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If in the figure above AM=1.6, DM=1.2 and CB=3. If AC and DB are perpe
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04 Jun 2015, 04:41
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62% (02:45) correct 38% (02:57) wrong based on 133 sessions
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If in the figure above AM=1.6, DM=1.2 and CB=3. If AC and DB are perpendicular, what is the length of CM? A. 0.8 B. 1.6 C. 1.8 D. 2.4 E. 2.8 KUDOS for a concise and creative Answer
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If in the figure above AM=1.6, DM=1.2 and CB=3. If AC and DB are perpe
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04 Jun 2015, 06:13
reto wrote: Attachment: The attachment T8783.png is no longer available If in the figure above AM=1.6, DM=1.2 and CB=3. If AC and DB are perpendicular, what is the length of CM? A. 0.8 B. 1.6 C. 1.8 D. 2.4 E. 2.8 KUDOS for a concise and creative Answer Answer: Option
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Re: If in the figure above AM=1.6, DM=1.2 and CB=3. If AC and DB are perpe
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04 Jun 2015, 06:24
I found a nice way; since the lines are AC and DB are perpendicular, this means that the triangles are proportionate to each other and that they are right triangles. Since we know sides AM and DM we can use pythagorem theorem to solve for AD
1.2 = 6/5 1.8 = 8/5
6/5*6/5 + 8/5*8/5 = 4 which is a perfect square and solves for 2 So AD is 2, we know that CB is proportionate to AD. CB is 1 1/2 x more than AD So CM must be 1 1/2 times more than DM. 6/5 * 3/2 = 18/10 = 1.8 Therefore answer C



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Re: If in the figure above AM=1.6, DM=1.2 and CB=3. If AC and DB are perpe
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06 Jun 2015, 02:30
reto wrote: Attachment: T8783.png If in the figure above AM=1.6, DM=1.2 and CB=3. If AC and DB are perpendicular, what is the length of CM? A. 0.8 B. 1.6 C. 1.8 D. 2.4 E. 2.8 KUDOS for a concise and creative Answer Thanks for participating But how about just ballparking and POE this task? Since in Problem Solving Questions, Figures are usually drawn to scale: CM looks a bit longer than AM (1.6). Therefore POE A, B, D, E and choose C.



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Re: If in the figure above AM=1.6, DM=1.2 and CB=3. If AC and DB are perpe
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20 Dec 2016, 07:28
Ι took the two similar triangles ADM and BCM: AM/DM=CM/BM. Why am I wrong here?



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If in the figure above AM=1.6, DM=1.2 and CB=3. If AC and DB are perpe
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22 Dec 2016, 17:45
Could some confirm this for me, I couldn't make out the theory that GMATInsight was using in the blue ink, so I don't want to make this assumption if it only holds true here:
If two chords intersect in a circle, are the "vertical" arcs proportional?
I.e. for this example, given the chords AC and DB, are arcs AD and BC similar/proportionate to arcs CD and AB? I guess my question is: is this a theory, or does this proportional relationship happen to be the case because perhaps these chords are perpendicular (or some other relationship that I'm missing)?
Making this assumption, we could the ratio of the hypotenuses are equal to the ratio of the other angles/sides in a "reflective" manner.
EDIT: Another thought/question: Going off that theory, if these were NOT right triangles and they had provided an area for one of the triangles, along the lengths of two related sides, could we calculate the area of the other triangle? Like in this example, if the area of the smaller triangle were X, and they told us CB=3 and AD = 2, could we use that ratio of 3:2 to calculate the area of the larger triangle? I.e. would the area always equal X*(3/2)^2?



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Re: If in the figure above AM=1.6, DM=1.2 and CB=3. If AC and DB are perpe
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22 Dec 2016, 21:37
arven wrote: Ι took the two similar triangles ADM and BCM: AM/DM=CM/BM. Why am I wrong here? This is because the ratio of sides you took is not proper. You can use the below method. Triangles ADM and BCM are similar. Just take any two points at a time for each of the triangle(A,D from first triangle and corresponding B,C from second triangle and so on..) So, AD/BC = DM/CM = AM/BMCheck , the proportion should be DM/AM = CM/BM. But in your case, you have reversed the LHS to AM/DM. Hope this helps.



