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niteshwaghray
GMAT 1: 700 Q45 V40
Posts: 59
Updated on: 29 May 2017, 11:43
Originally posted by niteshwaghray on 29 May 2017, 10:11.
Last edited by Bunuel on 29 May 2017, 11:43, edited 1 time in total.
Last edited by Bunuel on 29 May 2017, 11:43, edited 1 time in total.
Edited the question.
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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:
65%
(hard)
Question Stats:
64% (02:59) correct
36%
(02:54)
wrong
based on 128
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In the figure above, Δ POQ and Δ AOB are both right-angled at point O. If BQ = OB and OA = AP, and the length of side AB is 5 units, what is the area of the shaded region?
(1) OA = OB + 1
(2) If a rectangle is drawn with sides of length 2QB and \(\frac{PA}{4}\), the area of the rectangle will be 6 square units
Attachment:
oQ6sNCK.png [ 33.11 KiB | Viewed 4785 times ]
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
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Data Sufficiency Forum
29 May 2017, 12:01
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In the figure above, Δ POQ and Δ AOB are both right-angled at point O. If BQ = OB and OA = AP, and the length of side AB is 5 units, what is the area of the shaded region?
(1) OA = OB + 1
OA^2 + OB^2 = AB^2
(OB + 1)^2 + OB^2 = 5^2
We can find OB, then OA. We also know that BQ = OB and OA = AP. So, we know everything about the figure and will be able to answer the question. Sufficient.
(2) If a rectangle is drawn with sides of length 2QB and \(\frac{PA}{4}\), the area of the rectangle will be 6 square units
2QB*PA/4 = 6;
Since BQ = OB and OA = AP, then 2OB*OA/4 = 6 --> OB*OA = 12 --> OA = 12/OB.
OA^2 + OB^2 = AB^2
(12/OB)^2 + OB^2 = 5^2
The same here: we can find OB, then OA. We also know that BQ = OB and OA = AP. So, we know everything about the figure and will be able to answer the question. Sufficient.
Answer: D.
05 May 2020, 05:42
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niteshwaghray
(1) sufic
ob=y, oa=y+1
\((y^2)+(y+1)^2=5^2,y^2+y-12=0,y=3\)
(2) sufic
2qb*pa/4=6, qb*pa=12, qb=ob=x, pa=oa=y, xy=12
\(x^2+y^2=25,x^2+(12/x)^2=25, x^4+144=25x^2\)
\(x^4-25x^2+144=0,(x^2-9)(x^2-16)=0,x=(4,3),y=(3,4)\)
ans (D)
