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Updated on: 16 Sep 2021, 02:58
Originally posted by ostrick5465 on 16 Sep 2021, 02:38.
Last edited by Bunuel on 16 Sep 2021, 02:58, edited 1 time in total.
Last edited by Bunuel on 16 Sep 2021, 02:58, edited 1 time in total.
Renamed the topic and edited the question.
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
24% (01:54) correct
76%
(02:28)
wrong
based on 67
sessions
History
Date
Time
Result
Not Attempted Yet
ABCD is rectangular. The area of BMC is 36 \(cm^2\). Calculate the area of AIB.
(1) The area of AID = the area of BIM
(2) The area of BMC = \(\frac{9}{16}\) the area of IMD.
Attachment:
HN Armsterdam 20-21.png [ 84.12 KiB | Viewed 2324 times ]
16 Sep 2021, 07:43
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ostrick5465
(1) The area of AID = the area of BIM
Does not give any dimension.
Insufficient
(2) The area of BMC = \(\frac{9}{16}\) the area of IMD.
Attachment:
75C01A3B-0BE3-40B7-98A2-AE03ED824C6F.jpeg [ 22.61 KiB | Viewed 2146 times ]
Now Area of AOB=Area of DOC
But area of AMC and BMC are equal as both are between the same two parallel lines and same base CM.
Area(DIM)+Area(BCM)= Area(DOM)+Area(ACM)
Area(DIM)+Area(BCM)= Area(DOM)+Area(OCMI)+Area(AOI)
Area(DIM)+Area(BCM)= Area(DOC)+Area(AOI)
Area(DIM)+Area(BCM)= Area(AOB)+Area(AOI)=Area(AIB)
Area of DIM = \(\frac{16}{9}\)*area of BCM=\(36*\frac{16}{9}=64\)
Area of AIB = 64+36 = 100
Sufficient
B
18 Sep 2021, 21:12
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ostrick5465
(1) The area of AID = the area of BIM
We can not calculate anything else.
=> Insufficient
(2) The area of BMC = \(\frac{9}{16}\) the area of IMD.
=> \(A(IMD) = 36 : \frac{9}{16} = 64 cm^2\)
\(A(ABM) = A(BCD) = \frac{1}{2} A(ABCD)\)
\(A(BMC)+A(IMD)+A(BIM)=A(BCD)=\frac{1}{2} A(ABCD)\) (*)
\(A(AIB)+A(BIM)=A(ABM)=\frac{1}{2} A(ABCD)\) (**)
(*) & (**) => \(A(AIB) = A(BMC)+A(IMD) = 36 + 64 = 100 cm^2\)
=> Sufficient
=> OA: B
