Can someone help me with this one?
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 are of the triangular region RCS?
1) The area of triangular region ABX is 32
2) the length of one of the altitudes pf triangle ABC is 8
Geometry
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Here's a diagram of the figure described.
Notice what I did, which is to use a trick that works well on many GMAT triangle questions.
Depending on the situation, you may be able to use any triangle that fits the description. In many situations everything has to work as long as the triangle fits all of the constraints. In this situation, I made the angle at A a right angle, because doing that does not change any key parameters of the question and it makes everything come out more easily.
Statement 1 tells us the area of triangle ABX. Notice that in the version of the figure that I drew, the height of ABX is half that of ABC, while they share the same base.
This means that the area of ABC is twice that of ABX. So the area of ABC is 2 x 32 = 64.
Now we are getting somewhere.
If the length of segment CR is 1/2 of that of segment CX which is 1/2 of that of segment CA, then CR is 1/4 CA.
So the height of triangle RCS is 1/4 of that of ABC.
The other thing you need to know is that triangles which share an angle and have two proportional sides that meet at that angle are similar triangles. This concept is rarely incorporated into GMAT questions.
In any case what this means is that since RC is 1/4 of AC and SC is 1/4 of BC, triangle RCS is similar to triangle ABC, with all dimensions of RCS being 1/4 of those of ABC.
Since we know the area of ABC is 64 and we know the height of RCS is 1/4 of that of ABC and the base of RCS is also 1/4 of that of ABC, we can calculate the area of RCS.
So Statement 1 is sufficient.
Statement 2 gives us one of the heights of triangle ABC. We can't calculate area using just the height.
So Statement 2 is insufficient.
The correct answer is A.
Notice what I did, which is to use a trick that works well on many GMAT triangle questions.
Depending on the situation, you may be able to use any triangle that fits the description. In many situations everything has to work as long as the triangle fits all of the constraints. In this situation, I made the angle at A a right angle, because doing that does not change any key parameters of the question and it makes everything come out more easily.
Statement 1 tells us the area of triangle ABX. Notice that in the version of the figure that I drew, the height of ABX is half that of ABC, while they share the same base.
This means that the area of ABC is twice that of ABX. So the area of ABC is 2 x 32 = 64.
Now we are getting somewhere.
If the length of segment CR is 1/2 of that of segment CX which is 1/2 of that of segment CA, then CR is 1/4 CA.
So the height of triangle RCS is 1/4 of that of ABC.
The other thing you need to know is that triangles which share an angle and have two proportional sides that meet at that angle are similar triangles. This concept is rarely incorporated into GMAT questions.
In any case what this means is that since RC is 1/4 of AC and SC is 1/4 of BC, triangle RCS is similar to triangle ABC, with all dimensions of RCS being 1/4 of those of ABC.
Since we know the area of ABC is 64 and we know the height of RCS is 1/4 of that of ABC and the base of RCS is also 1/4 of that of ABC, we can calculate the area of RCS.
So Statement 1 is sufficient.
Statement 2 gives us one of the heights of triangle ABC. We can't calculate area using just the height.
So Statement 2 is insufficient.
The correct answer is A.
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Perfect Scoring Tutor With Over a Decade of Experience
MartyMurrayCoaching.com
Contact me at [email protected] for a free consultation.
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
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 are of the 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
In the figure, if we let the area be a, the area of XYR=2a, BXY=4a,ABX=8a.
Condition 1 mentions 8a=32, so a=4, and as it is sufficient, the answer beocomes (A).
Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.
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 are of the 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
In the figure, if we let the area be a, the area of XYR=2a, BXY=4a,ABX=8a.
Condition 1 mentions 8a=32, so a=4, and as it is sufficient, the answer beocomes (A).
Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.
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We know that ∆RCY, ∆XCY and ∆ACB are similar.
Suppose the base and height of ∆ACB are b and h, respectively. Its area is thus bh/2.
Since ∆XCY has base b/2 and height h/2, its area is bh/8. Likewise, ∆RCY has base b/4 and height h/4, so its area is bh/32. If we can find bh/32, we're done.
S1 says that b*(h/2)/2 = 32. Dividing both sides by 8 gives bh/32 = 4; SUFFICIENT.
S2 gives us h (kinda), but not b; INSUFFICIENT.
Suppose the base and height of ∆ACB are b and h, respectively. Its area is thus bh/2.
Since ∆XCY has base b/2 and height h/2, its area is bh/8. Likewise, ∆RCY has base b/4 and height h/4, so its area is bh/32. If we can find bh/32, we're done.
S1 says that b*(h/2)/2 = 32. Dividing both sides by 8 gives bh/32 = 4; SUFFICIENT.
S2 gives us h (kinda), but not b; INSUFFICIENT.