The parallelogram shown has four sides of equal length. What is the

IMO D.

The parallelogram has four equal sides (x), if the angles is 60° for one, the opposite angle is 60° and the other two are 120° each (360-60-60/2).

Therefore, the parallelogram diagonal bisects every angle to form four different 30-60-90 rectangle triangles (ratio: \(x : 2x : x\sqrt{3}\)).

Longer diagonal = \(2x\sqrt{3}\)
Shorter diagonal = \(2x\)

Therefore the ratio is \(\frac{2x}{2x\sqrt{3}}\) = \(\frac{1}{\sqrt{3}}\)

Opposite angles of a parallegram are equal and sum of adjacent angles is 180 degree .
Each of the angles adjacent to 60 is 120 .
The shorter diagonal divies the parallelogram divides into 2 isosceles triangles .
Since all the sides of the parallelogram are equal , we get an equilateral triangle.
Therefore the shorter diagonal will be equal to length of sides of parallelogram = x

The longer diagonal divides parallelogram into 2 isoceles trianges with one angle 120 and the other 2 angles equal to 30 .

Since, the diagonals of a parallelogram are perpendicular bisectors ,the 2 diagonals divide the parallelogram into 4 - 30-60-90 triangles.

The sides of a 30-60- 90 triangle are in ratio 1:(3^(1/2)):2
Sin 60=(y/2)/x
Where y= length of the longer diagonal
=>(3^(1/2))/2=(y/2)/x
=>x/y=1/(3^(1/2))

Answer D

The figure must be a rhombus. Draw the diagonals on your scratch paper. You will now have two equiliteral triangles. The shorter diagonal is one side of such an equiliteral triangle. Let's say s=1.

To get the longer side of the diagonal of the rhombus, you have 2*height of the equiliteral triangle. If you split an equiliteral triangle into two parts, you will have two 30:60:90 triangles. The height of the equiliteral triangle if s = 1 is \(0.5*\sqrt{3}\) and the long diagonal will be \(2*0.5*\sqrt{3}\)

So you have \(1:\sqrt{3}\)

Answer D

Assume the length of each side = x

The other angles would be 60, 120 and 120.
Diagonal divides the parallelogram into half

Hence the shorter diagonal will make two equilateral triangles
Length of the shorter diagonal = length of side = x

For the longer diagonal.
The diagonal will divide the parallelogram in to isosceles triangles with angles 30, 120, 30
Dropping a perpendicular from top most point on to the diagonal, we have two triangles with angles 30, 60 and 90 with base as half the length of diagonal

Hypotenuse = x,
Cos 30 = base/x
base = (√3/2)*x

Hence the length of the diagonal = √3x

Ratio of the shorter to the longer diagonal = x : √3x = 1: √3

Correct Option: D

Lets assume that side = 1
now the shorter diagonal will belong to squeezed quadrilateral with one side = 1 - 1/2 = 1/2( being a 30-60-90 triangle on the right side) & one side = (3)^0.5/2
Longer diagonal will belong to enlarged quadrilateral with one side = 1 + 1/2 = 3/2 (being a 30-60-90 triangle but side will add) & one side = (3)^0.5/2

to find the diagonal we will apply Pythagoras to both rectangles
Squeezed Quadrilateral : ( 1/4 + 3/4) ^ 0.5 = 1
Enlarged Quadrilateral : ( 9/4 + 3/4 ) ^ 0.5 = 3^0.5

so the ratio= 1/ (3)^0.5

A novice question:

For parallelogram, i understand the rules of relationship for 30:60:90, opposite angels are congruent etc.

But how can we ascertain that when we split the parallelogram in to two parts , the 60 and 120 degrees angles on the vertices split in to exactly HALF? (i.e. 30 , 60 degrees)

Is that a math rule i've forgotten?

Please help thank you !

Usually diagonals of a parallelogram do not bisect the angles but here we have special kind of parallelogram - a rhombus, where the diagonals bisect the angles.

For more check here: math-polygons-87336.html

Hope it helps.

Attached is a visual that should help.

In a parallelogram, the opposite angles are equal. So angle opposite to 60 degrees is also 60.
Draw the shorter diagonal. Since the sides are equal, we see that we get two congruent equilateral triangles. The shorter diagonal is the side of each equilateral triangle and the longer diagonal is twice the altitude of each equilateral triangle.
We know that if the side of an equilateral triangle is s, its altitude is \(\sqrt{3}s/2\) and hence twice of its altitude is \(\sqrt{3}s\).

Required Ratio \(= s/\sqrt{3}s = 1/\sqrt{3}\)

A low level approach of assuming that each side is of length 1 and that we have two equilateral triangles joint together:

I get 1/root(2). Can somebody please help me to explain where I am wrong.

Here is my approach for the long diagonal. So I know that I have a triangle with the following facts: 1 Angle 120° and two times 30°. Two sides have length x therefore my third side (the long diagonal) must be x*root(2).

The sides are in the ratio \(1:1:\sqrt{2}\) only when the angles are 45-45-90. It is not the same when the angles are 30-30-120. There is no "hypotenuse" here.

Do the diagonals of a traditional parallelogram with 2 pairs of equal sides also bisect at 90 degrees?

Response:

No, they will not. Referring to the figure I provided, there are four (larger) triangles that have two black sides and one blue side (these triangles are obtained by joining two adjacent small triangles). For each of these triangles, one of the diagonals of the parallelogram is an angle bisector. In order for an angle bisector to also be a height, the triangle must be an isosceles triangle (i.e. the black sides of the triangles must have equal lengths). If we apply this reasoning to all four of the triangles, we see that the two diagonals can meet perpendicularly only if all the black sides have equal lengths.

Another way to see that this is not necessarily true is to imagine a parallelogram where one pair of sides is very short and the other pair of sides is very long. For such a parallelogram, it is clear that the diagonals do not meet at a 90 degree angle.

Another easy solution:

the 60 degree angle is cut in 2*30 through the Diagonal
then half of the short diagonal, half of the long diagonal and one side of the parallelogram form a rectangular triangle. There the tangent can be used:

tan(30°)= 0.5*shortDiagonal / 0.5*longDiagonal

so the tangent of 30° equals the sought ratio

tan(30°)= 3^(-1/2) => Answer D

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