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If in the figure above AM=1.6, DM=1.2 and CB=3. If AC and DB are perpe
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23 Dec 2016, 05:52
reto wrote: Attachment: T8783.png If in the figure above AM=1.6, DM=1.2 and CB=3. If AC and DB are perpendicular, what is the length of CM? A. 0.8 B. 1.6 C. 1.8 D. 2.4 E. 2.8 KUDOS for a concise and creative Answer Hi Another way... when two chords intersect, the product of the segment's is equalHere AM * MC = DM * MB..... 1.6MC=1.2MB......MB=4MC/3.... Take ∆CBM.... \(MC^2+MB^2=CB^2........(\frac{4MC}{3})^2+MC^2=3^2...........\frac{25MC^2}{9}=3^2..... \frac{5MC}{3}=3....... MC=9/5=1.8\) C
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Re: If in the figure above AM=1.6, DM=1.2 and CB=3. If AC and DB are perpe
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23 Dec 2016, 06:55
shashank1tripathi wrote: arven wrote: Ι took the two similar triangles ADM and BCM: AM/DM=CM/BM. Why am I wrong here? This is because the ratio of sides you took is not proper. You can use the below method. Triangles ADM and BCM are similar. Just take any two points at a time for each of the triangle(A,D from first triangle and corresponding B,C from second triangle and so on..) So, AD/BC = DM/CM = AM/BMCheck , the proportion should be DM/AM = CM/BM. But in your case, you have reversed the LHS to AM/DM. Hope this helps. I thought I just have to take Base1/Base2=height1/height2 So as I understand do I have to take the sides that are perpendicular to each other?



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Re: If in the figure above AM=1.6, DM=1.2 and CB=3. If AC and DB are perpe
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23 Dec 2016, 09:34
arven wrote: shashank1tripathi wrote: arven wrote: Ι took the two similar triangles ADM and BCM: AM/DM=CM/BM. Why am I wrong here? This is because the ratio of sides you took is not proper. You can use the below method. Triangles ADM and BCM are similar. Just take any two points at a time for each of the triangle(A,D from first triangle and corresponding B,C from second triangle and so on..) So, AD/BC = DM/CM = AM/BMCheck , the proportion should be DM/AM = CM/BM. But in your case, you have reversed the LHS to AM/DM. Hope this helps. I thought I just have to take Base1/Base2=height1/height2 So as I understand do I have to take the sides that are perpendicular to each other? I found out the theory in use: When two chords intersect, the product of one chords parts = the product of the other chord's parts. So AM*MC=DM*MB; Therefore, AM/DM = MB/MC; which basically says 2 sides of these triangles are similar. That, along with the similar vertical angle of 90 tells us the triangles are similar. If AM and DM are two parts of a 345, then we know MB and MC have to be parts of a 345. The previous ratio shows which is the 3 and which is the 4 (Numerators are "4s")



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Re: If in the figure above AM=1.6, DM=1.2 and CB=3. If AC and DB are perpe
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21 Aug 2017, 02:05
reto wrote: Attachment: T8783.png If in the figure above AM=1.6, DM=1.2 and CB=3. If AC and DB are perpendicular, what is the length of CM? A. 0.8 B. 1.6 C. 1.8 D. 2.4 E. 2.8 KUDOS for a concise and creative Answer /_DAC = /_DBC (angle subtended by same arc DC on the circle.) /_ADB = /_ADC (angle subtended by same arc AB on the circle.) /_AMD = /_BMC = 90 deg tria(AMD) ~ tria (BMC) AM/BM = DM/MC = AD/BC 1.6/BM = 1.2/MC = AD/3 AD = \(\sqrt{(AM^2 + DM^2)}\) =\(\sqrt{(1.6^2 + 1.2^2)}\) =2 1.2/MC = 2/3 MC = 1.8 Answer C



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Re: If in the figure above AM=1.6, DM=1.2 and CB=3. If AC and DB are perpe
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Re: If in the figure above AM=1.6, DM=1.2 and CB=3. If AC and DB are perpe
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