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Data Sufficiency. Geometry

Expert replies
Source: — Data Sufficiency |

by [email protected] » Fri Jul 05, 2013 7:37 pm
Hi Mario_87,

There's an upper-level Geometry rule that you need to know to solve this problem:

When dealing with a circle (or in this case, a semi-circle), if a triangle is made up of the circle's diameter and two other sides that touch the circle's circumference, then that triangle is A RIGHT TRIANGLE.

So, in addition to the two obvious right triangles that you can see, there is a third right triangle (the big one). Using the Pythagorean Theorem and some basic algebra, can you solve the problem now?

As an aside, this Multi-Shape Geometry question would be a non-factor to your score. You can get it wrong and still get an 800, so don't get too down on yourself if you find this question tough.

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
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by GMATGuruNY » Sat Jul 06, 2013 2:18 am
The posted question is similar to the following problem from the OG13:
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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