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27 May 2022, 01:30
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
67% (01:36) correct
33%
(01:40)
wrong
based on 49
sessions
History
Date
Time
Result
Not Attempted Yet
If \(QR=10\), what is the area of the triangular region \(PQR\) above?
(1) \(PS = SR = 5\)
(2) \(PQ = PR\)
Attachment:
1.jpg [ 4.16 KiB | Viewed 2333 times ]
27 May 2022, 02:36
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Statement 1: PS=SR=5
Consider QR to be diameter of a circle with radius 5, so that SQ, SR, SP are radius of this circle each having length = 5 units.
On this circle we can move point P at different locations and the angle QPR will always be 90 degree. (Angle bounded by diameter of a circle)
However, the area of each triangle would be different because moving Point P would change the length of PQ and PR.
Statement 1 is not sufficient.
Statement 2: PQ=PR
Given QR=10. Using Pythagoras theorem we can find out lengths of PQ and PR and thereby we can find area of this triangle.
Statement 2 is sufficient.
Answer: B
Consider QR to be diameter of a circle with radius 5, so that SQ, SR, SP are radius of this circle each having length = 5 units.
On this circle we can move point P at different locations and the angle QPR will always be 90 degree. (Angle bounded by diameter of a circle)
However, the area of each triangle would be different because moving Point P would change the length of PQ and PR.
Statement 1 is not sufficient.
Statement 2: PQ=PR
Given QR=10. Using Pythagoras theorem we can find out lengths of PQ and PR and thereby we can find area of this triangle.
Statement 2 is sufficient.
Answer: B
27 May 2022, 05:18
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(1)Only
Image that QR is the diameter of a circle with S as its center.
P can be any point on the circle except for the point Q and point R.
Triangle PQR varies with its area approaching zero in some cases.
Insufficient.
(2)Only
We should be able to compute value for PQ / PR, and then calculate the area.
Sufficient.
The answer is thus (B).
(If it is a PS question,
PR^2+PQ^2=100
PR^2=50
The area is 1/2*50=25)
Image that QR is the diameter of a circle with S as its center.
P can be any point on the circle except for the point Q and point R.
Triangle PQR varies with its area approaching zero in some cases.
Insufficient.
(2)Only
We should be able to compute value for PQ / PR, and then calculate the area.
Sufficient.
The answer is thus (B).
(If it is a PS question,
PR^2+PQ^2=100
PR^2=50
The area is 1/2*50=25)
