Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

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.

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.

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. _________________

If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

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; \(a^2+(qm)^2=(pq)^2\) \((pq)^2=a^2+2^2\)

\(b^2+(qm)^2=(qr)^2\) \((qr)^2=b^2+2^2\)

\((pq)^2+(qr)^2=(pr)^2\) \(a^2+2^2+b^2+2^2=(a+b)^2\) \(a^2+2^2+b^2+2^2=a^2+b^2+2ab\) \(2^2+2^2=2ab\) \(8=2ab\) \(ab=4\)---- This is the final equation we arrived at just by looking at the figure

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 10211 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]
21 Dec 2012, 21:20

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.

Re: If arc PQR above is a semicircle, what is the length of [#permalink]
22 Dec 2012, 04:18

Expert's post

anon1 wrote:

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.

On September 6, 2015, I started my MBA journey at London Business School. I took some pictures on my way from the airport to school, and uploaded them on...

When I was growing up, I read a story about a piccolo player. A master orchestra conductor came to town and he decided to practice with the largest orchestra...

I’ll start off with a quote from another blog post I’ve written : “not all great communicators are great leaders, but all great leaders are great communicators.” Being...