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Triangle

by sk8ternite » Tue Aug 18, 2009 1:48 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.

Official Guide #109, dont really understand the explanation
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by eustudent » Tue Aug 18, 2009 3:31 pm
OA is A.


from statement 1 you have the area of ABC. this is sufficient to answer the question.

from statement 2 you have the length of one of the verticies. From here you can probably find the corresponding vertice of RCS, but you can not find the surface area. insufficient
Two triangles are similar if:

they have
AAA
SSS
ASA

A- angle(all three angles are the same_
S- side.(the corresponding sides are in the same proportion)

from the problem you know that SC= 1/4 BC
RC=1/4 XC
the above two share the same angle ACB or RCS,
therefore they fit in the definition for similar triangles.
Since the two triangles are similar, their surface areas should be in the same ratio as their corresponding sides. Area of RCS= 1/4 Area of ABC= 8
Last edited by eustudent on Thu Sep 03, 2009 1:37 pm, edited 2 times in total.
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by yezz » Tue Aug 18, 2009 3:46 pm
eustudent wrote:OA is A.


from statement 1 you have the area of ABC. this is sufficient to answer the question.

from statement 2 you have the length of one of the verticies. From here you can probably find the corresponding vertice of RCS, but you can not find the surface area. insufficient
Two triangles are similar if:

they have
AAA
SSS
ASA

A- angle(all three angles are the same_
S- side.(the corresponding sides are in the same proportion)

from the problem you know that SC= 1/4 BC
RC=1/4 XC
the above two share the same angle ACB or RCS,
therefore they fit in the definition for similar triangles.
Since the two of the sides of triangle RSC = 1/4 of two of the sides of triangle ABC, the surface area of RSC should be also 1/4 of the area of ABC.
if the ratio of the sides is 1:4 thus a ... 16 not 1:4
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by eustudent » Tue Aug 18, 2009 7:22 pm
thanks a lot for the correction yezz!!
i totally forgot about that.

my test is in two weeks. i really need to pay more attention
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by sk8ternite » Tue Aug 18, 2009 8:03 pm
eustudent wrote:OA is A.


from statement 1 you have the area of ABC. this is sufficient to answer the question.

from statement 2 you have the length of one of the verticies. From here you can probably find the corresponding vertice of RCS, but you can not find the surface area. insufficient
Two triangles are similar if:

they have
AAA
SSS
ASA

A- angle(all three angles are the same_
S- side.(the corresponding sides are in the same proportion)

from the problem you know that SC= 1/4 BC
RC=1/4 XC
the above two share the same angle ACB or RCS,
therefore they fit in the definition for similar triangles.
Since the two triangles are similar, their surface areas should be in the same ratio as their corresponding sides. Area of RCS= 1/4 Area of ABC= 8
did you mean RC=1/4 AC?
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by BrianSmith » Tue Aug 18, 2009 11:58 pm
Re: first statement:

Another way is to draw ABC (without making assumptions about equal sides or anything); a median (line from one vertex to the middle of opposite side) divides a triangle to two triangles with half the original area, so:

ABC is divided by XB to create XBC, area 32/2=16
XBC is divided by XY to create XYC, area 16/2=8
XYC is divided by XS to create XSC, area 8/2=4
XSC is divided by SR to create RSC, arae 4/2=2
Sufficient.
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