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03 Sep 2020, 00:30
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B
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Difficulty:
85%
(hard)
Question Stats:
35% (01:22) correct
65%
(01:42)
wrong
based on 94
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A regular polygon is partially obscured by a box. (Dotted lines show that polygon is hidden behind the box.). What is the sum of the angles of the polygon?
(1) x°= y°
(2) x°+ y°= 72°
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paras7424
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03 Sep 2020, 01:46
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IMO B.
X+y+ the 2 angles above=360 as it is a quadrilateral.
Also since it is q regular polygon the 2 above angles will be equal.
Statement 1---> x=y=can be any value between 1 and 179 and accordingly the values of 2 above angles will vary. (Insufficient)
Statement 2---> x+y=72
The two above angles will be 144 each. Hence the polygon is a Decagon with sum of angles as 1440. (SUFFICIENT)
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X+y+ the 2 angles above=360 as it is a quadrilateral.
Also since it is q regular polygon the 2 above angles will be equal.
Statement 1---> x=y=can be any value between 1 and 179 and accordingly the values of 2 above angles will vary. (Insufficient)
Statement 2---> x+y=72
The two above angles will be 144 each. Hence the polygon is a Decagon with sum of angles as 1440. (SUFFICIENT)
Posted from my mobile device
yashikaaggarwal
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03 Sep 2020, 01:56
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Sum of angle of a polygon = (N-2)*180
Since its a regular polygon its all sides are same.
One angle = (n-2)*180/n --------------(1)
So upper two angles will be equal. We have find the value of N
The shown quadrilateral have the sum total of 180
1) X = Y.
Upper two angles are also equal.
Case 1: all angles are equal.
The polygon is a quadrilateral only. With sum of angles = 360
Case 2: lower two angles are bigger than upper two.
The polygon shape can vary with the angle sum.
(Insufficient)
2) x+y = 72
Upper two angles = 360-72
Angle A + Angle B = 288
Angle A = Angle B
2*Angle A = 288
Angle A = 144 ------------(2)
Putting value in equation 1
144 = (n-2)*180/n
144n = 180n-360
360 = 36n
N = 10
The polygon have 10 sides and the sum of angles = (10-2)*180
=> 1440
(Sufficient)
IMO B
Posted from my mobile device
Since its a regular polygon its all sides are same.
One angle = (n-2)*180/n --------------(1)
So upper two angles will be equal. We have find the value of N
The shown quadrilateral have the sum total of 180
1) X = Y.
Upper two angles are also equal.
Case 1: all angles are equal.
The polygon is a quadrilateral only. With sum of angles = 360
Case 2: lower two angles are bigger than upper two.
The polygon shape can vary with the angle sum.
(Insufficient)
2) x+y = 72
Upper two angles = 360-72
Angle A + Angle B = 288
Angle A = Angle B
2*Angle A = 288
Angle A = 144 ------------(2)
Putting value in equation 1
144 = (n-2)*180/n
144n = 180n-360
360 = 36n
N = 10
The polygon have 10 sides and the sum of angles = (10-2)*180
=> 1440
(Sufficient)
IMO B
Posted from my mobile device
03 Sep 2020, 02:34
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A regular polygon is partially obscured by a box. (Dotted lines show that polygon is hidden behind the box.). What is the sum of the angles of the polygon?
(1) x°= y°
(2) x°+ y°= 72°
Each angle of Regular polygon is equal.
Statement 1) x°= y° does not give any clue about angle of regular polygon
Insufficient.
Statement 2) x°+ y°= 72°
x°+ y° + 2 Equal Angle of regular polygon = 360
2 Equal Angle of regular polygon = 360 - x°+ y°
= 360-72
= 288
Each Angle of regular polygon = 144
Now we have one angle of regular polygon. We can calculate no of Sides of regular polygon.
Sufficient.
Each angle of N-sided regular polygon = 180 (n-2) / n
144= 180 (n-2) / n
After solving we get n = 10.
IMO Answer is B
(1) x°= y°
(2) x°+ y°= 72°
Each angle of Regular polygon is equal.
Statement 1) x°= y° does not give any clue about angle of regular polygon
Insufficient.
Statement 2) x°+ y°= 72°
x°+ y° + 2 Equal Angle of regular polygon = 360
2 Equal Angle of regular polygon = 360 - x°+ y°
= 360-72
= 288
Each Angle of regular polygon = 144
Now we have one angle of regular polygon. We can calculate no of Sides of regular polygon.
Sufficient.
Each angle of N-sided regular polygon = 180 (n-2) / n
144= 180 (n-2) / n
After solving we get n = 10.
IMO Answer is B
