[email protected] wrote:Hi gmmattesttaker2,

I'm going to give you a couple of hints and let you try this question again:

1) Look for "hidden" right triangles. Notice how x = 60 degrees? Think about WHERE you could find right triangles inside and OUTSIDE of the given shape.
2) Since the 6-sided shape is equilateral, what does that mean about the 6 sides?
3) Line segment BD is OUTSIDE the shape. Try drawing that into the picture, then cut that line IN HALF.

GMAT assassins aren't born, they're made,
Rich

Hi Rich,

I was wondering if length BD is also included in the perimeter of ABCDEF and similarly if AE would also be included in the perimeter of ABCDEF?

Thanks,
Sri

gmattesttaker2 wrote:Hello,

Can you please assist with this:

In the figure above, polygon ABCDEF is equilateral and has perimeter 42. If AB is
parallel to ED, AF is parallel to BC, EF is parallel to CD, and x equals 60
degrees, what is the length of line segment BD (not shown)?

OA: [spoiler]7.sqroot(3)[/spoiler]

Hi Sri,
Given: X = 60 Degree and each side of length 7
We need to find Length of BD.
Length of BD = Length of AE.

Draw a line Joining B and E and it Intersects Line FC, lets keep that point as G.
Now AE = AG+GE

Consider triangle AFG.
It forms a 30:60:90 Triangle.
Whose sides will be in the ration x:√3X:2X
we know that side for the angle 90 is 7
hence 2X = 7 => X = 7/2
we need AG which is formed by 60 degree angle,
So AG = (7/2)*√3

Similarly do for another triangle, FGE
you will end up GE = (7/2)*√3

Hence AE = (7/2)*√3+ (7/2)*√3 = 7√3

BD = 7√3

Hope it helps you.

Regards,
Uva.

gmattesttaker2 wrote:Hello,

Can you please assist with this:

In the figure above, polygon ABCDEF is equilateral and has perimeter 42. If AB is
parallel to ED, AF is parallel to BC, EF is parallel to CD, and x equals 60
degrees, what is the length of line segment BD (not shown)?

OA: [spoiler]7.sqroot(3)[/spoiler]

Join BD, drop CQ perpendicular on BD, this would create two identical 30-60-90 triangles in which BQ = QD = 3.5√3 by property; hence BD = 7√3.