P. Therefore,
Sum of the angles =
n
2
[
2
a
+
(
n
−
1
)
d
]
n2[2a+(n−1)d] where
a
≥
20
a≥20 and
d
≥
20
d≥20

Since we need to find maximum value of n, put minimum value for
a
a and
d
d.
i.e., take
a
=
20
a=20 and
d
=
20
d=20.

Then sum of the angles
=
n
2
[
2
×
20
+
(
n
−
1
)
20
]
=
20
n
2
[
2
+
(
n
−
1
)
]
=
10
n
(
n
+
1
)
=n2[2×20+(n−1)20]=20n2[2+(n−1)]=10n(n+1)

Therefore,
10
n
(
n
+
1
)
=
(
n
−
2
)
180
n
(
n
+
1
)
=
(
n
−
2
)
18
n
2
+
n
=
18
n
−
36
n
2
−
17
n
+
36
=
0
n
=
17
±
√
(
−
17
)
2
−
4
×
1
×
36
2
×
1
=
17
±
√
289
−
144
2
=
17
±
√
145
2
(∵takling maximum value)
≈
14.52
10n(n+1)=(n−2)180n(n+1)=(n−2)18n2+n=18n−36n2−17n+36=0n=17±(−17)2−4×1×362×1=17±289−1442=17±1452(∵takling maximum value)≈14.52

Since number of sides must be a positive integer, maximum value of n = 14

Is this solution Helpfull? Yes (1) | No (12)  

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