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In a rhombus ABCD, AC = 3 units and angle ABC = 120°. What is the area

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In a rhombus ABCD, AC = 3 units and angle ABC = 120°. What is the area  [#permalink]

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New post 17 Jul 2019, 04:39
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In a rhombus ABCD, AC = 3 units and angle ABC = 120°. What is the area of ABCD? (in unit\(^2\))

A. \(3\sqrt{3}/8\)
B. \(3\sqrt{3}/4\)
C. \(3\sqrt{3}/2\)
D. \(3\sqrt{3}\)
E. \(6\sqrt{3}\)
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Re: In a rhombus ABCD, AC = 3 units and angle ABC = 120°. What is the area  [#permalink]

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New post 17 Jul 2019, 05:26
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Dillesh4096 wrote:
In a rhombus ABCD, AC = 3 units and angle ABC = 120°. What is the area of ABCD? (in unit\(^2\))

A. \(3\sqrt{3}/8\)
B. \(3\sqrt{3}/4\)
C. \(3\sqrt{3}/2\)
D. \(3\sqrt{3}\)
E. \(6\sqrt{3}\)


Rhombus:
1.Diagonals are perpendicular bisector of each other.
2. Adjacent angles are supplementary.
AC= 3
Diagonals bisect each other at P(say) AC, BD
AP+PC=AC
AP= 3/2
IN TRIANGLE -ABP
As adjacent angles are supplementary angle
angle at A= 60
angle ABC=120
as Diagonals are perpendicular bisector of each other
in triangle ABP( 30-60-90)
We know AP=3/2
BP=square root 3/2
area= (AC X BD)/2
=(3 X 2 BP)/2
= 3 square root 3/2
C
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Re: In a rhombus ABCD, AC = 3 units and angle ABC = 120°. What is the area  [#permalink]

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New post 17 Jul 2019, 19:02
In triangle ABC
\(\frac{AB}{sin30}\)=\(\frac{AC}{sin120}\)
AB=\(\sqrt{3}\)

Area=\(AB^2 sin120\) = \(\sqrt{3}^2*\sqrt{3}/2\)= \(3\sqrt{3}/2\)

Dillesh4096 wrote:
In a rhombus ABCD, AC = 3 units and angle ABC = 120°. What is the area of ABCD? (in unit\(^2\))

A. \(3\sqrt{3}/8\)
B. \(3\sqrt{3}/4\)
C. \(3\sqrt{3}/2\)
D. \(3\sqrt{3}\)
E. \(6\sqrt{3}\)
Intern
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Re: In a rhombus ABCD, AC = 3 units and angle ABC = 120°. What is the area  [#permalink]

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New post 08 Aug 2019, 04:07
Bunuel please help with this. I'm getting 4.5 root3
Got it buy 30-60-90 triangle. Where am i wrong?

Posted from my mobile device
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Re: In a rhombus ABCD, AC = 3 units and angle ABC = 120°. What is the area  [#permalink]

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New post 14 Aug 2019, 01:47
satya2029 wrote:
Dillesh4096 wrote:
In a rhombus ABCD, AC = 3 units and angle ABC = 120°. What is the area of ABCD? (in unit\(^2\))

A. \(3\sqrt{3}/8\)
B. \(3\sqrt{3}/4\)
C. \(3\sqrt{3}/2\)
D. \(3\sqrt{3}\)
E. \(6\sqrt{3}\)


Rhombus:
1.Diagonals are perpendicular bisector of each other.
2. Adjacent angles are supplementary.
AC= 3
Diagonals bisect each other at P(say) AC, BD
AP+PC=AC
AP= 3/2
IN TRIANGLE -ABP
As adjacent angles are supplementary angle
angle at A= 60
angle ABC=120
as Diagonals are perpendicular bisector of each other
in triangle ABP( 30-60-90)
We know AP=3/2
BP=square root 3/2
area= (AC X BD)/2
=(3 X 2 BP)/2
= 3 square root 3/2
C

Can someone please help on how to get from 3/2 (60°) to square root 3/2 (30°)? For the 30-60-90 Triangle
GMAT Club Bot
Re: In a rhombus ABCD, AC = 3 units and angle ABC = 120°. What is the area   [#permalink] 14 Aug 2019, 01:47
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In a rhombus ABCD, AC = 3 units and angle ABC = 120°. What is the area

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