Data Sufficiency

Is there an easier way to solve task nr. 119 ? I mean, it is very simple when one looks at the solution in the OG 13, however, it is very time consuming. Also, if you see the problem for a first time, it is rather unlikely to start inversting your 120 seconds in a hope that there will be a connection between X^2 ,Y^2 and (a+b)^2?

Also, is it the same, if I draw the triangle from task 113 with C on the top/ edge. In the solution it's drawn with B on the top and B down on the right side. This way it looks a little easier to comprehend.

Last question, again for task 113. Are we really expected to draw an additional height in order to solve the problem or is there a short cut?

Thank you in advance, guys.

In triangle ABC, point X is the midpoint of side AC and point Y is the midpoint of side BC. If point R is the midpoint of line segment XC and if point S is the midpoint of line segment YC, what is the area of triangular region RCS ?
(1) The area of triangular region ABX is 32.
(2) The length of one of the altitudes of triangle ABC is 8.

Triangles RCS and ABC:
Side RC = 1/4(AC).
Side SC = 1/4(BC).
The two triangles share angle BCA.

Triangles with a shared angle (BCA) formed by corresponding sides in the same proportion (RC:AC = 1:4, SC:BC = 1:4) are SIMILAR.
Thus, triangle RCS is similar to triangle ABC.
In similar triangles, corresponding bases and heights are in the same proportion as corresponding sides.
Thus, the base of triangle RCS is 1/4 the base of triangle ABC, and the height of triangle RCS is 1/4 the height of triangle ABC.

Area of triangle ABC = (1/2)bh.
Area of triangle RCS = (1/2)*1/4(b)*1/4(h) = (1/16)(1/2)bh.
Thus, the area of triangle RCS is 1/16 the area of triangle ABC.

Question rephrased: What is the area of triangle ABC?

Statement 1: ABX = 32.

In triangle ABX, AX = 1/2(AC).
In other words, the base of triangle ABX is 1/2 the base of triangle ABC.
Triangles ABX and ABC share height BZ.
Since AX = 1/2(AC), and the two triangles have the same height, ABX = 1/2(ABC).
Thus, the area of triangle ABC = 64, and the area of triangle RCS = (1/16)(64) = 4.
SUFFICIENT.

Statement 2: height = 8.
No way to determine the area of triangle ABC or of triangle RCS.
INSUFFICIENT.

The correct answer is A.

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

(1) a = 4
(2) b = 1

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.



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

Thank you very much Mitch. really appreciate your help