|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
firas92
Current Student
Joined: 16 Jan 2019
Last visit: 02 Dec 2024
Posts: 616
Given Kudos: 142
Location: India
Concentration: General Management
Schools: Tepper '23 (M$) Foster '23 (D)
GMAT 1: 740 Q50 V40
WE:Sales (Other)
Products:
19 Jun 2019, 07:25
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
51% (02:30) correct
49%
(02:11)
wrong
based on 145
sessions
History
Date
Time
Result
Not Attempted Yet
If the quadrilateral PQRS can be completely inscribed in a circle C with PR passing through the centre of the circle O, then what is the measure of the area of the quadrilateral PQRS?
1. PQ = QR and PR is 4 units long.
2. PS = 2√2
1. PQ = QR and PR is 4 units long.
2. PS = 2√2
19 Jun 2019, 11:48
Kudos
Bookmarks
Information from Question: Since PR passes through center of the circle, PR is the diameter of the circle and Q and S are right angles (angles in semicircles are right angles).
Information from the first statement: Since PR is known, diameter of circle is known. Also, PQ=QR, implying that angle QPR = angle QRP = 45 degrees. Hence, area of triangle PQR is known. No information is given about triangle PSR. Hence, statement (1) is not sufficient.
Information from the second statement: Length of PS is given. No other information about triangle PSR and triangle PQR. Hence, statement (2) is not sufficient.
Both (1) and (2): Area of triangle PQR is known from statement (1).
From statement (2), \(PS = PR/\sqrt{2}\). Hence, angle SPR = 45 degrees. Now, area of triangle PSR is also known. Area of quadrilateral is the sum of area of triangle PQR and that of triangle PSR.
So, both (1) and (2) are needed and are sufficient for finding area of the quadrilateral.
Information from the first statement: Since PR is known, diameter of circle is known. Also, PQ=QR, implying that angle QPR = angle QRP = 45 degrees. Hence, area of triangle PQR is known. No information is given about triangle PSR. Hence, statement (1) is not sufficient.
Information from the second statement: Length of PS is given. No other information about triangle PSR and triangle PQR. Hence, statement (2) is not sufficient.
Both (1) and (2): Area of triangle PQR is known from statement (1).
From statement (2), \(PS = PR/\sqrt{2}\). Hence, angle SPR = 45 degrees. Now, area of triangle PSR is also known. Area of quadrilateral is the sum of area of triangle PQR and that of triangle PSR.
So, both (1) and (2) are needed and are sufficient for finding area of the quadrilateral.
20 Jun 2019, 04:00
Kudos
Bookmarks
Although the question asks us to find out the area of the quadrilateral, we can break it down to finding the areas of two triangles which the quadrilateral comprises of.
Note: A quadrilateral which is completely inscribed in a circle, i.e. all its vertices are on the circumference of the circle, is known as a cyclic quadrilateral.
In a cyclic quadrilateral, the opposite angles are supplementary.
Let us draw the figure corresponding to the question data:
20th June 2019 - Reply 3 - 1.JPG [ 16.98 KiB | Viewed 3822 times ]
Since PR passes through the centre of the circle O, it represents the diameter. Therefore, PR divides the quadrilateral into two right angled triangles (angle subtended by the diameter at the circumference is always 90 degrees), PQR and PSR.
Triangle PQR is right angled at Q, so PQ and QR are the perpendicular sides. The area of triangle PQR = ½ * PQ * QR.
Triangle PSR is right angled at S, so PS and SR are the perpendicular sides. The area of triangle PSR = ½ * PS * SR.
The sum of these triangles is equal to the area of the quadrilateral. Any data that helps us find out the areas of these triangles will be sufficient data.
From statement I alone, we gather that triangle PQR is an isosceles right angled triangle whose hypotenuse PR is 4 units in length.
20th June 2019 - Reply 3 - 2.JPG [ 17.84 KiB | Viewed 3736 times ]
Therefore, PQ = QR = \(\frac{4}{√2}\) = 2√2. We can calculate the area of triangle PQR using this data. However, we do not have any information about triangle PSR in statement I. So, statement I alone is insufficient.
Answer options A and D can be eliminated. Possible answers are B, C or E.
From statement II alone, we know PS = 2√2.
20th June 2019 - Reply 3 - 3.JPG [ 17.53 KiB | Viewed 3695 times ]
This is insufficient to even find the area of triangle PSR, let alone area of triangle PQR. Clearly, it is insufficient to find the area of the quadrilateral.
Answer option B can be eliminated.
Combining statement I and II, we can calculate the area of triangle PQR.
20th June 2019 - Reply 3 - 4.JPG [ 19.85 KiB | Viewed 3675 times ]
Since we know PR and PS, we can calculate QS and therefore, the area of triangle PSR. Adding up the two areas will give us the area of the quadrilateral. Combined data is sufficient.
The correct answer option is C.
Remember that, when you are trying to solve the question using Statement II alone, you do not know anything about PR. So, you need to make a conscious effort to not use PR = 4 along with Statement II, because PR = 4 is given in statement I. Until you combine the two statements, you can’t use both together.
Hope this helps!
Note: A quadrilateral which is completely inscribed in a circle, i.e. all its vertices are on the circumference of the circle, is known as a cyclic quadrilateral.
In a cyclic quadrilateral, the opposite angles are supplementary.
Let us draw the figure corresponding to the question data:
Attachment:
20th June 2019 - Reply 3 - 1.JPG [ 16.98 KiB | Viewed 3822 times ]
Since PR passes through the centre of the circle O, it represents the diameter. Therefore, PR divides the quadrilateral into two right angled triangles (angle subtended by the diameter at the circumference is always 90 degrees), PQR and PSR.
Triangle PQR is right angled at Q, so PQ and QR are the perpendicular sides. The area of triangle PQR = ½ * PQ * QR.
Triangle PSR is right angled at S, so PS and SR are the perpendicular sides. The area of triangle PSR = ½ * PS * SR.
The sum of these triangles is equal to the area of the quadrilateral. Any data that helps us find out the areas of these triangles will be sufficient data.
From statement I alone, we gather that triangle PQR is an isosceles right angled triangle whose hypotenuse PR is 4 units in length.
Attachment:
20th June 2019 - Reply 3 - 2.JPG [ 17.84 KiB | Viewed 3736 times ]
Therefore, PQ = QR = \(\frac{4}{√2}\) = 2√2. We can calculate the area of triangle PQR using this data. However, we do not have any information about triangle PSR in statement I. So, statement I alone is insufficient.
Answer options A and D can be eliminated. Possible answers are B, C or E.
From statement II alone, we know PS = 2√2.
Attachment:
20th June 2019 - Reply 3 - 3.JPG [ 17.53 KiB | Viewed 3695 times ]
This is insufficient to even find the area of triangle PSR, let alone area of triangle PQR. Clearly, it is insufficient to find the area of the quadrilateral.
Answer option B can be eliminated.
Combining statement I and II, we can calculate the area of triangle PQR.
Attachment:
20th June 2019 - Reply 3 - 4.JPG [ 19.85 KiB | Viewed 3675 times ]
Since we know PR and PS, we can calculate QS and therefore, the area of triangle PSR. Adding up the two areas will give us the area of the quadrilateral. Combined data is sufficient.
The correct answer option is C.
Remember that, when you are trying to solve the question using Statement II alone, you do not know anything about PR. So, you need to make a conscious effort to not use PR = 4 along with Statement II, because PR = 4 is given in statement I. Until you combine the two statements, you can’t use both together.
Hope this helps!
