Latest Elitmus Aptitude Question SOLUTION: A regular hexagon is there. The mid points of the sides were joined and formed another hexagon.Then What is the percentage reduction in area. Asked 13 sept......

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(#M40153704) ELITMUS Numerical Ability question Geometry Keep an EYE

A regular hexagon is there. The mid points of the sides were joined and formed another hexagon.Then What is the percentage reduction in area.
Asked 13 sept......

Asked In Elitmus Malaya Prasad Lenka (8 years ago)

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mohit  (8 years ago)

Let ABCDEF be the regular hexagon with sides AB and ED parallel. P and Q are mid points of AF and EF resp.
AE=root(3) * s (s is side of hexagon)
then PQ =(root(3) * s)/2 mid point theorem
Now u know side of inner hexagon
use formula area of hexagon = (3 * root(3))2 * s^2
and find diff in area of two hexagon multiply by 100 and divide the product by area of bigger hexagon
you will get the answer.
Ans 25%

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jai shankar pandey  (8 years ago)

area of the bigger hexagon=6*(root(3)/4)*a^2
for the base of reduce hexagon=2*(root(3)/4)
side of reduce hexagon =root(3)/2
area of reduce hexagon=6*(root(3)/4)*((root(3)*a)/2)^2=(9*root(3)*a^2)/8
diffrece between area=(3*root(3)*a^2)/8
%reduction=(diffrence beetween area)*100/area of bigger hexagon=25%

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