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ABCDEF is a regular hexagon with its vertices as A, B, C, D, E , F in

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ABCDEF is a regular hexagon with its vertices as A, B, C, D, E , F in  [#permalink]

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New post 13 Jul 2018, 10:54
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ABCDEF is a regular hexagon with its vertices as A, B, C, D, E , F in this order. What is the area of triangle BDF?

(1) BDF is an equilateral triangle.

(2) AB = 6 units.
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Re: ABCDEF is a regular hexagon with its vertices as A, B, C, D, E , F in  [#permalink]

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New post 13 Jul 2018, 12:00
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amanvermagmat wrote:
ABCDEF is a regular hexagon with its vertices as A, B, C, D, E , F in this order. What is the area of triangle BDF?

(1) BDF is an equilateral triangle.

(2) AB = 6 units.


Since ABCDEF is a regular hexagon, hence the distance between any two alternate vertices are same. So BDF is an equilateral triangle.

Question stem:- Area of triangle BDF?

St1:- BDF is an equilateral triangle.
No new info, it is already inferred from the given data in the question.
Insufficient to determine area without any info on side or altitude of the triangle.

St2:-AB = 6 units
So,\(FB=6\sqrt{3}\) (The equilateral triangle inside a regular hexagon of side 'a' has it's side as \(\sqrt{3}*a)\)
So, area of the equilateral triangle DEF can be determined when length of side is known.
\(Area=(\sqrt{3}/4)*FB^2\)

So, sufficient.

Ans (B)
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Re: ABCDEF is a regular hexagon with its vertices as A, B, C, D, E , F in  [#permalink]

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New post 13 Jul 2018, 22:09
PKN wrote:
amanvermagmat wrote:
ABCDEF is a regular hexagon with its vertices as A, B, C, D, E , F in this order. What is the area of triangle BDF?

(1) BDF is an equilateral triangle.

(2) AB = 6 units.


Since ABCDEF is a regular hexagon, hence the distance between any two alternate vertices are same. So BDF is an equilateral triangle.

Question stem:- Area of triangle BDF?

St1:- BDF is an equilateral triangle.
No new info, it is already inferred from the given data in the question.
Insufficient to determine area without any info on side or altitude of the triangle.

St2:-AB = 6 units
So,\(FB=6\sqrt{3}\) (The equilateral triangle inside a regular hexagon of side 'a' has it's side as \(\sqrt{3}*a)\)
So, area of the equilateral triangle DEF can be determined when length of side is known.
\(Area=(\sqrt{3}/4)*FB^2\)

So, sufficient.

Ans (B)


Can You please explain the assumption of statement 2 [i]" The equilateral triangle inside a regular hexagon of side 'a' has its side as √3∗a "??
:(
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Re: ABCDEF is a regular hexagon with its vertices as A, B, C, D, E , F in  [#permalink]

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New post 13 Jul 2018, 22:41
amanvermagmat wrote:
ABCDEF is a regular hexagon with its vertices as A, B, C, D, E , F in this order. What is the area of triangle BDF?

(1) BDF is an equilateral triangle.

(2) AB = 6 units.


See the attached figure..
    1. regular hexagon, so all sides will be equal.
    2. total measure of angle = (n-2)*180=(6-2)*180=4*180, therefore each angle = \(\frac{4*180}{6}=120\)
    3. also BDF will be an equilateral triangle
    4. to find area of an equialteral triangle, we require the side


(1) BDF is an equilateral triangle.

BDF will always be an equilateral triangle, so statement I does not add anything new
hence insufficient

now
(2) AB = 6 units.

Draw a perpendicular on DF from E
angle OED will be HALF of angle DEF, so \(\frac{120}{2}=60\)
thus OED is 30:60:90 triangle and sides OE:OD:DE are in ratio 1:\(\sqrt{3}\):2
Now AB=DE=6, so OE:OD:DE=1:\(\sqrt{3}\):2=OE:OD:6
so OD=\(\sqrt{3}*\frac{6}{2}=3\sqrt{3}\)
also side of equilatera triangle BDF = DF=2*OD=2*3\(\sqrt{3}\)
since we know the side, we can find the area
sufficient

B
Attachments

1.png
1.png [ 5.5 KiB | Viewed 523 times ]


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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: ABCDEF is a regular hexagon with its vertices as A, B, C, D, E , F in  [#permalink]

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New post 13 Jul 2018, 22:42
swagatamoitra wrote:
PKN wrote:
amanvermagmat wrote:
ABCDEF is a regular hexagon with its vertices as A, B, C, D, E , F in this order. What is the area of triangle BDF?

(1) BDF is an equilateral triangle.

(2) AB = 6 units.


Since ABCDEF is a regular hexagon, hence the distance between any two alternate vertices are same. So BDF is an equilateral triangle.

Question stem:- Area of triangle BDF?

St1:- BDF is an equilateral triangle.
No new info, it is already inferred from the given data in the question.
Insufficient to determine area without any info on side or altitude of the triangle.

St2:-AB = 6 units
So,\(FB=6\sqrt{3}\) (The equilateral triangle inside a regular hexagon of side 'a' has it's side as \(\sqrt{3}*a)\)
So, area of the equilateral triangle DEF can be determined when length of side is known.
\(Area=(\sqrt{3}/4)*FB^2\)

So, sufficient.

Ans (B)


Can You please explain the assumption of statement 2 [i]" The equilateral triangle inside a regular hexagon of side 'a' has its side as √3∗a "??
:(


Dear swagatamoitra,

As I have already stated in my explanation, the cited statement is not an assumption rather a derivation obtained from the given data.

The derivation is enclosed herewith for your easy reference. Please note that the diagram is not drawn to scale.

Hope it helps.

P.S:- It saves time if properties and derivations are always alive with you. Otherwise, you can deduce always.
Attachments

Hexagon.jpg
Hexagon.jpg [ 196.62 KiB | Viewed 521 times ]


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Re: ABCDEF is a regular hexagon with its vertices as A, B, C, D, E , F in &nbs [#permalink] 13 Jul 2018, 22:42
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