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Is there any way to solve this besides applying Algebra?

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Source: — Data Sufficiency |
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by srcc25anu » Sat Jun 08, 2013 6:40 am
We know PR is the diameter. Any triangle drawn with PR as a side and 3rd point (Q) on the circumference of a circle is a right angled triangle at point Q.
If we draw a perpendicular from Q on the diameter, say point T, then QT^2 = a * b
Now we know QT = 2 so QT^2 = 4
St1: a = 4
Then we can find out b = 4 / 4 = 1
Diameter = a + b = 5
Sufficient

St2: b =1
Again we can find out a as 4 / 1 = 4
So Diameter = a + b = 5
Sufficient

Hence answer is D.
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by [email protected] » Sat Jun 08, 2013 8:07 am
srcc25anu wrote:We know PR is the diameter. Any triangle drawn with PR as a side and 3rd point (Q) on the circumference of a circle is a right angled triangle at point Q.
If we draw a perpendicular from Q on the diameter, say point T, then QT^2 = a * b
Now we know QT = 2 so QT^2 = 4
St1: a = 4
Then we can find out b = 4 / 4 = 1
Diameter = a + b = 5
Sufficient

St2: b =1
Again we can find out a as 4 / 1 = 4
So Diameter = a + b = 5
Sufficient

Hence answer is D.


Not clear as to how you got QT^2 = a * b
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by Brent@GMATPrepNow » Sat Jun 08, 2013 9:19 am
[email protected] wrote:If arc PQR above is a semicircle, what is the length of
diameter PR ?
(1) a=4
(2) b=1
Image
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 our free video: https://www.gmatprepnow.com/module/gmat- ... cy?id=1103

This technique can save a lot of time, however, it is difficult to explain in text only (which is why I encourage you to watch our free video).

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: If 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

Cheers,
Brent
Last edited by Brent@GMATPrepNow on Tue Jun 11, 2013 5:02 am, edited 1 time in total.
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by GMATGuruNY » Sun Jun 09, 2013 4:32 am
Image

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

(1) a = 4
(2) b = 1
Approach 1:
Clearly, the two statements combined are sufficient to determine length of PR.
Eliminate E.

Answer choice C is WAY TOO EASY.
If the correct answer is C, then 100% of test-takers will answer the question correctly, rendering the problem pointless.
Eliminate C.

Given the symmetry of the figure:
If statement 1 by itself is sufficient (implying that a=4 is sufficient to determine that b=1), then statement 2 by itself must also be sufficient (implying that b=1 is sufficient to determine that a=4).
Thus, each statement by itself must be sufficient.

The correct answer is D.

Approach 2:
An INSCRIBED ANGLE is formed by two chords.
Thus, angle PQR is an inscribed angle.
An inscribed angle that intercepts the diameter is a RIGHT ANGLE.
Thus, angle PQR is a right angle, implying that triangle PQR is a RIGHT TRIANGLE.

Image

In the figure above, PS is a height drawn through right angle PQR.
A height drawn through the right angle of a triangle forms SIMILAR TRIANGLES.
Proof:
If angle QPR = x and angle PQS = y, then x+y = 90.
Since angle PQR = 90, angle SQR = 90-y = x.
Since angle QSR = 90, angle SQR = x, and x+y=90, angle QRP = y.
Thus, all three triangles -- PQS, QRS and PQR -- have the SAME COMBINATION OF ANGLES, as shown in the figure above:
x - y - 90.
Triangles that have the same combination of angles are SIMILAR.

The legs of similar triangles are in the SAME RATIO.
Thus, in all 3 triangles:
(leg opposite x) : (leg opposite y) = (leg opposite x) : (leg opposite y).
In triangle PQS, (leg opposite x) : (leg opposite y) = 2/a.
In triangle QRS, (leg opposite x) : (leg opposite y) = b/2.
Since the two ratios are equal, we get:
2/a = b/2
ab = 4.

Statement 1: a=4
Since ab=4, b=1, implying that PR = 4+1 = 5.
SUFFICIENT.

Statement 2: b=1
Since ab=4, a=4, implying that PR = 4+1 = 5.
SUFFICIENT.

The correct answer is D.

Problems that test the same concept:

https://www.beatthegmat.com/inscribed-tr ... 74152.html
https://www.beatthegmat.com/length-of-th ... 71979.html
https://www.beatthegmat.com/geo-question ... nta-14-649
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