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Arc AED above is a semicircle. EC is the height to the base AD. The ar

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Arc AED above is a semicircle. EC is the height to the base AD. The ar  [#permalink]

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New post 27 Oct 2014, 08:17
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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  [#permalink]

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New post 27 Oct 2014, 19:02
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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
T8910x.png [ 22.56 KiB | Viewed 3766 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  [#permalink]

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New post 27 Oct 2014, 19:09
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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  [#permalink]

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New post 17 Dec 2014, 13:41
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  [#permalink]

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New post 17 Dec 2014, 18:13
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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  [#permalink]

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New post 17 Dec 2014, 20:31
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  [#permalink]

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New post 18 Dec 2014, 03:39
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 ... -the-gmat/
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Arc AED above is a semicircle. EC is the height to the base AD. The ar  [#permalink]

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New post 18 Dec 2014, 04:01
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  [#permalink]

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New post 19 Dec 2014, 04:52
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  [#permalink]

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New post 06 Sep 2017, 00:01
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  [#permalink]

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Re: Arc AED above is a semicircle. EC is the height to the base AD. The ar   [#permalink] 28 Jan 2019, 08:46
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