Bunuel wrote:
A pentagon with 5 sides of equal length and 5 interior angles of equal measure is inscribed in a circle. Is the perimeter of the pentagon greater than 26 centimeters?
(1) The area of the circle is 16π square centimeters.
(2) The length of each diagonal of the pentagon is less than 8 centimeters.
DS75271.01
OG2020 NEW QUESTION
Please find the diagram attached herewith.
Lets see why option B is also sufficient.
Triangle ABC and Triangle BCF and Trinagle FDE are 36 , 36 , 108 degree triangle.
( I explained the reason in the diagram attached )
So I am keeping it brief here.
Angle BAC = 108 degree ( interior angle of a pentagon)
Since AB = AC (sides of a regular pentagon are equal ) , Angle ABC = angle ACB = (180 - 108) / 2 = 72 / 2 = 36 degree
ANgle BDE = 108 degree ( interior angle of a pentagon)
Angle DBE = Angle DFB = (180 degree - angle BDE ) / 2 = (180 - 108 )/ 2 = 72 / 2 = 36 degree
Angle CBE = 108 - ( angle ABC + angle DBE) = 108 - 72 = 36 degree
Similarly , Angle BCD = 36 degree
ANgle BFC = 180 - ( CBE + BCD) = 180 - ( 36 + 36 ) = 108 degree
Say , The length of each diagonal = c
So BE = c
Now Triangle ABC and traingle BCF are congruent .( because they have two angles equal and one side common So ASA condition satisifes)
ANgle ABC = ANgle CBE = 36 degree
Angle ACB = angle BCF = 36 degree
SIde BC common
So Triangle ABC and traingle BCF are congruent.
SO the corresponding sides AB = BF
and AC = CF
Say AB = a
So , BF = a
Now , FE = BE - BF = c - a
SImilarly FD = c - a
DE = a
Now Triangles ABC and triangle FDE are similar. ( All the angles are same for them...36 , 36 , 108 )
So corresponding sides are of same proportion...
So , c/a = a / c-a
so c^2 - a*c = a^2
For the maximum length of the diagonal i.e 8 cm,
64 - 8*a = a^2
Now Say , side of the pentagon = 5 cm
it does not satisfy the equation above as (64 - 8 * 5 ) is smaller than 5^2 = 25 cm
So 'a' has to a bit smaller than 5 cm.
Now if the diagonal is less than 8 cm , then it will be even less than 64 . And side i.e the value of 'a' will be even smaller.
Even if the side was 5 cm , the perimeter of the pentagon would have been 5 *5 = 25 cm , which is smaller than 26 cm.
If the sides are even smaller , the perimeter will be even smaller and hence less than 26 cm.
Hence , Option B is also sufficient.
Please give me KUDO s if you liked my explanation.
Attachments
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