Latest Elitmus Aptitude Question SOLUTION: Given a regular hexagon ABCDEF whose area is 42sq m,than find the area of a pentagon made by joining the verticesABCEF. 35,49/root3,

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(#M40040665) ELITMUS Numerical Ability question geometry Keep an EYE

Given a regular hexagon ABCDEF whose area is 42sq m,than find the area of a pentagon made by joining the verticesABCEF.
35,49/root3,

Asked In Elitmus priyanka kumari (9 years ago)

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Read Solution (15) :

harmender  (9 years ago)

ans is 35
since regular hexagon comprise of 6 equilateral triangles of equal areas,area of each triangle 7 sqm
area of pentagon abcef = area of hexagon abcdef - area of triangle edc i.e 7
so ans comes out to be 35

Is this solution Helpfull? Yes (36) | No (12)  

swati kunj  (9 years ago)

area of hexagon= 42sqm
so regular hexagon has 6 equal triangles.
so area of triangle cde= 42/6= 7
now area of pentagonal=42-7=35sqm

Is this solution Helpfull? Yes (9) | No (5)  

Mithun Kumar Kesari  (9 years ago)

ANS= 49/(ROOT 3)
Formula used, area of regular polygon= ((n*s^2)/4)*cot(180/n)
where, n= no. of sides and s= length of side

Is this solution Helpfull? Yes (7) | No (5)  

Rahul Sharma  (9 years ago)

Area of Hexagon= 6*[root(3)/4] a^2 where a=side of hexagon
so, 42=6*[root(3)/4]a^2
a^2=28/root(3)
thus Area of pentagon= 1.72 * a^2=1.72*28/root(3)=49/root(3)

Is this solution Helpfull? Yes (7) | No (2)  

Alok Pandey  (9 years ago)

dont be confused guys the correct answer will be 35

the formula which u r using like area=ns^2 *cot 180/n is for irregular polygons ok. but here question is regular polygon.
plz go through the correct propeties of polygon.

Is this solution Helpfull? Yes (3) | No   

JANARTHANAN RAM  (8 years ago)

Ans: 35 m^2

Area of Regular hexagon=(3(root3)/2)*a^2= 42 m^2
a=root(28) / (3 ^ 0.25)---------(1)
We should cut the Triangle CDE.
CDE is two sided equal triangle.
Sin 60=(CE/2) / a
CE=b=root28 * 3^0.25-------(2)
Area of isosceles triangle(CDE)(with sides a,a,b)
A=(b/4) * [root( (4*a^2) - b^2) ]------(3)
Using (1) , (2) & (3) A= 7 m^2
Required Area=42-7=35 m^2

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sanchit  (9 years ago)

answer is 35

Is this solution Helpfull? Yes (2) | No (6)  

Rayagouda   (9 years ago)

since area of hexagon consist of 6 equilateral triangle, therefore the area of pentagon is
42-(42/6)=35

Is this solution Helpfull? Yes (1) | No (3)  

Pranshu Preet Kaur  (9 years ago)

Hexagon is comprised of 6 equilateral triangles. Itz Area is given to be 42 sq m i.e each triangle is having area=7sq m.
Hence required area of pentagon formed is 42-7=35sq m.

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

kundan prasad  (8 years ago)

Ans: 35

Is this solution Helpfull? Yes (1) | No   

Vikram  (9 years ago)

answer is 31.5 because hexagon divided into 8 parts...then pentagon having 6 parts(triangles)=(42/8)6=31.5

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swati priya  (9 years ago)

area of reg hexa=6*(Rt3/4)a^2=42
a^2=4*42/6rt3

area of pentagon=5*(rt3/4)a^2=35

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Pankaj Kumar  (8 years ago)

area of hexagonal=(square root three)(side square)(divided by four)=42
=>square of side =56*(square root of three)(divided by 6).
Now. area to be discarded = 7
then Ans- is 35

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akash  (8 years ago)

the answer is first correct because the REGULAR HEXAGON .

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Supriya Mahanto  (6 years ago)

Only restriction given is that base of the cone should be on one of the face of cuboid. Since there are no other restrictions, any value of the base radius is possible such as r=1, 2, 0.2, 0.01 etc because all such bases can be on one of the face of cuboid. Theoretically, infinite number of cones can be cut.

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