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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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Bunuel

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) \(15 \sqrt{3}\)

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Source: Chili Hot GMAT

The correct answer is C.
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∆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

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?



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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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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