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Arc AED above is a semicircle. EC is the height to the base AD. The ar
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27 Oct 2014, 09:17
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63% (03:02) correct 37% (03:30) wrong based on 147 sessions
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Arc AED above is a semicircle. EC is the height to the base AD. The area of triangle AEC is 96. If CE=12, what is the area of triangle CDE? A. 48 B. 54 C. 72 D. 84 E. 150
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Re: Arc AED above is a semicircle. EC is the height to the base AD. The ar
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27 Oct 2014, 20:02
Answer = B = 54 Area \(\triangle ECA = 96 = \frac{1}{2} * AC * 12\) AC = 16 Arc AED is a semicircle means AD is the diameter of the circle EA & ED are touching the end points of diameter, which means \(\angle AED = 90^{\circ}\) (This is by property of circle)Refer diagram below which shows various angles of both the triangles Attachment:
T8910x.png [ 22.56 KiB  Viewed 4667 times ]
Note that \(\triangle ECA\) and \(\triangle DCE\) are similar triangles. Now that they are similar, then there corresponding sides are also proportional \(\triangle\) ECA Base = 16 & Height = 12 \(\triangle\) DCE Base = 12; so Height = \(12 * \frac{12}{16} = 9\) Area \(\triangle DCE = \frac{1}{2} * 12 * 9 = 54\)
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Re: Arc AED above is a semicircle. EC is the height to the base AD. The ar
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27 Oct 2014, 20:09
minhphammr wrote: Arc AED above is a semicircle. EC is the height to the base AD. The area of triangle AEC is 96. If CE=12, what is the area of triangle CDE? A. 48 B. 54 C. 72 D. 84 E. 150 For detailed explanation, above post can be referred. Shortcut: Area AEC = 96; so AC = 16 Area shaded region = \(96 * \frac{12}{16} = 54\) Answer = B
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Re: Arc AED above is a semicircle. EC is the height to the base AD. The ar
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17 Dec 2014, 14:41
PareshGmat wrote: Answer = B = 54 Area \(\triangle ECA = 96 = \frac{1}{2} * AC * 12\) AC = 16 Arc AED is a semicircle means AD is the diameter of the circle EA & ED are touching the end points of diameter, which means \(\angle AED = 90^{\circ}\) (This is by property of circle)Refer diagram below which shows various angles of both the triangles Attachment: T8910x.png Note that \(\triangle ECA\) and \(\triangle DCE\) are similar triangles. Now that they are similar, then there corresponding sides are also proportional \(\triangle\) ECA Base = 16 & Height = 12 \(\triangle\) DCE Base = 12; so Height = \(12 * \frac{12}{16} = 9\) Area \(\triangle DCE = \frac{1}{2} * 12 * 9 = 54\) Is it a general property that both triangles are going to be similar?



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Re: Arc AED above is a semicircle. EC is the height to the base AD. The ar
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17 Dec 2014, 19:13
nachobioteck wrote: PareshGmat wrote: Answer = B = 54 Area \(\triangle ECA = 96 = \frac{1}{2} * AC * 12\) AC = 16 Arc AED is a semicircle means AD is the diameter of the circle EA & ED are touching the end points of diameter, which means \(\angle AED = 90^{\circ}\) (This is by property of circle)Refer diagram below which shows various angles of both the triangles Attachment: T8910x.png Note that \(\triangle ECA\) and \(\triangle DCE\) are similar triangles. Now that they are similar, then there corresponding sides are also proportional \(\triangle\) ECA Base = 16 & Height = 12 \(\triangle\) DCE Base = 12; so Height = \(12 * \frac{12}{16} = 9\) Area \(\triangle DCE = \frac{1}{2} * 12 * 9 = 54\) Is it a general property that both triangles are going to be similar? When corresponding angles are same (irrespective of dimensions), both triangles are similar In this case, YES (Refer the diagram above for angle breakup)
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Re: Arc AED above is a semicircle. EC is the height to the base AD. The ar
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17 Dec 2014, 21:31
minhphammr wrote: Attachment: T8910x.png Arc AED above is a semicircle. EC is the height to the base AD. The area of triangle AEC is 96. If CE=12, what is the area of triangle CDE? A. 48 B. 54 C. 72 D. 84 E. 150 The 2 triangles are similar and some relations can come handy AE=AC*AD DE=CD*AD and CE^2=AC*CD we know CE=12 and Area of Triangle AEC=96 so AC=96*2/12 or AC =16 So CD=CE^2/AC or CD=9 Then area of triangle CDE =1/2*9*12=54
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Re: Arc AED above is a semicircle. EC is the height to the base AD. The ar
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18 Dec 2014, 04:39
nachobioteck wrote: Is it a general property that both triangles are going to be similar? This post gives you some figures which should make you think about similar triangles. Check them out. http://www.veritasprep.com/blog/2014/03 ... thegmat/
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Arc AED above is a semicircle. EC is the height to the base AD. The ar
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18 Dec 2014, 05:01
WoundedTiger wrote: The 2 triangles are similar and some relations can come handy
AE=AC*AD DE=CD*AD
I doubt the above 2 are true...the dimension of LHS is m while of RHS is m^2
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Re: Arc AED above is a semicircle. EC is the height to the base AD. The ar
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19 Dec 2014, 05:52
Arc AED above is a semicircle. EC is the height to the base AD. The area of triangle AEC is 96. If CE=12, what is the area of triangle CDE?
A. 48 B. 54 C. 72 D. 84 E. 150
The area of triangle AEC is \(96 = \frac{1}{2}(AC)*(CE)\) or \(AC = \frac{(96*2)}{12}=16\)
\((EC)^2=AC*CD\)
\(CD= (12)^2/16=9\)
Area of the triangle CDE = \(\frac{1}{2}(EC)*(CD)\) = \(\frac{(12*9)}{2}=54\)
Answer: B



