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If arc PQR above is a semicircle, what is the length of diameter PR ? [#permalink]
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13 Dec 2012, 07:16
26
67
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.
Re: If arc PQR above is a semicircle, what is the length of diameter PR ? [#permalink]
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22 Dec 2012, 06:04
Walkabout wrote:
Attachment:
The attachment Semicircle2.png is no longer available
If arc PQR above is a semicircle, what is the length of diameter PR ?
(1) a = 4 (2) b= 1
Another approach that could be implemented in thsi question is: Since there is a perpendicular drawn to the hypotenuese, therefore the two triangles that are formed must be similar to each other and to the larger one.
So if one side of a triangle reduces by a certain ratio, the other side must also reduce. In the diagram attached, if one considers any of the statement then he will be able to find out the other side. Consider statement 1) a=4
Look into the diagram. In the middle triangle, "a" or PI=4. We are given with the fact that IQ=2. Now in the smallest triangle, the corresponding side of PI=IQ. IQ=2. Therefore the factor with which PI has reduced is 2. Therefore other side must also reduce by the same factor. Hence IR=1. Sufficient
Statement 2) b=1. "b" is the corresponding side of IQ. So IQ , in the middle traingle, has reduced by a factor of 2. In the smallest triangle IQ=2. Therefore PI must be 4. Sufficient.
Attachments
geometry solution.png [ 13.23 KiB | Viewed 29421 times ]
Re: If arc PQR above is a semicircle, what is the length of diameter PR ? [#permalink]
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29 Mar 2014, 03:55
Can anybody explain why the area of the triangle PQR : Which is a right angle triangle in the figure above at Q, is not 1/2*QR*PQ ? Assuming this triangle was drawn without the semi-circle and if I slightly redraw the triangle keeping the base at QR, and PQ becomes the height and PR is the hypotenuse, then isnt the area of the triangle 1/2*QR*PQ?
Why is it that in these type of triangles which are drawn in such manner, that the hypotenuse is the base, the height is drawn from one vertex to another and then area is calculated?
This question is just to douse this silly doubt lingering in my head.
Re: If arc PQR above is a semicircle, what is the length of diameter PR ? [#permalink]
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29 Mar 2014, 03:58
sudeeptasahu29 wrote:
Can anybody explain why the area of the triangle PQR : Which is a right angle triangle in the figure above at Q, is not 1/2*QR*PQ ? Assuming this triangle was drawn without the semi-circle and if I slightly redraw the triangle keeping the base at QR, and PQ becomes the height and PR is the hypotenuse, then isnt the area of the triangle 1/2*QR*PQ?
Why is it that in these type of triangles which are drawn in such manner, that the hypotenuse is the base, the height is drawn from one vertex to another and then area is calculated?
This question is just to douse this silly doubt lingering in my head.
The area of triangle PQR IS 1/2*PQ*QR but it's ALSO 1/2*PR*QT.
_________________
Re: If arc PQR above is a semicircle, what is the length of diameter PR ? [#permalink]
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16 Jul 2014, 23:12
1
Bunuel wrote:
If arc PQR above is a semicircle, what is the length of diameter PR?
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
(1) a = 4. Sufficient.
(2) b = 1. Sufficient.
Answer: D.
1/2*QT*PR=1/2*QP*QR => PR = (QP*QR)/2
With this equation, how statement 1 or 2 is being judged. Will you please fill up the gap?
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Re: If arc PQR above is a semicircle, what is the length of diameter PR ? [#permalink]
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17 Jul 2014, 08:46
2
3
musunna wrote:
Bunuel wrote:
If arc PQR above is a semicircle, what is the length of diameter PR?
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
(1) a = 4. Sufficient.
(2) b = 1. Sufficient.
Answer: D.
1/2*QT*PR=1/2*QP*QR => PR = (QP*QR)/2
With this equation, how statement 1 or 2 is being judged. Will you please fill up the gap?
It's better to use ratios for this question.
For (1) use the following ratio: PT/QT = QT/RT --> 4/2 = 2/RT --> RT = 1 --> PR = PT + RT = 4 + 1 = 5.
For (2) use the the same ratio: PT/QT = QT/RT --> PT/2 = 2/1--> PT = 4 --> PR = PT + RT = 4 + 1 = 5.
Re: If arc PQR above is a semicircle, what is the length of diameter PR ? [#permalink]
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02 Jun 2015, 01:02
Bunuel wrote:
Attachment:
Semicircle2.PNG
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.
we do not need to know this property. just make an equation in which b is a unknown number . we can solve
we can not remember this property in the test room.
the og explanation of this problem is good enough.
there are 5 or 6 person who following me whenever I go out of my house. I feel not safe. those persons prevent me from looking for the British I met in Halong bay. before I worked for a janapese company
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visit my facebook to help me. on facebook, my name is: thang thang thang
Re: If arc PQR above is a semicircle, what is the length of diameter PR ? [#permalink]
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10 May 2016, 06:04
1
Do you seriously think that question will ask what is a+b = ? So that , from stat (1) + (2) you can say a + b = 5 ; GMAT will never ask addition of two numbers , so cut out the option C and try to get the soln from A and from B alone.
