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Since it is a PS question, I would pick D on exam.
sides are equal so it most probably a rhomb. if i recall correctly, the area of a rhomb is half the product of its diagonals? d1d2 = 18*2 = 36. perhaps diagonals are equal in lenght (sqrt 36 = 6).

so we have two equilateral traingles. where angles are equal to 60 degrees.
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A too.

We know that Area= base * height. If the Area=18, and base = 6, height must be 3, that is, the angle must be 30º.
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Given each side is 6 and area is 18. We can find the altitude, ie: alt = 18/6 = 3.
Now we have a triangle with one side as 6, the alt as 3, and the base which we don't know.

Using Pythagoras theorem, 6^2 = 3^2 + Base^2 => Base = 3V3
So, the sides are now 3*1, 3*V3, 3*2 => 1:V3:2, which is a 30-60-90 triangle.

Hence the angle that we know is 30, which forms a side of the parallelogram.
Ans: A.
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6*6*sin(alpha) = 18 -> sin(alpha) = 1/2 -> alpha = 30 -> A
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Each side of a certain parallelogram has length 6. if the area of the parallelogram is 18.
which of the following is the measure of one of its angles?
A. 30’ B. 45’ C. 60’ D. 90’ E.120’

Equal sides of a parallegrame means it's a rhombus -> angle should be either 60 or 120?
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Ans is 30deg

area of a parallelogram = b*h

b*h=18
h=18/6=3

sin theta =opp/hyp= 3/6

theta = sin inv of 1/2=30 deg
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A=b*h or 1/2 d1*d2, if all 4 sides are equal then this is a rhombus, which is basically a tilted square. How can you have a height of 3 and area of 18? Shouldnt height be 6 and area 36? Is this question correctly worded?
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A=b*h or 1/2 d1*d2, if all 4 sides are equal then this is a rhombus, which is basically a tilted square. How can you have a height of 3 and area of 18? Shouldnt height be 6 and area 36? Is this question correctly worded?

If you have a rhombus with sides b and c, the area will only be b*c if the rhombus is a square. Otherwise the area will be less than b*c. To find the area of a rhombus (or any other parallelogram), you need to multiply the base and the height. If your parallelogram is 'tilted' (i.e. if the angles are not all 90 degrees), then the height is definitely not one of the sides. In this example, the rhombus is pretty seriously tilted to get an area as small as 18.
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I have a question here. how are we considering the art of the parallelogram to be a right triangle?? Can someone please explain
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I have a question here. how are we considering the art of the parallelogram to be a right triangle?? Can someone please explain

HI longhaul123

Can you be more specific with your query. If you drop a perpendicular on the base of the parallelogram then it will form a right angles triangle with one of the side of the parallelogram as its hypotenuse.
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Ans is clearly A--- see the attached image

A= base x Height
6xH = 18
H=3

Sin A = 3/6 = 1/2
A=30

Answer is A
Attachments

IMG_0031.JPG
IMG_0031.JPG [ 957.94 KiB | Viewed 11125 times ]

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We can work on options to get to the answer quickly,
Option C and option E can be eliminated easily as if one angle is 60 one of the other angle has to be 120.
Option D is not possible as if one angle is 90 all angles will be 90, thus it becomes a square with side 6 and area of square is 6*6 = 36. Not matching.
Out of option A and option B, working on option A is easier, thus even if someone doesnt know how to work on this answer can still guess it correct.

Now working on option A
we can draw a diagonal and break parallelogram into two triangles of equal area. One angle of triangle is 30 and sides are 6. Using formula
area of triangle = \(\frac{1}{2}\) * a*b*sin(x)
area of one triangle becomes \(\frac{1}{2}\) * 6*6*sin(30) = \(\frac{1}{2}\) * 6*6*\(\frac{1}{2}\) = 9
Thus total area = 9+9 =18
Thus option A
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Use Heron's formula here: absinθ=area. Given a=b=6, so 36sinθ=18 or sinθ=0.5 or θ=30.
Although this formula is not advised for most of the quadrilaterals, since parallelogram is mentioned, I used it.
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