In the figure above, PQ is a diameter of circle O, PR = SQ, and ΔRST

Question

https://gmatclub.com/forum/download/file.php?id=50468 In the figure above, PQ is a diameter of circle O, PR = SQ, and ΔRST is equilateral. If the length of PQ is 2, what is the length of RT ? A. \frac{1}{2} B. \frac{1}{\sqrt{3}} C. \frac{\sqrt{3}}{2} D. \frac{2}{\sqrt{3}} E. \sqrt{3} PS57402.01 2019-09-21_1817.png

BrentGMATPrepNow · Accepted Answer

GIVEN: The length of PQ is 2
In other words, the DIAMETER = 2
From this, we can conclude that the RADIUS =   1  
So, we can add this information to our diagram:
 https://i.imgur.com/WQ0yIiN.png

Since ΔRST is equilateral, we know the altitude (TO) will be a  perpendicular  bisector of side RS
Also, since ΔRST is equilateral, we know that ∠ORT = 60°, which also means ∠RTO = 30°
 https://i.imgur.com/YR7oSGV.png

At this point, we can see that  ΔTRO  is a special 30-60-90 triangle. 
If we compare  ΔTRO  with the  BASE 30-60-90 triangle , we can create an equation by comparing   corresponding sides

We can write:  x / 2  =  1 / (√3)

Multiply both sides by 2 to get: x = 2/(√3)

Answer: D

MikeScarn · Answer

Diameter equals 2. So our radius is 1.

We can draw a line down from T to O, which would = radius = 1

Splitting our Equilateral Triangle into two 30,60,90 triangles.

Property of 30, 60, 90 Triangle:  The sides are always in ratio 1: \sqrt{3} : 2

So our ratio is   x : 1 : 2x

Solve for x  ----> x *  \sqrt{3}  = 1 ------> x =  \frac{1}{\sqrt{3}}

So our side lengths are:    \frac{1}{\sqrt{3}}  : 1 :  \frac{2}{\sqrt{3}}

RT=TS=ST= \frac{2}{\sqrt{3}}

Archit3110 · Answer

OT = hieght = √3/2 *x = radius

√3/2 *x*2= 2
x= 2/√3
IMO D  \frac{2}{\sqrt{3}}

satya2029 · Answer

OT=1(HALF OF THE DIAMETER)
Half of an equilateral triangle is 30-60-90 triangle.
in 30-60-90 triangle sides are a:a√3:2a ratio
here a√3=1
2a=RT=2/√3
D:)

ckadams · Answer

In the figure above, PQ is a diameter of circle O, PR = SQ, and ΔRST is equilateral. If the length of PQ is 2, what is the length of RT ?

I am a little confused... I thought that the gmat would want the denominator rationalized e.g. multiplied by \frac{\sqrt{3}}{\sqrt{3}}

gmatway · Answer

I din't understand what clarity you want . The question is straight forward it says what is the value of RT we know RT = RS =ST . So lets find RS .
We know height of equilateral triangle is root3/2 (side) . So 1 = root 3/2 hence side = RS =RT =2/root 3 .

ash124 · Answer

We can solve the question using area formula as well

lets take 'a' as a side of the eq. triangle

1/2 * base * height = Area of eq. Triangle

1/2 * a * 1 = √3/4 * a*a

solving the equation 
a = 2/√3

Basshead · Answer

Since PQ = diameter = 2, we can conclude TO = 1

RST is an equilateral triangle; if we draw a line from T to O, we will have two 30:60:90 triangles.

The ratio for a 30:60:90 triangle is  1:\sqrt{3}:2 .

1 = 1/\sqrt{3} 
 2 = 2/\sqrt{3}

Answer is D.

vickytvn · Answer

BrentGMATPrepNow
What is the use of data given PR= SQ in this question? Is it to confirm that Line from T to O will be median thus perpendicular bisector in equilateral triangle

BrentGMATPrepNow · Answer

That's correct. 
In other words, the equilateral triangle is centered on diameter PQ.

Hoozan · Answer

BrentGMATPrepNow   KarishmaB  based on what logic can we say that the radius will be the altitude i.e. the angle TOQ = 90°?

What if we have a circle with Centre O such that point T is not exactly above O i.e. angle TOQ ≠ 90°. In that case, we cannot use the 30-60-90 property.

BrentGMATPrepNow · Answer

Great question.

Since PR = SQ, we know that the equilateral triangle is centered on the diameter PQ (in other words 0 is in the middle of side RS).
So, the altitude of equilateral triangle RTS will pass through the center (since the altitude of an equilateral triangle is the perpendicular bisector of its base)

KarishmaB · Answer

PO = OQ (Both radii) Screenshot 2022-03-12 at 12.31.54.png We also know that PR = SQ. Then the leftover parts (RO and OS) are equal. So RO = OS and O is right in the middle of the base of the equilateral triangle. Since it is an equilateral triangle, it will be symmetrical and the point T will be right above O. So OT will be the altitude of the triangle RST. OT is also the radius so it is 1. OTS gives us a 30-60-90 triangle so OT : TS is in the ratio sqrt(3) : 2. Since OT = 1, TS = 2/sqrt(3) Answer (D)

tomercahn · Answer

I solved the problem a bit differently..

We know that PO=1 (hence TO=1 and PT=\sqrt{2}).
From these we can conclude that RT will be between 1 and \sqrt{2}.

bumpbot · Answer

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In the figure above, PQ is a diameter of circle O, PR = SQ, and ΔRST

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In the figure above, PQ is a diameter of circle O, PR = SQ, and ΔRST

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