Re: If arc PQR above is a semicircle, what is the length of diameter PR ? [#permalink]
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31 Jul 2016, 01:01
1
Walkabout wrote:
Attachment:
Semicircle.png
If arc PQR above is a semicircle, what is the length of diameter PR ?
(1) a = 4 (2) b = 1
I did this in a different way and still got the same answer, So if you look at the property of a semicircle, then if the triangle is formed using the diameter of the semicircle the the angle subtended by the angle formed at the arc is 90
Now given that Pythagorus theorem for the bog triangle PQR, \(PQ^2=QR^2+PR^2\); \(PR^2= (a+b)^2\) Now let PQ be x and QR be y then \(x^2+y^2= (a+b)^2\)---eq1
Smaller triangles PQa \(x^2=a^2+4\); QRb \(y^2= b^2+4\)--- eq2,3 now eq 2+3 \(x^2+y^2=a^2+b^2+8\) equate the same with eq 1 and get rid of x and y
\((a+b)^2=a^2+b^2+8\) \(a^2+b^2+2ab=a^2+b^2+8\)
2ab=8 ab=4
Now the question is actually asking if ab=4 what is a+b
knowing either the value of a or b we will know the other variable because of the multiplication constraint and get a+b
Re: If arc PQR above is a semicircle, what is the length of diameter PR ? [#permalink]
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08 Feb 2017, 00:12
3
1
Since angle PQR is inscribed in a semicircle,it is a right triangle. PQR is divided into two right triangles by the vertical line from Q to side PR. Let PQ=x and QR=y
The larger right triangle has hypotenuse \(x\), so \(x^2 = 4 + a^2\) The smaller right triangle has hypotenuse \(y\) , so \(y^2 = 4 + b^2\)
From triangle PQR \((a+b)^2 = x^2 + y^2\), by substituting \((a+b)^2= 4+a^2 + 4+b^2\) \(a^2 + 2ab + b^2 = 8 + a^2 +b^2\) \(2ab=8 ; ab =4\)
1) if a =4 and ab=4; b must be 1 SUFFICIENT 2) if b =1 and ab=4; a must be 4 SUFFICIENT
Re: If arc PQR above is a semicircle, what is the length of diameter PR ? [#permalink]
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15 Aug 2017, 07:00
Hi! How do we know if PQR is a right triangle?
Thanks!
Bunuel wrote:
Attachment:
Semicircle2.PNG
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.
Re: If arc PQR above is a semicircle, what is the length of diameter PR ? [#permalink]
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15 Aug 2017, 09:25
RodrigoMi wrote:
Hi! How do we know if PQR is a right triangle?
Thanks!
Bunuel wrote:
Attachment:
Semicircle2.PNG
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.
(1) a = 4. Sufficient.
(2) b = 1. Sufficient.
Answer: D.
Please check the highlighted part in the solution.
_________________
Re: If arc PQR above is a semicircle, what is the length of diameter PR ? [#permalink]
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15 Aug 2017, 10:05
1
D clearly
Let the perpendicular to PR be M
using similar triangle Triangle PQM~ Triangle QRM using AAA property and Also Triangle PQR is right angle at Q since angle contained by diameter to the circle is 90.
PQ/QR = PM/QM = QM/RM
statement 1 is sufficient to find RM. statement 2 is sufficient to find PM.
therefore D
_________________
Give Kudos for correct answer and/or if you like the solution.
We can answer this question without performing any calculations. Instead, we can use some visualization.
Important point: For geometry DS questions, we are typically checking to see whether the statements "lock" a particular angle or length into having just one value. This concept is discussed in much greater detail in the video below.
Target question: What is the length of diameter PR? We want to check whether the statements lock this side into having just 1 possible length.
Given: Arc PQR above is a semicircle. This means that angle PQR is 90 degrees (an important property of circles)
Statement 1: a = 4 If a = 4, then we now have the lengths of 2 sides of a right triangle. So, we could apply the Pythagorean Theorem to find the length of side PQ. Since we can find the lengths of all 3 sides of that right triangle, there is only 1 triangle in the universe with those lengths. In other words, statement 1 "locks" the left-hand triangle into exactly 1 shape. This means that the angle QPR is locked into one angle. In turn, angle QRP is locked into one angle So, all three angles of triangle PQR are locked. Plus we could determine the length of side PQ. All of this tells us that statement 1 locks triangle PQR into 1 and only 1 triangle, which means there must be only one possible value for the length of side PR. Since we could (if we chose to perform the necessary calcations) answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: b = 1 If b = 1, then we now have the lengths of 2 sides of a right triangle (the small triangle on the right-hand side). So, we could apply the Pythagorean Theorem to find the length of side QR. Since we can find the lengths of all 3 sides of that right triangle, there is only 1 triangle in the universe with those lengths. In other words, statement 2 "locks" the small triangle (on the right side) into exactly 1 shape. This means that the angle PRQ is locked into one angle. In turn, angle QPR is locked into one angle So, all three angles of triangle PQR are locked. All of this tells us that statement 2 locks triangle PQR into 1 and only 1 triangle, which means there must be only one possible value for the length of side PR. Since we could (if we chose to perform the necessary calcations) answer the target question with certainty, statement 2 is SUFFICIENT
Answer: D
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If arc PQR above is a semicircle, what is the length of diameter PR ?
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