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Re: Arc AED above is a semicircle. EC is the height to the base AD. The ar
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06 Sep 2017, 01:01
minhphammr wrote: Attachment: T8910x.png Arc AED above is a semicircle. EC is the height to the base AD. The area of triangle AEC is 96. If CE=12, what is the area of triangle CDE? A. 48 B. 54 C. 72 D. 84 E. 150 In triangle ACE , 1/2*AC* CE = 96 > AC = 16 Now triangle CED ~ triangle CAE CE/CD = CA/CE CE^2 = CA * CD (If you can remember this property of the right angles triangle, it will be highly helpful to solve the problem in a quick manner.) 12^2 = 16 * CD CD = 9 Area CAE = 1/2 * 9 * 12 = 54 Answer B
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Re: Arc AED above is a semicircle. EC is the height to the base AD. The ar
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14 Jul 2019, 06:20
Hi everyone! Does anybody know, whether figures in Problem Solving part are drawn on a scale? I have seen explanation of this task in Economist tutor, it is written there almost of the figures are on scale (if there is no some special signs like 'figure above is not on scale') and they uses a strategy calles 'ballpark', explaning this answer. So, could we trust scale of figures in questions like this? Thanks!



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Re: Arc AED above is a semicircle. EC is the height to the base AD. The ar
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14 Jul 2019, 06:22
plk wrote: Hi everyone! Does anybody know, whether figures in Problem Solving part are drawn on a scale? I have seen explanation of this task in Economist tutor, it is written there almost of the figures are on scale (if there is no some special signs like 'figure above is not on scale') and they uses a strategy calles 'ballpark', explaning this answer. So, could we trust scale of figures in questions like this? Thanks! OFFICIAL GUIDE:Problem SolvingFigures: All figures accompanying problem solving questions are intended to provide information useful in solving the problems. Figures are drawn as accurately as possible. Exceptions will be clearly noted. Lines shown as straight are straight, and lines that appear jagged are also straight. The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero. All figures lie in a plane unless otherwise indicated. Data Sufficiency:Figures:• Figures conform to the information given in the question, but will not necessarily conform to the additional information given in statements (1) and (2). • Lines shown as straight are straight, and lines that appear jagged are also straight. • The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero. • All figures lie in a plane unless otherwise indicated. Hope it helps.
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Re: Arc AED above is a semicircle. EC is the height to the base AD. The ar
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14 Jul 2019, 06:27
Bunuel wrote: plk wrote: Hi everyone! Does anybody know, whether figures in Problem Solving part are drawn on a scale? I have seen explanation of this task in Economist tutor, it is written there almost of the figures are on scale (if there is no some special signs like 'figure above is not on scale') and they uses a strategy calles 'ballpark', explaning this answer. So, could we trust scale of figures in questions like this? Thanks! OFFICIAL GUIDE:Problem SolvingFigures: All figures accompanying problem solving questions are intended to provide information useful in solving the problems. Figures are drawn as accurately as possible. Exceptions will be clearly noted. Lines shown as straight are straight, and lines that appear jagged are also straight. The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero. All figures lie in a plane unless otherwise indicated. Data Sufficiency:Figures:• Figures conform to the information given in the question, but will not necessarily conform to the additional information given in statements (1) and (2). • Lines shown as straight are straight, and lines that appear jagged are also straight. • The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero. • All figures lie in a plane unless otherwise indicated. Hope it helps. Thank you very much!



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Re: Arc AED above is a semicircle. EC is the height to the base AD. The ar
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14 Jul 2019, 10:31
Solved through a longer method: Since ArcAED is a semicircle ∆AED is right angled triangle at ∠E. ∴ AD^2 = AE^2 + ED^2 And AD^2 = (AC + CD)^2 = AC^2 + CD^2 + 2*AC*CD → Eqn ① AE^2 + ED^2 = EC^2 + AC^2 + EC^2 + CD^2 → Eqn ② Now Equating Eqn ① & ② we have AC^2 + CD^2 + 2*AC*CD = EC^2 + AC^2 + EC^2 + CD^2 Solving gives us 2*AC*CD = 2*EC^2 → Eqn ③ Also we have Area ∆AEC = 96 and EC = 12, ½*EC*AC = 96 AC = (2*96)/12 = 16 → Eqn ④ Now from Eqn ③ & ④ CD = ((12)^2)/16 = 9 → Eqn ⑤ Now Area ∆CDE = ½*EC*CD and using Eqn ⑤ = ½*12*9 = 54 Answer (B)
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