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11 Jun 2021, 08:36
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
54% (02:42) correct
46%
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wrong
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The interior angles of a convex polygon are in arithmetic progression. The smallest angle is \(120^{\circ}\) and the common difference is \(5^{\circ}\). Find the number of sides of the polygon if the polygon has more than 10 sides.
A. 15
B. 16
C. 17
D. 18
E. 20
A. 15
B. 16
C. 17
D. 18
E. 20
11 Jun 2021, 09:26
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Solution:
We know sum of interior angle for any polygon: \((n-2)180^o\)
It is given that the smallest angle is 120^o the common difference is 5^o.
So, the sum of angles will be: \(\frac{n}{2}[2a+(n-1)d] = \frac{n}{2}(2 * 120 + (n-1)5)\)
Thus, \((n-2)180^o = \frac{n}{2}(2 * 120 + (n-1)5)\)
\(⇒ 240n+5n^2-5n=360n-720\)
\(⇒ n^2-25n+144=0\)
\(⇒ n^2-16n-9n+144=0\)
\(⇒ (n-16)(n-9)=0\)
Thus the value of n can be 16 or 9.
But the question says that polygon has more than 10 sides, therefore \(n=16\)
Hence the right answer is Option B.
We know sum of interior angle for any polygon: \((n-2)180^o\)
It is given that the smallest angle is 120^o the common difference is 5^o.
So, the sum of angles will be: \(\frac{n}{2}[2a+(n-1)d] = \frac{n}{2}(2 * 120 + (n-1)5)\)
Thus, \((n-2)180^o = \frac{n}{2}(2 * 120 + (n-1)5)\)
\(⇒ 240n+5n^2-5n=360n-720\)
\(⇒ n^2-25n+144=0\)
\(⇒ n^2-16n-9n+144=0\)
\(⇒ (n-16)(n-9)=0\)
Thus the value of n can be 16 or 9.
But the question says that polygon has more than 10 sides, therefore \(n=16\)
Hence the right answer is Option B.
18 Jun 2021, 00:41
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A quicker way to use the answer choices and the evenly spaced set of numbers to your benefit is to list out the first 11 terms in the arithmetic progression:
120–125—130—135—140–145—150–155–160–165—170
Rule: for an evenly spaced set the Sum of the values is equal to = (Median) * (Count of Terms)
Rule: the SUM of interior angles for any N sided polygon is = 180 * (N - 2)
Where N= a positive integer number of sides that must be 3 or greater
Therefore, whatever the SUM of the interior angles is, the following equation must be satisfied:
(Median of Angle measures) * (Count of Angles/Sides = N) = 180 * (N -2)
or
(Median) * (N) = 180 * (N - 2)
Since the SUM of the interior angles will be equal to a positive integer, this SUM must be divisible by 180 ———> and from the factor foundation rule, this means the SUM must be divisible by the factors of 180: such as 9, 4, and 5
(E) 20 sides
Median = (10th Value + 11th Value)
N = 20
SUM = (165 + 170)/2 * 20
= 335 * 10
This sum is not divisible by 9, thus it can not be divisible by 180
Eliminate E
(D) 18 sides
(160 + 165)/2 * 18
325 * 9
The SUM is not divisible by 4, this it can not be divisible by 180
(C) 17
(160) * 17
Since the sum is not divisible by 9, it can not be divisible by 180
(B) 16
(155 + 160)/2 * 16
315 * 8 ——-> prime factorization = (2)^3 * (3)^2 * (5) * (7)
Prime Factorization of 180 = (2)^2 * (3) * (5)^2
Since the exponents are equal to or greater than the exponents of 180’s Prime Factorization, the SUM of the interior Angles of 16 sides given this Arithmetic Progression will be a valid sum
(B) 16
*note* you can check that (A) doesn’t work in the same way
Posted from my mobile device
120–125—130—135—140–145—150–155–160–165—170
Rule: for an evenly spaced set the Sum of the values is equal to = (Median) * (Count of Terms)
Rule: the SUM of interior angles for any N sided polygon is = 180 * (N - 2)
Where N= a positive integer number of sides that must be 3 or greater
Therefore, whatever the SUM of the interior angles is, the following equation must be satisfied:
(Median of Angle measures) * (Count of Angles/Sides = N) = 180 * (N -2)
or
(Median) * (N) = 180 * (N - 2)
Since the SUM of the interior angles will be equal to a positive integer, this SUM must be divisible by 180 ———> and from the factor foundation rule, this means the SUM must be divisible by the factors of 180: such as 9, 4, and 5
(E) 20 sides
Median = (10th Value + 11th Value)
N = 20
SUM = (165 + 170)/2 * 20
= 335 * 10
This sum is not divisible by 9, thus it can not be divisible by 180
Eliminate E
(D) 18 sides
(160 + 165)/2 * 18
325 * 9
The SUM is not divisible by 4, this it can not be divisible by 180
(C) 17
(160) * 17
Since the sum is not divisible by 9, it can not be divisible by 180
(B) 16
(155 + 160)/2 * 16
315 * 8 ——-> prime factorization = (2)^3 * (3)^2 * (5) * (7)
Prime Factorization of 180 = (2)^2 * (3) * (5)^2
Since the exponents are equal to or greater than the exponents of 180’s Prime Factorization, the SUM of the interior Angles of 16 sides given this Arithmetic Progression will be a valid sum
(B) 16
*note* you can check that (A) doesn’t work in the same way
Posted from my mobile device
