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Re: If arc PQR above is a semicircle, what is the length of [#permalink]
13 Dec 2012, 06:16

6

This post received KUDOS

Expert's post

21

This post was BOOKMARKED

Attachment:

Semicircle2.PNG [ 4.95 KiB | Viewed 9024 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.

Re: If arc PQR above is a semicircle, what is the length of [#permalink]
22 Dec 2012, 05:04

Expert's post

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 8525 times ]

Re: If arc PQR above is a semicircle, what is the length of [#permalink]
22 Dec 2012, 05:05

Bunnel can you explain the below part little elaboarately

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.

I really dont understand the concept _________________

"Giving kudos" is a decent way to say "Thanks" and motivate contributors. Please use them, it won't cost you anything

Re: If arc PQR above is a semicircle, what is the length of [#permalink]
14 Jan 2013, 10:38

Marcab wrote:

Walkabout wrote:

Attachment:

Semicircle2.png

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.

Little complex for me... don u think bunuel's method is easier? _________________

I've failed over and over and over again in my life and that is why I succeed--Michael Jordan Kudos drives a person to better himself every single time. So Pls give it generously Wont give up till i hit a 700+

Re: If arc PQR above is a semicircle, what is the length of [#permalink]
06 Mar 2014, 09:16

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If arc PQR above is a semicircle, what is the length of [#permalink]
29 Mar 2014, 02:55

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 [#permalink]
29 Mar 2014, 02:58

Expert's post

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. _________________

If arc PQR above is a semicircle, what is the length of [#permalink]
16 Jul 2014, 22:12

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? _________________

______________ KUDOS please, if you like the post or if it helps "Giving kudos" is a decent way to say "Thanks"

Re: If arc PQR above is a semicircle, what is the length of [#permalink]
17 Jul 2014, 07:46

Expert's post

1

This post was BOOKMARKED

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 [#permalink]
05 Aug 2014, 20:38

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 [#permalink]
02 Jun 2015, 00:02

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

If arc PQR above is a semicircle, what is the length of diameter PR? [#permalink]
22 Jul 2015, 17:48

This is a new question (DS 131) from the 2016 edition of the Official Guide.

Note that there are three right triangles total (small, medium, large). A right angle is marked at the bottom of the medium triangle, but there is another in the large as well (esoteric geometry rule: all triangles inscribed in semicircles are 90 degrees--knowing this is vital or you cannot solve the question), and of course in the small triangle (straight line = 180 degrees).

Pythagorean theorem time!

1) Given a, we can solve for PQ in terms of a. 2) Given b, we can solve for QR in terms of b.

This means we have two variables, a and b. a + b = PR, so given either a or b we can solve for PR.

If arc PQR above is a semicircle, what is the length of [#permalink]
22 Jul 2015, 18:43

mcelroytutoring wrote:

This is a new question (DS 131) from the 2016 edition of the Official Guide.

Note that there are three right triangles total (small, medium, large). A right angle is marked at the bottom of the medium triangle, but there is another in the large as well (esoteric geometry rule: all triangles inscribed in semicircles are 90 degrees--knowing this is vital or you cannot solve the question), and of course in the small triangle (straight line = 180 degrees).

Pythagorean theorem time!

1) Given a, we can solve for PQ in terms of a. 2) Given b, we can solve for QR in terms of b.

This means we have two variables, a and b. a + b = PR, so given either a or b we can solve for PR.

1) a = 4. sufficient

2) b = 1. sufficient

The answer is D.

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