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If arc PQR above is a semicircle, what is the length of [#permalink]
 20 Sep 2005, 15:54
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If arc PQR above is a semicircle, what is the length of diameter PR? (1) a=4 (2) b=1 OPEN DISCUSSION OF THE QUESTION IS HERE: if-arc-pqr-above-is-a-semicircle-what-is-the-length-of-144057.html
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Last edited by Bunuel on 22 Dec 2012, 05:18, edited 2 times in total.
OA added.
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D
each is enough because angles on a straight line adds up to 180, so since one of the angles is right angle (90), then either a or b can be manipulated to get the length of the diameter
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Folaa3, can you explain your solution? I didn't get it. I will choose C but then it will make this problem the simplest of all.
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Yes, C is obviously sufficient but it seems almost too obvious. However, I cannot see how either a or b alone are sufficient.
We know both triangles are right-angle triangles. That much is obvious. But they're not necessarily similar, nor do we have the radius or the center or any information that allows us to use the angles or the arcs to calculate it.
Basically, what I'm trying to say is, I don't get it either.
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D.
let
pq = x
x^2 = 4 + a^2 ---- (i)
similarly, if qr = y
y^2 = 4 + b^2 ---- (ii)
as triangle pqr is right angle traingle,
x^2 + y^2 = (a + b) ^2 or
4 + a ^2 + 4 + b^2 = a^2 + b^2 + 2*a*b or
ab = 4
given a or b we can find the other one.
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there are many possible right triangles, you can think of connecting any
point on arc pqr with p and r, you will have one such triangle. But as soon
as you decide how far away it is from p or r (on the diameter line), you have it fixed. now given two sides of a right triangle, you can get the third side and the angle, therefore the diameter is solvable with either a or b known.
D
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OA is D) ! IMO the best way to solve it is to think. ask yourself the question: can i draw this figure ? yes i can ! i know that the larger triangle is right. so i draw the longer leg that is 4 and i draw the shorter leg that is 2. i draw the hypothenuse that connects the legs. the vertex that lies on the circle is right as well. so i can draw the shorter triangle. at the end i just measure the diameter. it is...??? it doesn matter ! it only matters that i can get the value of it. i dont need more that 15s to come to this conclusion without drawing any eq`s on my scratch paper.
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how do you know triangle BQR is right angle?
that would help me understand why D is the answer...
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fresinha12 wrote: how do you know triangle BQR is right angle?
that would help me understand why D is the answer...
angle BQR is not right.
QBR, PBQ & PQR are.
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PQR is a rt angle triangle becos any triangle drawn in a semicircle is 90 degrees.
The rest follows.
GA
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if arc pqr above is a semicircle, what is the length of diameter pr ? 1) a=4 2) b=1 What information have we got from the figure: Diameter, (pr)=a+b Let's call the point on which perpendicular line from point q intersects with the Diameter "pr" as m. qm=2 qmp and qmr are two right triangles with hypotenuse as pq and qr respectively. Another thing that is not explicitly mentioned here is; Inscribed angle subtended by a diameter is always a right angle. Diameter pr is making an angle at point q on the circle; it must be a 90 degree angle. So; Triangle rqp is a right angle triangle. Let's make some useful equations out of the information we got so far; Using pythagoras theorem; a^2+(qm)^2=(pq)^2(pq)^2=a^2+2^2b^2+(qm)^2=(qr)^2(qr)^2=b^2+2^2(pq)^2+(qr)^2=(pr)^2a^2+2^2+b^2+2^2=(a+b)^2a^2+2^2+b^2+2^2=a^2+b^2+2ab2^2+2^2=2ab8=2abab=4---- This is the final equation we arrived at just by looking at the figure (1) a=4; ab=4; b=1; Diameter, pr=a+b=4+1=5. Sufficient. (2) b=1; ab=4; a=4; Diameter, pr=a+b=4+1=5. Sufficient. Ans: "D"
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christoph wrote: if arc pqr above is a semicircle, what is the length of diameter pr ?
