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705-805 (Hard)|   Geometry|      
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ostrick5465
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BM = MC; CN = 3 x NA; SAEN = 27 cm2. What is the area of ABC? 07 Sep 2021, 03:56
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75% (hard)

Question Stats:

49% (03:08) correct 51% (03:04) wrong based on 35 sessions

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BM = MC; CN = 3 x NA; The area of AEN is 27 \(cm^2\). What is the area of ABC?

A. \(216\) \(cm^2\)
B. \(324\) \(cm^2\)
C. \(208\) \(cm^2\)
D. \(276\) \(cm^2\)
E. \(278\) \(cm^2\)
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A
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BM = MC; CN = 3 x NA; SAEN = 27 cm2. What is the area of ABC? 07 Sep 2021, 12:49
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In the picture we see that triangle AKN is similar to triangle CMN.

CN=3*NA -> BM=MC=3*AK

area CMN = \(3^2\) * area AKN (realtion between sides is 3, so relation between areas is \(3^2\))

We also see that triangle BEM is similar to triangle AEK.

area BEM = \(3^2\) * area AEK

We also know that:

area AEN = area AKN + area AEK = 27

area ABC = area CMN + area BEM - area AEN = 9*area AKN + 9*area AEK - area AEN = 9*(area AKN + area AEK) - area AEN = 9*area AEN - area AEN = 8*area AEN = 8*27 = 216

IMO A
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BM = MC; CN = 3 x NA; SAEN = 27 cm2. What is the area of ABC? 07 Sep 2021, 13:36
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andreagonzalez2k
In the picture we see that triangle AKN is similar to triangle CMN.

CN=3*NA -> BM=MC=3*AK

area CMN = \(3^2\) * area AKN (realtion between sides is 3, so relation between areas is \(3^2\))

We also see that triangle BEM is similar to triangle AEK.

area BEM = \(3^2\) * area AEK

We also know that:

area AEN = area AKN + area AEK = 27

area ABC = area CMN + area BEM - area AEN = 9*area AKN + 9*area AEK - area AEN = 9*(area AKN + area AEK) - area AEN = 9*area AEN - area AEN = 8*area AEN = 8*27 = 216

IMO A
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No where is given ME perpendicular to BC , so how do u assume AKN is similar to CMN (so u cant say AK is perpendicular to EN) , only one angle is same , we are not sure of other two .
Seems some additional info is needed in question.
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BM = MC; CN = 3 x NA; SAEN = 27 cm2. What is the area of ABC? 07 Sep 2021, 13:46
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Gaurav2896

No where is given ME perpendicular to BC , so how do u assume AKN is similar to CMN (so u cant say AK is perpendicular to EN) , only one angle is same , we are not sure of other two .
Seems some additional info is needed in question.
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Yes, Nowhere is given ME perpendicular to BC so we cannot assume AKN is similar to CMN, but the question supplies enough information to solve.
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BM = MC; CN = 3 x NA; SAEN = 27 cm2. What is the area of ABC? 07 Sep 2021, 14:10
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Gaurav2896
andreagonzalez2k
In the picture we see that triangle AKN is similar to triangle CMN.

CN=3*NA -> BM=MC=3*AK

area CMN = \(3^2\) * area AKN (realtion between sides is 3, so relation between areas is \(3^2\))

We also see that triangle BEM is similar to triangle AEK.

area BEM = \(3^2\) * area AEK

We also know that:

area AEN = area AKN + area AEK = 27

area ABC = area CMN + area BEM - area AEN = 9*area AKN + 9*area AEK - area AEN = 9*(area AKN + area AEK) - area AEN = 9*area AEN - area AEN = 8*area AEN = 8*27 = 216

IMO A

No where is given ME perpendicular to BC , so how do u assume AKN is similar to CMN (so u cant say AK is perpendicular to EN) , only one angle is same , we are not sure of other two .
Seems some additional info is needed in question.
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I have not assumed that ME is perpendicular to BC as I don't use Pythagorean theorem in any step. I only assume that the line I draw (AK) is parallel to BC. And YES, even if AK is not perpendicular to EN but AK is parallel to BC, AKN is similar to CMN (angles KNA and MNC are equal and also angles NAK and NCM, so the third angle must be equal)
https://en.wikipedia.org/wiki/Intercept_theorem
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BM = MC; CN = 3 x NA; SAEN = 27 cm2. What is the area of ABC? 07 Sep 2021, 14:20
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Gaurav2896

No where is given ME perpendicular to BC , so how do u assume AKN is similar to CMN (so u cant say AK is perpendicular to EN) , only one angle is same , we are not sure of other two .
Seems some additional info is needed in question.

Yes, Nowhere is given ME perpendicular to BC so we cannot assume AKN is similar to CMN, but the question supplies enough information to solve.
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You are wrong, if AK is parallel to BC even if AK is not perpendicular to EM and EM is not perpendicular to BC, AKN is similar to CMN.
You can see the explication here:
https://en.wikipedia.org/wiki/Intercept_theorem
In the fourth figure you have this exact case.
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BM = MC; CN = 3 x NA; SAEN = 27 cm2. What is the area of ABC? 07 Sep 2021, 22:02
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Connect E to C, B to N.
Attachment:
HN Armsterdam loi giai.jpg
HN Armsterdam loi giai.jpg [ 8.58 KiB | Viewed 2044 times ]

We have CN=3NA

=> area CEN = 3 * area AEN = 27 * 3 = 81 \(cm^2\)

BM = MC
=> area EBM = area ECM; area NBM = area NCM

While area BEN = area EBM - area NBM, and area CEN = area ECM - area NCM
=> area BEN = area CEN = 81 \(cm^2\)

=> area BAN = area BEN - area AEN = 81 - 27 = 54 \(cm^2\)

CN=3NA => AC = 4 NA
=> area ABC = 4 area BAN = 54 * 4 = 216 \(cm^2\)
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BM = MC; CN = 3 x NA; SAEN = 27 cm2. What is the area of ABC? 15 Oct 2023, 11:01
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