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21 Jul 2020, 11:00
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Q. In the regular pentagon given in the figure, what is X?
A. \( 24^{\circ} \)
B. \( 36^{\circ} \)
C. \( 42^{\circ} \)
D. \( 45^{\circ} \)
E. \( 72^{\circ} \)
A. \( 24^{\circ} \)
B. \( 36^{\circ} \)
C. \( 42^{\circ} \)
D. \( 45^{\circ} \)
E. \( 72^{\circ} \)
Attachments
Pentagon.png [ 10.53 KiB | Viewed 3189 times ]
yashikaaggarwal
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21 Jul 2020, 11:18
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Interior angle of a regular Pentagon = (5-2)*180
=> 3*180 => 540
=> One interior angle of a Pentagon = 540/5 = 180
=> Angle A = Angle B = Angle C = Angle D = Angle E = 108
In triangle ABC
AB = BC
And Angle B = 108
The other two angles = 180-108/2
=> 72/2 = 36 = Angle BAC = Angle ACB
Similarly in triangle, CDE
Angle ECD = Angle DEC = 36
Angle C = X + Angle ECD + Angle ACB
108 = X + 36 + 36
108-72 = X
X = 36
Answer is B
Posted from my mobile device
=> 3*180 => 540
=> One interior angle of a Pentagon = 540/5 = 180
=> Angle A = Angle B = Angle C = Angle D = Angle E = 108
In triangle ABC
AB = BC
And Angle B = 108
The other two angles = 180-108/2
=> 72/2 = 36 = Angle BAC = Angle ACB
Similarly in triangle, CDE
Angle ECD = Angle DEC = 36
Angle C = X + Angle ECD + Angle ACB
108 = X + 36 + 36
108-72 = X
X = 36
Answer is B
Posted from my mobile device
Attachments
IMG_20200721_234209.jpg [ 46.54 KiB | Viewed 3137 times ]
21 Jul 2020, 11:23
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Sum of interior angles of the pentagon = (n-2)*\(180^{\circ}\) [where n=5]
= 3*\(180^{\circ}\) = \(540^{\circ}\)
So, Each interior angle of pentagon = \(\frac{540^{\circ}}{5}\) = \(108^{\circ}\)
In regular pentagon angle x will be \(\frac{1}{3}rd\) of each interior angle of the pentagon = \(\frac{1}{3} * 108^{\circ}\)
= \(36^{\circ}\)
If we don't want to use the property then we can solve it using geometry too
Refer Below Image for nomenclature
Consider \(\triangle ABC\), we know that the two sides AC and BC are equal so its an isosceles \(\triangle\)
=> \(\angle a\) = \(\angle b\)
a + b + \(\angle C\)= \(180^{\circ}\)
=> b + b = \(180^{\circ}\) - \(108^{\circ}\) [ as a=b and \(\angle C\) = Interior angle of Pentagon = \(108^{\circ}\) ]
=> 2b = \(72^{\circ}\)
=> b = \(36^{\circ}\)
Using similar logic d = \(36^{\circ}\)
a + x + d = Interior angle of pentagon = \(108^{\circ}\)
=> x = \(108^{\circ}\) - \(36^{\circ}\) - \(36^{\circ}\) = \(36^{\circ}\)
So, Answer will be B
Hope it helps!
= 3*\(180^{\circ}\) = \(540^{\circ}\)
So, Each interior angle of pentagon = \(\frac{540^{\circ}}{5}\) = \(108^{\circ}\)
In regular pentagon angle x will be \(\frac{1}{3}rd\) of each interior angle of the pentagon = \(\frac{1}{3} * 108^{\circ}\)
= \(36^{\circ}\)
If we don't want to use the property then we can solve it using geometry too
Refer Below Image for nomenclature
Consider \(\triangle ABC\), we know that the two sides AC and BC are equal so its an isosceles \(\triangle\)
=> \(\angle a\) = \(\angle b\)
a + b + \(\angle C\)= \(180^{\circ}\)
=> b + b = \(180^{\circ}\) - \(108^{\circ}\) [ as a=b and \(\angle C\) = Interior angle of Pentagon = \(108^{\circ}\) ]
=> 2b = \(72^{\circ}\)
=> b = \(36^{\circ}\)
Using similar logic d = \(36^{\circ}\)
a + x + d = Interior angle of pentagon = \(108^{\circ}\)
=> x = \(108^{\circ}\) - \(36^{\circ}\) - \(36^{\circ}\) = \(36^{\circ}\)
So, Answer will be B
Hope it helps!