1) a=4
2) b=1 Attachment: 
untitled.PNG [ 4.95 KiB | Viewed 3589 times ]
If arc PQR above is a semicircle, what is the length of diameter PR? You should know the following properties to solve this question: • A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle.So, as given that PR is a diameter then angle PQR is a right angle. • Perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle. Thus, the perpendicular QT divides right triangle PQR into two similar triangles PQT and QRT (which are also similar to big triangle PQR). Now, in these three triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles). For example: QR/PR=QT/PQ=TR/QR. This property (sometimes along with Pythagoras) will give us the following: if we know ANY 2 values from PR, PQ, QR, PT, QT, TR then we'll be able to find other 4. We are given that QT=2 thus to find PR we need to know the length of any other line segment. Also in such kind of triangles might be useful to equate the areas to find the length of some line segment, for example area of PQR=1/2*QT*PR=1/2*QP*QR (for more check: triangles-106177.html, geometry-problem-106009.html, mgmat-ds-help-94037.html, help-108776.html) (1) a = 4. Sufficient. (2) b = 1. Sufficient. Answer: D.
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+1 kudos  wooooooow! thank you bunuel!  your solution is very prefect and useful and nice and good and excellent and ...
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Similar triangle approach, t be the intersection point of perpendicular from Q. thus QT is altitude for triangle PQR. PQ = 20^1/2, Hence pq/pr = pt/pq [base/ hypotenuse] thus we get pr = 20/4 = 5 . A sufficient. QR= 5^1/2 , Hence qr/pr = rt/qr thus we get pr = 5. B sufficient. Hence D.
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Last edited by amit2k9 on 04 May 2011, 00:25, edited 2 times in total.
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a^2 + 4 = pq^2 b^2 + 4 = qr^2 (a+b)^2 = pq^2 + qr^2 1/2(a+b) * 2 = 1/2 * pq * qr The diameter pr = a+b (1) pq = root(20) (a+b)^2 = 20 + qr^2 2(a+b) = root(20)*qr => 4(a+b)^2 = 20 * qr^2 => (a+b)^2 = 20 + (a+b)^2/5 => (a+b)^2 = 5/4 * 20 = 25 => a+b = 5 (2) b^2 + 4 = qr^2 qr = root(5) (a+b)^2 = pq^2 + 5 2(a+b) = root(5)*pq 4(a+b)^2 = 5 * pq^2 (a+b)^2 = 4(a+b)^2/5 + 5 => (a+b)^2/5 = 5 => (a+b) = 25 Answer - D
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Re: if arc pqr above is a semicircle, what is the length of [#permalink]
28 Oct 2012, 03:25
Bunuel and Others, Remembering that QT^2 = PT * RT will help us get the soln in the minutest amount of time.. 
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Re: If arc PQR above is a semicircle, what is the length of [#permalink]
01 Nov 2012, 02:44
I solved this problem by using a property of right triangle that I learnt at high school: PQ^2 = PT * PR = a * PRQR^2 = RT * PR = b* PRFrom (1): I can calculate the value of PQ, then of PR ==> suff From (2): I can calculate the value of RT, then of PR ==> suff
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Re: If arc PQR above is a semicircle, what is the length of [#permalink]
21 Dec 2012, 22:20
I don't understand why we can assume that angle PQR is 90 degrees.
If PQR was 90 degrees, the answer should be obviously D because of the similar triangles rule. However, as Q moves about the circle, that angle changes degrees.
And we are not given information as to what that angle may be, so I chose C. I thought this was a trap question from GMAT to make us want to assume that angle PQR was 90 degrees.
Can anyone help here please.
Why can we assume angle PQR is 90. that is the question. thanks.
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Re: If arc PQR above is a semicircle, what is the length of [#permalink]
22 Dec 2012, 05:18
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Re: If arc PQR above is a semicircle, what is the length of
[#permalink]
22 Dec 2012, 05:18
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