Latest Elitmus Aptitude Question SOLUTION: A circle inscribes a Square ABCD. E,F,G.H are mid-points of Sides AB,BC,CD and DA respectively. I is a point on EF such that IF=1/3 EF. What is the ratio of area of circle to area

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(#M40038312) ELITMUS Numerical Ability question Asked on 3 August Keep an EYE

A circle inscribes a Square ABCD. E,F,G.H are mid-points of Sides AB,BC,CD and DA respectively. I is a point on EF such that IF=1/3 EF. What is the ratio of area of circle to area of triangle HGI.

Asked In Elitmus Amit Toor (9 years ago)

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Amit Toor  (9 years ago)

@PRANSHU
as i can see you have let square side equal to a then by pythagoras theoram
diameter d= sqroot(a^2+a^2)= root2*a
radius r= d/2 which will be a/root2 not a/2.
so answer is 2pi:1

Is this solution Helpfull? Yes (24) | No (3)  

Pranshu Chaurasia  (9 years ago)

let side of square= a and a/c to question E,F,G,H are the mid points so EF=FG=GH=HA= a/square root 2.
In triangle IHG,the height of triangle= a/square root 2 and base of triangle is also a/square root 2.
Now, the circle radius= a/2.
Ratio= [pi*(a/2)^2]/[(1/2)*(a/square root 2)*(a/square root 2)]
Ratio= pi:1

Is this solution Helpfull? Yes (14) | No (14)  

Amit Toor  (9 years ago)

Alternate way:
let AB=BC=CD=DA=a
then height & base of triangle HGI will be equal to a.
area of triangle= 1/2*a^2
In Square AEOH (O is center of circle)
AO=EH=a (Since Diagnols of Square will be equal)
AO is clearly radius.
so r=a
area of circle pi*r^2= Pi*a^2
ratio is : (Pi*a^2)/(1/2*a^2) =>2pi:1

Is this solution Helpfull? Yes (7) | No (6)  

Amit Toor  (9 years ago)

Let EF=FG=HG=3a

then base of triangle HGI is 3a and height is also 3a
So are 1/2*base*height=9/2*a^2

Now Using pythagoras theorem we can easily find side of square.

and then again applying pythagoras we find diagonal of square which is diameter using that area of circle can be find easily.

using above calculations we will get
radius=3a
area of circle will be pi*9^a^2

ratio will be
(pi*9^a^2)/(9/2*a^2)=2pi

Is this solution Helpfull? Yes (4) | No (14)  

Mohit Anand  (9 years ago)

let the side of square be =2a
so the radius of circle would be =a
therefore area of circle is =pie*a*a
EF and HG are two parallel lines , therefore area of triangle between them common base would have equal area,

hence area of triangles EHG and IHG are equal, which is=(1/2)*EH*HG
HG=EH=root2*a
area is=a*a

therefore ratio is=(pie*a*a)/(a*a)
=pie

Is this solution Helpfull? Yes (4) | No (4)  

manoj kumar jha  (9 years ago)

first draw the diagram according to question.....
now,let sides of bigger square be 2a
so,sides of smaller square be a
so,IF = 1/3a and IE = 2/3a
now area of triangle HGI=area of square EFGH - area of triangle EHI -area of GFI
=EH*EH-1/2*EI*EH-1/2*IF*FG
=a2 - 1/2*2a/3*a - 1/2*1a/3*a(area of triangle = 1/2*base*height)
=a2/2
now area of circle=pi*r*r
and r=sqrt((2a)2+(2a)2)/2
=(root2)a
so area of circle=pi*(root2)a*(root2)a=2*pi*a2
so area of circle:area of TRIangle GHI=2*pi*a2:a2/2
=4PI:1

Is this solution Helpfull? Yes (4) | No (7)  

akanksha  (8 years ago)

diameter will b root(2)a where a is side of the square.so area of circle will be (a^2 pi)/2
now area of triangle will be=(area of square EFGH- (area of tri HIE + area of tri GFI)) which will b a^2/4
so ratio=2 pi:1

Is this solution Helpfull? Yes (4) | No   

arjun kumar jha  (9 years ago)

@amit u r right... I have solved this and anser is 2pi....

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

Pranshu Chaurasia  (9 years ago)

@ AMIT
I thought that the circle is inside the square that's why i did that calculation..If circle is circumscribe the square then your answer is right.

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ritesh kumar  (9 years ago)

@Pranshu Chaurasia u r ryt....

Is this solution Helpfull? Yes (2) | No   

Hitesh Sharma  (9 years ago)

Mohit Anand's solution is most efficient except the fact that he has mistakenly reversed the required ratio of respective figures.
He has used the property of triangles and has simplified the solution to a great extent.
I am seeing that his answer is not supported much on this forum but actually it is short and correct.
So dnt get confuse, refer that approach.

Is this solution Helpfull? Yes (2) | No   

Atik  (8 years ago)

Please drw diagram it will help you to understand.
1st outer squ side =a;
2nd circle radius =a/2;
now we can find EF=a/root(2);
draw square of EFGH;
HGI is half of inner squre;
innr squ side is a/root(2);
innr square area=(a^2)/2;
area of inner triangle=(a^2)/4;
area of circle =pi (a^2)/4;
ratio= 1:pi;

Is this solution Helpfull? Yes | No (2)  

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