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655-705 (Hard)|   Geometry|      
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Bunuel
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Two equilateral triangles are contained in square whose side length is 22 Apr 2019, 01:52
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Difficulty:

65% (hard)

Question Stats:

50% (02:59) correct 50% (03:17) wrong based on 36 sessions

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Two equilateral triangles are contained in square whose side length is \(2\sqrt 3\). The bases of these triangles are the opposite side of the square, and their intersection is a rhombus. What is the area of the rhombus?


(A) 3/2

(B) \(\sqrt{3}\)

(C) \(2\sqrt{2}-1\)

(D) \(8\sqrt{3}-12\)

(E) \(\frac{4\sqrt{3}}{3}\)
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D
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Two equilateral triangles are contained in square whose side length is 22 Apr 2019, 19:10
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Bunuel
Two equilateral triangles are contained in square whose side length is \sqrt{2\sqrt 3}. The bases of these triangles are the opposite side of the square, and their intersection is a rhombus. What is the area of the rhombus?


(A) 3/2

(B) \(\sqrt{3}\)

(C) \(2\sqrt{2}-1\)

(D) \(8\sqrt{3}-12\)

(E) \(\frac{4\sqrt{3}}{3}\)
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Hi Bunuel, Please confirm the side length of the square.
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Two equilateral triangles are contained in square whose side length is 23 Apr 2019, 01:19
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PKN
Bunuel
Two equilateral triangles are contained in square whose side length is \sqrt{2\sqrt 3}. The bases of these triangles are the opposite side of the square, and their intersection is a rhombus. What is the area of the rhombus?


(A) 3/2

(B) \(\sqrt{3}\)

(C) \(2\sqrt{2}-1\)

(D) \(8\sqrt{3}-12\)

(E) \(\frac{4\sqrt{3}}{3}\)

Hi Bunuel, Please confirm the side length of the square.
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____________________________
Fixed. Kudos for noticing!
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neha283
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Two equilateral triangles are contained in square whose side length is 17 May 2019, 04:25
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I tried with (diagonal 1 * diagonal 2 )/2 . One of the diagonal's length will be equal to side of square, i.e, 2 \sqrt{\(3\)}. I know its wrong to assume the other diagonal will be one-third of the side of square. What is the correct approach for this question?
Tried looking at subtracting some areas from square to get rhombus but could not find a solution.
Can someone post a solution for this?
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Two equilateral triangles are contained in square whose side length is 17 May 2019, 22:36
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Refer the enclosed diagram.

GH=\(2\sqrt{3}-1-1=2\sqrt{3}-2\)

IGH & JGH are symmetrical equilateral triangles.

Area of the rhombus IGJH=2* area of equilateral triangle IGH or JGH=\(2*\frac{\sqrt{3}}{4}\)*\((2\sqrt{3}-2)^2\)=\(8\sqrt{3}-12\)

Ans. (D)
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Rhombus.JPG
Rhombus.JPG [ 28.62 KiB | Viewed 3481 times ]

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Two equilateral triangles are contained in square whose side length is 18 May 2019, 01:47
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PKN
Refer the enclosed diagram.

GH=\(2\sqrt{3}-1-1=2\sqrt{3}-2\)

IGH & JGH are symmetrical equilateral triangles.

Area of the rhombus IGJH=2* area of equilateral triangle IGH or JGH=\(2*\frac{\sqrt{3}}{4}\)*\((2\sqrt{3}-2)^2\)=\(8\sqrt{3}-12\)

Ans. (D)
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Hi PKN
The question says the triangles should be equilateral, triangle DIC and AJB are not equilateral, all 3 sides should be \(2\sqrt{3}\)
IMO, the figure should look something like this(Attached)
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Figure 1.png
Figure 1.png [ 15.76 KiB | Viewed 3414 times ]

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Two equilateral triangles are contained in square whose side length is 18 May 2019, 04:12
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neha283
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Refer the enclosed diagram.

GH=\(2\sqrt{3}-1-1=2\sqrt{3}-2\)

IGH & JGH are symmetrical equilateral triangles.

Area of the rhombus IGJH=2* area of equilateral triangle IGH or JGH=\(2*\frac{\sqrt{3}}{4}\)*\((2\sqrt{3}-2)^2\)=\(8\sqrt{3}-12\)

Ans. (D)

Hi PKN
The question says the triangles should be equilateral, triangle DIC and AJB are not equilateral, all 3 sides should be \(2\sqrt{3}\)
IMO, the figure should look something like this(Attached)
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Which triangles are the equilateral triangles here?
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