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P and Q are two different points on line l
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31 Aug 2026, 08:16
31 Aug 2026, 08:16
Expert Reply
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Question Stats:
50% (03:40) correct
50% (02:07) wrong based on 2 sessions
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P and Q are two different points on line l , and R is a point not on line l such that triangle P Q R is isosceles with P R=Q R . Point T is the midpoint of segment P Q , and the distance from T to R is 6 .
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
The perimeter of triangle P Q R |
28 |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
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Re: P and Q are two different points on line l
[#permalink]
04 Sep 2026, 10:54
04 Sep 2026, 10:54
Expert Reply
Solution
Since \(PR=QR\), triangle \(PQR\) is isosceles, and the midpoint \(T\) of the base \(PQ\) makes \(RT\) perpendicular to \(PQ\).
So we have a right triangle with:
\(RT=6\)
Let \(PT=TQ=x\). Then:
\(PR=QR=\sqrt{x^2+6^2}=\sqrt{x^2+36}\)
Thus the perimeter is
\(2x+2\sqrt{x^2+36}\)
We need to compare this with 28. Set the perimeter equal to 28:
\(2x+2\sqrt{x^2+36}=28 x+\sqrt{x^2+36}=14\sqrt{x^2+36}=14-x\)
Squaring:
\(x^2+36=196-28x+x^2 28x=160 x=\frac{40}{7}\)
This is possible, so the perimeter can equal 28. But \(P\) and \(Q\) are only specified as different points, so \(x\) is not fixed. The perimeter can be less than, equal to, or greater than 28.
For example:
If \(x=1,\) perimeter \(\approx 14.6<28.\)
If \(x=10\), perimeter \(\approx 48.3>28.\)
Answer: D
Since \(PR=QR\), triangle \(PQR\) is isosceles, and the midpoint \(T\) of the base \(PQ\) makes \(RT\) perpendicular to \(PQ\).
So we have a right triangle with:
\(RT=6\)
Let \(PT=TQ=x\). Then:
\(PR=QR=\sqrt{x^2+6^2}=\sqrt{x^2+36}\)
Thus the perimeter is
\(2x+2\sqrt{x^2+36}\)
We need to compare this with 28. Set the perimeter equal to 28:
\(2x+2\sqrt{x^2+36}=28 x+\sqrt{x^2+36}=14\sqrt{x^2+36}=14-x\)
Squaring:
\(x^2+36=196-28x+x^2 28x=160 x=\frac{40}{7}\)
This is possible, so the perimeter can equal 28. But \(P\) and \(Q\) are only specified as different points, so \(x\) is not fixed. The perimeter can be less than, equal to, or greater than 28.
For example:
If \(x=1,\) perimeter \(\approx 14.6<28.\)
If \(x=10\), perimeter \(\approx 48.3>28.\)
Answer: D
gmatclubot
Re: P and Q are two different points on line l [#permalink]
04 Sep 2026, 10:54
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