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

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Director
Joined: 20 Jul 2017
Posts: 636
Location: India
Concentration: Entrepreneurship, Marketing
WE: Education (Education)
In a rhombus ABCD, AC = 3 units and angle ABC = 120°. What is the area  [#permalink]

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17 Jul 2019, 04:39
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61% (01:47) correct 39% (02:48) wrong based on 51 sessions

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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. $$6\sqrt{3}$$
Intern
Joined: 10 Dec 2017
Posts: 27
Location: India
Re: In a rhombus ABCD, AC = 3 units and angle ABC = 120°. What is the area  [#permalink]

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17 Jul 2019, 05:26
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. $$6\sqrt{3}$$

Rhombus:
1.Diagonals are perpendicular bisector of each other.
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
Director
Joined: 19 Oct 2018
Posts: 791
Location: India
Re: In a rhombus ABCD, AC = 3 units and angle ABC = 120°. What is the area  [#permalink]

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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
Joined: 11 May 2019
Posts: 12
Re: In a rhombus ABCD, AC = 3 units and angle ABC = 120°. What is the area  [#permalink]

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08 Aug 2019, 04:07
Got it buy 30-60-90 triangle. Where am i wrong?

Posted from my mobile device
Manager
Joined: 20 Apr 2019
Posts: 82
Re: In a rhombus ABCD, AC = 3 units and angle ABC = 120°. What is the area  [#permalink]

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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.
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
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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