If arc PQR above is a semicircle, what is the length of diameter PR ?

Better you learn some relations which hold true for such triangle.
One which has 2 right angles as here in attachment.
As per that

4 = a*b
a+b=?
Now.
A) a = 4 so b =1
B) b=1 so a = 4

OA 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.

Hope it's clear.

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.

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

hence D

Happy GMATing!

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

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

RELATED VIDEOS

Hi Bunuel , I understand that the triangles PQT and QRT are similar but how do you say <QRT = <PQT and <QPT = <TQR ? Why not <QRT= <TQR? Please help me understand...

<RPQ + <QRP = 90.
<RPQ + <PQT = 90.
Thus <QRP = <PQT

Similarly:
<RPQ + <QRP = 90.
<RQT + <QRP = 90.
Thus <RPQ = <RQT.

1. line from the center of the circle bisects the chord
2. If two chords intersect inside a circle, then the product of the lengths of the segments of one chord equals the product of the lengths of the segments of the other chord.

From (1) and (2), we get a*b=2*2=4

Statement 1- a=4, we get b=1
Statement 2- b=1, we get a=4

D

hi Bunuel, chetan2u

The first thing that came to my mind while approaching is similar triangles, but I didn't clearly understand how to write down the proportions as you did, QT/PQ = TR/QR, I undetstand that once we have the proportions pythagoras can be used in the smaller triangle and b can be found.
How do you understand the proportionS QT/PQ = TR/QR So I thought the base of the bigger triangle is twice it's height then the base of the smaller one should be twice it's height, that would mean b =4, but that's clearly incorrect

Write down the triangles in terms of the similar angles.
90-red arc-blue arc
So TPQ is similar to TQR
From here you can get the pair of similar sides.
TPQ~TQR
1)TP~TQ
2)TQ~TR
3)PQ~QR

From 2 and 3…2/3 will give you
\(\frac{TQ}{PQ}=\frac{TR}{QR}\)

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Bunuel KarishmaB BrentGMATPrepNow or other experts, please help me understand the question.

Unable to understand how did we get any of the equations in the explanatory section.

Using similarity between triangles PQR & QTR, I'm getting the following:
QR/PR = QT/PQ = TR/QR

This makes no sense to me and I'm lost!

You did realise that QTR is similar to PQR. Another thing to note is that PTQ is also similar to PQR (angle P is the same and they both are right triangles).
This means QTR is similar to PTQ (because both are similar to PQR)

So QT/PT = TR/TQ = 2/a = b/2
ab = 4

Then either statement alone is sufficient.

PQR is a right angle. Since this is a DS question and we just have to know that it is possible to find the diameter without really calculating the exact values, one can just say that knowing the second leg (the first one for each triangle is equal to 2) of either triangle, PQT or QRT (I refer to the Bunuel's picture) and using trigonometric functions one is able to find the diameter.

Statement 1. Tangent PQT = a/2 = 4/2 = 2. Knowing this (and that the angle has to be in the range from 0 to 90) we can find the exact angle value, let's say x. Then, we can find the value of the angle RQT = 90 - x = y. Knowing the angle and one leg we can again use the tangent function to find b: Tangent RQT = b/2. Thus we show that we are able to find b (which we need to find the diameter) and that the statement is sufficient. The same applies to the second statement.

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