Latest Elitmus Aptitude Question SOLUTION: A parallelogram having an acute angle of 30 degrees whose area is equal to the perimeter and the sides are whole numbers then number of such parallelograms possible are a) 0

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(#M40017678) ELITMUS Numerical Ability question Keep an EYE

A parallelogram having an acute angle of 30 degrees whose area is equal to the perimeter and the sides are whole numbers then number of such parallelograms possible are
a) 0 b) 1 c) 2 d) 3

Asked In Elitmus Nikhil (11 years ago)

Unsolved Read Solution (8)

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Mukesh Kr Singh  (11 years ago)

let the paralleogram consist of two triangle..
the area=(1/2)*ab*sin(30) [a and b are the sides]
=1/4ab
now area of parallelogram=1/2ab
peimeter=2a+2b
so acc. to qstn
2a+2b=1/2ab

so it can be written as
(a-4)(b-4)=16
Thus a−4 and b−4 must have product 16.
we may assume that a≤b
since a−4 and b−4 cannot both be negative.
So a−4=1, b−4=16, or a−4=2, b−4=8, or a−4=4, b−4=4.
So ansr is 3 possible parallelogram

Is this solution Helpfull? Yes (30) | No (12)  

jayavardhan  (11 years ago)

area of parllgrm=a*b sin(30) =1/2 ab
perimeter= 2(a+b)
1/2 ab=2(a+b) --> 4a+4b=ab -->a(b-4) =4b
a=4b/(b-4)
then(a,b) -->(20,5) (8,8) (5,20)
(8,8) --> is a square is a parlelogram but no side will be 30 degrees
so only (20,5) and (5,20) are possible
only 2 parallelograms are possible

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

AVINASH KUMAR  (11 years ago)

nikhil please
send me
avinashsingh56@gmail.com

Is this solution Helpfull? Yes (1) | No (9)  

ankita  (11 years ago)

send me too nikhil at ankita.apt@gmail.com

Is this solution Helpfull? Yes (1) | No (4)  

Shirlin Joseph  (11 years ago)

area of parallelogram=1/2 ab, perimeter=2(a+b)
so 2a+2b=1/2 ab
so,, (a-4)(b-4)=16
area is parallelogram is 1/2(a-4)(b-4)=8

therrefore possible combinations to obtain product of sides 8 should be 1,8 and 2,4.. answer 2 parallelograms.

Is this solution Helpfull? Yes (1) | No (4)  

farjana sayyed  (10 years ago)

square is also a parallelogram

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

Ankit Saun  (11 years ago)

hey nikhil if you have more questions then please post it.Or mail me at ankitsaun.thekiller@gmail.com will put it

Is this solution Helpfull? Yes | No (3)  

shubham verma  (8 years ago)

Given that perimeter of the parallelogram is equal to its area.
So we have: 2(l + b) = l * h --------------- (1)
From the diagram we have: sin300 = (h/b) => 1/2 = h/b
=> h = b/2 --------------- (2)
From eq (1) and (2):
2(l + b) = l * b/2
=> (l + b)/(l*b) = 1/4
=> l * b = 4(l + b)
If you are familiar with graph you can see that the the expression on the left hand side is the equation of a hyperbola and the expression on the right hand side matches the equation of a straight line. Till (l, b) = (8, 8) the graph of hyperbola will be below the graph of the straight line after this point the graph of hyperbola will always be above the graph of the straight line. So we can have just one solution for this.

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