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OG 12 DS 109

by navalpike » Thu Jun 25, 2009 6:48 pm
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.

This has been answered at the link provided below but none of the explanations were really helpful, at least not to me. Can someone show a clearer step by step method on how to approach this problem?

Your help is much appreciated.

https://www.beatthegmat.com/in-triangle-abc-t18137.html
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by rahulg83 » Fri Jun 26, 2009 2:31 am
Answer indeed is A...
I confess that i searchrd for few theorems in Google. It's been a long time since school B-)

Statement 1 Any line joining a vertex and mid point of opposite side divides the triangle into two parts of equal area. Hence, Area of ABX is equal to 1/2 of Area of triangle ABC
Therefore Area of triangle ABC=64

Now any line segment joining mid points of two sides of a triangle is parallel to the third side (and also half the third side's length, though that info not needed here)

So YX || AB || RS

Now triangle CSR is similar to triangle CYR and triangle CBA (try drawing a figure, you'll understand better)


CS:CY:CB=CR:CX:CA=SR:YX:BA=1:2:4 (Properties of similar triangles, and from info given in question)

Easily you can calculate that area of triangle CSR=One-fourth of triangle CBA..Sufficient

Statement 2 doesn't give much Info apart from one of the altitude. We don't know rest of the sides, or their ratio, or area of any of the regions..Insufficient
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by cubicle_bound_misfit » Fri Jun 26, 2009 5:59 am
Answer is A.

it is tough to make you understand without a figure

but think AC = 4 inches
BC = 8 inches

ax = 2 inches RC = . 5 inches
SC = 2 inches.

now as ABX and RSC are similar triangle

AX^2/RC^2 = area of ABX/Area of RSC hence A wins.
B can be any height if height on particular side was known, it could be answered.
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by Domnu » Fri Jun 26, 2009 8:03 am
The answer should be A. Here's why:

Look at triangles ABX and BCX. They both have the same height and base (AX = XC). This means that they have the same area, so ABC has area 64. Now, RSC is similar to ABC (there's a 1/4 scale factor), so the area of RSC can be computed.

If you're interested in what the area is, just take the scale factor squared and multiply by the area of ABC. This is 64 * (1/4)^2 = 4.

Generally, if you have similar figures and you know the scale factor from one to the other, to find lengths, you multiply by the scale factor, to find areas, multiply by scale factor squared, and for volumes, scale factor cubed, etc.
Have you wondered how you could have found such a treasure? -T
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by priyankm » Sat May 02, 2015 5:13 am
Domnu wrote:The answer should be A. Here's why:

Look at triangles ABX and BCX. They both have the same height and base (AX = XC). This means that they have the same area, so ABC has area 64. Now, RSC is similar to ABC (there's a 1/4 scale factor), so the area of RSC can be computed.

If you're interested in what the area is, just take the scale factor squared and multiply by the area of ABC. This is 64 * (1/4)^2 = 4.

Generally, if you have similar figures and you know the scale factor from one to the other, to find lengths, you multiply by the scale factor, to find areas, multiply by scale factor squared, and for volumes, scale factor cubed, etc.
Thats an excellent explanation. I somewhere knew you could you use the similar side to similar area ratio theorem, but could not comprehend the triangles. Scratched my head over 2 videos and few forums, but your explanation was Bang on target, easy and fluid. Thanks a ton! :D
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