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In the equilateral triangle below, each side has a length of 4 units.

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In the equilateral triangle below, each side has a length of 4 units.  [#permalink]

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New post 05 Dec 2014, 07:41
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Tough and Tricky questions: Geometry.



In the equilateral triangle below, each side has a length of 4 units. If PQ has a length of 1 unit and TQ is perpendicular to PR, what is the area of region QRST?
Image


A) \(\frac{1}{3} \sqrt{3}\)

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

C) \(\frac{7}{2} \sqrt{3}\)

D) \(4 \sqrt{3}\)

E) \(15 \sqrt{3}\)


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Source: Chili Hot GMAT

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Re: In the equilateral triangle below, each side has a length of 4 units.  [#permalink]

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New post 05 Dec 2014, 10:45
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Sorry for not having pictures, but hopefully the explanation will suffice.

The area of the big triangle (Triangle PRS) can be calculated by the area of an equilateral triangle, A = side^2*sqrt(3)/4. This yields 4*sqrt(3) for the big triangle.

Now for the smaller triangle, let's name this Triangle PQT with angles p, q and t (just to abbreviate). We know angle p = 60 degrees (it's one of the angles in the large equilateral triangle), angle q = 90 degrees as QT is perpendicular to PR, which means that angle t = 30 degrees. From this information, we have a 30:60:90 triangle with sides in the ratio of 1:sqrt(3):2. From this, the area of Triangle PQT = 1/2*PQ*QT = 1/2*1*sqrt(3) = sqrt(3)/2.

Now if we subtract the area of Triangle PQT from the area of Triangle PRS, we get the area of the shape QRST, which is 4sqrt(3) - 0.5sqrt(3) = 3.5*sqrt(3).

Choice C
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Re: In the equilateral triangle below, each side has a length of 4 units.  [#permalink]

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New post 05 Dec 2014, 12:54
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In the equilateral triangle PRS, each side has a length of 4 units.

Therefore, area of the equilateral triangle PRS = \(4^2*\sqrt{3}/4 = 4 * \sqrt{3}\)

In the right angle triangle \(\triangle PQT\), PQ has a length of 1 unit and TQ is perpendicular to PR.

Also, in the right angle triangle \(\triangle\) PQT, the angle QPT and angle PTQ are \(60^{\circ}\) and \(30^{\circ}\) respectively.

Therefore using the property of 30-60-90 triangle, \(PT = 2\) and \(QT = \sqrt{3}\)

So, area of the triangle \(\triangle\) PQT= \(\sqrt{3}\)/2

Area of the region QRST = Area of equilateral triangle PRS - Area of the right angle triangle PQT

=\(\sqrt{3}*{4}- \sqrt{3}/2\) \(= \sqrt{3}* 7/2\)

Answer: C
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Re: In the equilateral triangle below, each side has a length of 4 units.  [#permalink]

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New post 08 Dec 2014, 06:10
Bunuel wrote:

Tough and Tricky questions: Geometry.



In the equilateral triangle below, each side has a length of 4 units. If PQ has a length of 1 unit and TQ is perpendicular to PR, what is the area of region QRST?
Attachment:
2014-12-05_1839.png


A) \(\frac{1}{3} \sqrt{3}\)

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

C) \(\frac{7}{2} \sqrt{3}\)

D) \(4 \sqrt{3}\)

E) \(15 \sqrt{3}\)

Kudos for a correct solution.

Source: Chili Hot GMAT


The correct answer is C.
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Re: In the equilateral triangle below, each side has a length of 4 units.  [#permalink]

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New post 08 Dec 2014, 19:10
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Answer = C) \(\frac{7}{2} \sqrt{3}\)

Attachment:
2014-12-05_1839.png
2014-12-05_1839.png [ 6.19 KiB | Viewed 2973 times ]


Area of equilateral triangle \(= \frac{\sqrt{3}}{4} 4^2 = 4\sqrt{3}\)

Area of Right Angle triangle (30-60-90) (Sides \(1 - \sqrt{3} - 2\)) \(= \frac{1}{2} * 1 * \sqrt{3} = \frac{1}{2} \sqrt{3}\)

Area of shaded region \(= 4\sqrt{3} - \frac{1}{2} \sqrt{3} = \frac{7}{2} \sqrt{3}\)
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Re: In the equilateral triangle below, each side has a length of 4 units.  [#permalink]

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New post 31 Mar 2019, 02:12
∆QTP ; 30:60:90
side PT= 2 and QT = √3; area of ∆QTP = √3/2
area of ∆PRS; 4√3
so area of side QRTS = 4√3-√3/2
IMO C
\(\frac{7}{2} \sqrt{3}\)

Bunuel wrote:

Tough and Tricky questions: Geometry.



In the equilateral triangle below, each side has a length of 4 units. If PQ has a length of 1 unit and TQ is perpendicular to PR, what is the area of region QRST?
Image


A) \(\frac{1}{3} \sqrt{3}\)

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

C) \(\frac{7}{2} \sqrt{3}\)

D) \(4 \sqrt{3}\)

E) \(15 \sqrt{3}\)


Kudos for a correct solution.

Source: Chili Hot GMAT

Attachment:
2014-12-05_1839.png
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Re: In the equilateral triangle below, each side has a length of 4 units.  [#permalink]

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New post 10 Jun 2019, 05:45
Area of the quadrilateral=area of the triangle PRS- area of the triangle PQT

By applying the the rule of 1:sqrt3:2 in the 30:60:90 triangle PQT, we get the height QT=sqrt3.

now the area of PRS- area of PQT
4 * sqrt3 - sqrt3/2
(7* sqrt3)/2
Choice C
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Re: In the equilateral triangle below, each side has a length of 4 units.   [#permalink] 10 Jun 2019, 05:45
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