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## Need to understand the logic please

This topic has 2 expert replies and 3 member replies
shibsriz@gmail.com Master | Next Rank: 500 Posts
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#### Need to understand the logic please

Wed Jun 05, 2013 4:03 am
In triangle ABC, point Xis the midpoint of sideAC and
point Yis the midpoint of side BC. If point Ris the
midpoint of line segment XC and if point S is the
midpoint of line segment YC, what is the area of
triangular region RCS1
(1) The area of triangular region ABX is 32.
(2) The length of one of the altitudes of triangle ABC
is 8

Nina1987 Senior | Next Rank: 100 Posts
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Sat Feb 13, 2016 6:01 am
The question is testing properties of similar triangles here. There 3 ways to tell if the triangles are similar:

1. AAA (angle angle angle)
All three pairs of corresponding angles are the same.
(We actually need two angles really:)

2. SSS in same proportion (side side side)
All three pairs of corresponding sides are in the same proportion

3. SAS (side angle side)
Two pairs of sides in the same proportion and the included angle equal.

It is this 3rd property that is quite handy here. Two triangles ABC and RSC are similar since RC/AC=SC/BC=1/4 and they include the same angle THETA (as shown in the figure)

1) Tells us that Area(ABX) =32. Since height of triangle ABC is same that of triangle ABX and the base is twice, area (ABC) = 32*2= 64. Since ABC and RSC are similar and their sides are in the ration of 1:4 their ares will be in the ration of 1:16. Sufficient to solve.
2) Tells us one of the altitudes is 8. Since it does not tell which one, insufficient!

GaneshMalkar Senior | Next Rank: 100 Posts
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Sun Jun 09, 2013 5:32 am
shibsriz@gmail.com wrote:
Hey Mitch, what I dont understand is that how u got the altitude BZ, i.e ABZ is a rigth angle triangle!

Thanks
Shiba
BZ is the perpendicular drop from the point B to the extended base AC which is the height or altitude of triangle ABC.

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Matt@VeritasPrep GMAT Instructor
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Wed Jun 05, 2013 2:52 pm
I tried to illustrate this (picture attached), but here's the basic logic:

If a line runs through the midpoints of two sides of a triangle, that line is PARALLEL to the third side of the triangle. So XY is parallel to AB, and RS is parallel to XY.

Since AB, XY, and RS are parallel to one another, they make the same angles where they meet sides AC and BC, respectively, so we have three similar triangles: ABC, XYC, and RSC. Since RC = 1/4 AC and SC = 1/4 BC, all the dimensions of RSC are 1/4 the respective dimensions of ABC, including the base and height. So if the area of ABC is bh/2, the area of RSC is ((b/4)(h/4))/2, or bh/32, and all we need to solve the problem is the area of ABC.

S1 tells us ABX has an area of 32. From this we can get the area of ABC - SUFFICIENT.
S2 gives us an altitude of ABC, but not the length of its corresponding base. We aren't able to get the area of ABC - INSUFFICIENT.

(Problems requiring an understanding of midsegments are uncommon on the GMAT, so I wouldn't sweat this one too much. For more practice on this, check out this link:
http://www.regentsprep.org/Regents/math/geometry/GP10/MidLineL.htm)

Enroll in a Veritas Prep GMAT class completely for FREE. Wondering if a GMAT course is right for you? Attend the first class session of an actual GMAT course, either in-person or live online, and see for yourself why so many students choose to work with Veritas Prep. Find a class now!

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GMATGuruNY GMAT Instructor
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Thu Jun 06, 2013 3:19 am
Quote:
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.

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shibsriz@gmail.com Master | Next Rank: 500 Posts
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Fri Jun 07, 2013 7:15 am
Hey Mitch, what I dont understand is that how u got the altitude BZ, i.e ABZ is a rigth angle triangle!

Thanks
Shiba

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