mirzohidjon wrote:
Hi, Bunuel,
I really appreciate your effort to help us. Finally after your explanation, I got the answer (since I imagined what kind of parallelogram the problem was talking about.
But, based on what assumptions, you deduce that parallelogram needs to be the way you described (diagonal equal to height)?
Attachment:
PointLatticeParallelograms_1000.gif
If the area of a parallelogram is 100, what is the perimeter of the parallelogram?
Given: \(Area=base*height=100\). Q: \(P=2b+2l=?\) (b - base, l - leg )
(1) The base of the parallelogram is 10 --> \(base=height=10\). Infinite variations are possible. Look at the diagram (let the distance between two horizontal and vertical points be 10): all 4 parallelograms have \(base=height=10\) but they have different perimeter. Not sufficient. Side notes: \(l\geq{10}\), when \(l=10=h\) we would have the square (case #3 on the diagram) and \(P=40\) (smallest possible perimeter), maximum value of perimeter is not limited.
(2) One of the angles of the parallelogram is 45 degrees. Clearly insufficient. But from this statement height BX and AX will make isosceles right triangle: \(height=BX=AX\).
(1)+(2) As from 2 we have that \(height=BX=AX\) and from (1) we have that \(base=height=10\) --> \(AX=base=AD=10\) --> X and D coincide (case #4 on the diagram) --> leg (AB) becomes hypotenuse of the isosceles right triangle with sides equal to 10 --> \(AB=10\sqrt{2}\) --> \(P=20+20\sqrt{2}\). Sufficient.
Answer: C.
Attachment:
m10-31.png
Hope it's clear.
We know from 2 that AD=BD, hence and hence we can find out base of the //gm , and we can find of the other side using pythagoras on sin45. hence finding the perimeter. I am unable to understand that how is B insufficient?