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29 Aug 2019, 06:40
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C
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Difficulty:
65%
(hard)
Question Stats:
62% (02:45) correct
38%
(02:36)
wrong
based on 107
sessions
History
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Not Attempted Yet
In the figure above, if PQ = x, is the area of triangle PQR equal to the area of PST?
(1) \(PT = \sqrt{2}*PR\)
(2) The triangle PST is isosceles.
Attachment:
image288.png [ 11.67 KiB | Viewed 3060 times ]
Updated on: 02 Sep 2019, 09:10
Originally posted by TestPrepUnlimited on 02 Sep 2019, 08:37.
Last edited by TestPrepUnlimited on 02 Sep 2019, 09:10, edited 2 times in total.
Last edited by TestPrepUnlimited on 02 Sep 2019, 09:10, edited 2 times in total.
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Bunuel
Analyzing the question:
We are comparing two areas of right triangles so the height and base are important. ½*PQ*PR gives the area of triangle PQR and ½*PT*PS for triangle PST so we will try to look out for those lengths. We will likely represent everything in terms of x due to PQ = x and ST = x. Note we can drop the ½ for both since we are comparing them.
Statement 1:
Both PT and PR are unrelated to x so we can’t fill in anything for them yet with this information, insufficient.
Statement 2:
PST is an isosceles right triangle, this confirms we have the 90 - 45 - 45-degree shape triangle and this means the ratio of the lengths are \(1: 1: \sqrt{2}\). Since \(ST = x\), we have \(PS = PT = \frac{x}{\sqrt{2}}\) from that ratio. Although it is nice to have the area of PTS in terms of x, we still don’t know PR in terms of x. Therefore this is still insufficient. An important note here is that if we knew \(QP <= PS\), this would be sufficient as the graph proves \(PR < PT\). Therefore triangle PST would have a bigger area and they would not have equal area.
Combined:
Since \(PT = \frac{x}{\sqrt{2}}\), \(PR = \frac{PT}{\sqrt{2}} = \frac{x}{2}\). We have all lengths in terms of a ratio of x. The areas are both a ratio of \(x^2\) so this is sufficient, as we’re just comparing their coefficients now. If we work out the math, these two triangles have same area but that is unnecessary to calculate.
Answer: C
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02 Sep 2019, 08:59
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Bunuel
Given: PQ = x
Asked: Is the area of triangle PQR equal to the area of PST?
Area of triangle PQR = 1/2 * QP * PR = 1/2 * x * PR
Area of triangle PST = 1/2 * SP * PT
(1) \(PT = \sqrt{2}*PR\)
\(PT = \sqrt{2}* PR\)
NOT SUFFICIENT
(2) The triangle PST is isosceles.
PS = PT = x/\sqrt{2}
Area of triangle PST \(= 1/2 * (x/\sqrt{2})^2 = x^2/4\)
But area of triangle PQR is unknown
NOT SUFFICIENT
Combining (1) & (2)
(2) The triangle PST is isosceles.
\(PS = PT = x/\sqrt{2}\)
Area of triangle PST \(= \frac{1}{2} * (x/\sqrt{2})^2 = \frac{x^2}{4}\)
(1) \(PT = \sqrt{2}*PR\)
\(PT = \sqrt{2}* PR\)
\(x/\sqrt{2} = \sqrt{2}* PR\)
PR = x/2
Area of triangle PQR \(= \frac{1}{2}* x * \frac{x}{2} = \frac{x^2}{4}\)
Area of triangle PQR = Area of triangle PST
IMO C
