Latest Elitmus Aptitude Question SOLUTION: perimeter of a regular hexagon ABCDEF is 42.find the area of ABDEF polygon option are : a) 49 b)49 root 3 c) 36 d

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(#M40160089) ELITMUS Numerical Ability question Keep an EYE

perimeter of a regular hexagon ABCDEF is 42.find the area of ABDEF polygon


option are : a) 49
b)49 root 3
c) 36
d) 36 root3/2

asked today

Asked In Elitmus Sahil Tanwar (7 years ago)

Unsolved Read Solution (11)

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sangeeta kumari  (7 years ago)

none of the answers is correct .there is some problem with the question,i also think they have asked for quad ABED ,then only 49 root 3 could be the answer

Is this solution Helpfull? Yes (7) | No (1)  

palak tandon  (6 years ago)

perimeter=42,which is equal to the sum of side.
here a=6
so a=42/6=7
area of hexagon=(3root3*a^2)/2
(3root3*49)/2

Is this solution Helpfull? Yes (5) | No (1)  

premanshu ankit  (6 years ago)

Let ABCDEF be a regular hexagon.
Given that perimeter is 42.
therefore,side will be 42/6 = 7.
Area of regular hexagon = (3*root(3)a^2)/2 = (3*root(3)*49)/2
For regular hexagon, BC = a, AD = 2a, BD = root(3)*a = 7*root(3).
In isisceles traingle BCD, BC = CD = 7 and BD = root(3)*7
From Pythagoras th. height will be 7/2.
So Area of traingle BCD = base*height/2 = 49*root(3)/4.
So area of ABDEF = ABCDEF - BCD = ((49*3*root(3)/2) - (49*root(3))/4)) = 49*root(3)*5/4

Is this solution Helpfull? Yes (5) | No (1)  

manoj kumar  (7 years ago)

Actually polygon ABDEF, there are 4 equilateral triangle with side 7.
Then the option B is correct, 49 root3

Is this solution Helpfull? Yes (3) | No (7)  

Pradosh kumar  (7 years ago)

quesion was asked that to find the area of ABDE only not F.

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Triyambak Gupta  (6 years ago)

36
Req Area=Area of hexagon -Area of Triangle BCD
Angle BCD=120
Area of triangle =1/2*Product of two sides *Angle between them
1/2*Root 3/2*24/Root 3=6
Required Area =42-6=36

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Amit  (7 years ago)

the diagram ABDE would be a parllelogram
so area will be
1/2*d1*d2*sin(theta)=1/2*14*14*root(3)/2=49root3

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deeksha sinha  (6 years ago)

Answer: b) 49 root 3
Explanation:
6*a=42
therefore a=7
we are asked to find the area of ABDEF leaving the node C.
hence we will have only 4 equilateral triangles instead of 6.
therefore, required area= 4* (root3/4) *a*a
req. area= 49 root3
:P

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Triyambak Gupta  (6 years ago)

Sorry for the previous post answer would be 35 .In place of 24/Root3 there would be 28/root3.And here Area is equal to 42 not perimeter.

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Sloka  (6 years ago)

Answer - 5/4* 49root(3)
(a=7)
Area of regular hexagon = (3 root(3))/2 * a^2 = (3 root(3))/2 * 49
Draw perpendicular from Vertex C to line Segment BD
Now, Using Trigonometry find BD, which is equivalent to 7root(3)
Area of rectangle ABDE = 49root(3)
Area of two triangle i.e. AFE & BCD = Area of Hexagon - Area of rectangle ABDE
= 49 root(3)/2
Area of two triangle = 49 root(3)/4
Area of polygon ABDEF = (1+1/4) 49 root(3) = 5/4 49root(3)

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  (5 years ago)

none of the options is correct.
the area of ABDEF is 245 root 3 by 4 squints.

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