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# Geometry 2

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Manager
Joined: 19 Aug 2009
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26 Sep 2009, 03:22
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Q.
In a triangle ABC, point D is on side AB and pt E is on side AC, such that BCED is a trapezium. DE : BC = 3:5. Calculate the ratio of the triangle ADE and trapezium BCED.

1. 3:4
2. 9:16
3. 3:5
4. 9:25
5. 5:9

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Manager
Joined: 27 Oct 2008
Posts: 185

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Re: Geometry 2 [#permalink]

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26 Sep 2009, 12:33
1
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I think the answer is 9/16. OA pls.

My method is, area of tri(ADE)/area of tri(ABC) = DE^2 / BC^2 since tri(ADE) ~ tri(ABC)
thus
area of tri(ADE)/area of tri(ABC) = 9/25
from this we have
area of tri(ABC)/area of tri(ADE) = 25/9
we also know that area of tri(ABC) = area of tri(ADE) + area of trap(BCDE)
so substituting

(area of tri(ADE) + area of trap(BCDE))/area of tri(ADE) = 25/9
rearranging
1 + area of trap(BCDE)/area of tri(ADE) = 25/9
taking 1 to other side
area of trap(BCDE)/area of tri(ADE) = (25/9) - 1 = 16/9

since question is asking for area of tri(ADE)/area of trap(BCDE) = 9/16

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Manager
Joined: 15 Sep 2009
Posts: 131

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Re: Geometry 2 [#permalink]

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27 Sep 2009, 01:48
Certainly it shud be 9:16

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Manager
Joined: 19 Aug 2009
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Re: Geometry 2 [#permalink]

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28 Sep 2009, 14:32
Pls explain this step again-

My method is, area of tri(ADE)/area of tri(ABC) = DE^2 / BC^2 since tri(ADE) ~ tri(ABC)

ok...understood triangles are similar..but confusion over how u reached DE^2/ BC^2...pls elaborate this step

srivas wrote:
I think the answer is 9/16. OA pls.

My method is, area of tri(ADE)/area of tri(ABC) = DE^2 / BC^2 since tri(ADE) ~ tri(ABC)
thus
area of tri(ADE)/area of tri(ABC) = 9/25
from this we have
area of tri(ABC)/area of tri(ADE) = 25/9
we also know that area of tri(ABC) = area of tri(ADE) + area of trap(BCDE)
so substituting

(area of tri(ADE) + area of trap(BCDE))/area of tri(ADE) = 25/9
rearranging
1 + area of trap(BCDE)/area of tri(ADE) = 25/9
taking 1 to other side
area of trap(BCDE)/area of tri(ADE) = (25/9) - 1 = 16/9

since question is asking for area of tri(ADE)/area of trap(BCDE) = 9/16

Kudos [?]: 284 [0], given: 46

Manager
Joined: 04 Sep 2009
Posts: 53

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WE 1: Real estate investment consulting
Re: Geometry 2 [#permalink]

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28 Sep 2009, 15:49
1
KUDOS
It is 9/16

I approached via algebra (this is for the ones not familiar with the properties of similar triangles ):

Let the height of ADE = x, height of BCDE = y

Then:

S(BCDE) = (3+5)/2 *y = 4y
S(ABC) = 5*(x+y)/2

3x/2 + 4y = 5(x+y)/2
3x+8y=5x+5y
x=1.5y

Now the ratio of ADE to BCDE = 3x/2 / 4y
3*1.5y/2 * 1/4y
4.5y/8y
4.5/8
9/16

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Manager
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Re: Geometry 2 [#permalink]

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28 Sep 2009, 15:52
1
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I agree that the answer is B: 9:16 .

Please note though that the triangles DO NOT have to be similar. However, since this case applies to all trapezium, triangle pairs, you can simplify the problem by assuming ADE is similar to ABC. I took a slightly different approach. Since there are no restrictions given on Triangle ABC, I decided to create one that simplifies the problem (see attached image).

I assume that ABC is a right triangle, and that DE is parallel to BC so that ADE is similar to ABC. Therefore,

Area(Trapezium) = Area(ABC) - Area(ADE)

Area(ADE) = 3 * 3 / 2 = 9/2
Area(ABC) = 5 * 5 / 2 = 25/2

Area(Triangle) = 9/2
Area(Trapezium) = 25/2 - 9/2 = 16/2 = 8

Area(Triangle) : Area(Trapezium) = 9/2 : 8 = 9 : 16
Attachments

triangle.JPG [ 10.83 KiB | Viewed 3665 times ]

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Manager
Joined: 25 Mar 2009
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Re: Geometry 2 [#permalink]

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28 Sep 2009, 22:00
1
KUDOS
[quote="virtualanimosity"]Pls explain this step again-

My method is, area of tri(ADE)/area of tri(ABC) = DE^2 / BC^2 since tri(ADE) ~ tri(ABC)

ok...understood triangles are similar..but confusion over how u reached DE^2/ BC^2...pls elaborate this step

Attachment:

untitled.JPG [ 9.57 KiB | Viewed 3611 times ]

ADK & ABH are similar triangle too:

Tri ADE & ABC are similar triangle:

=AK*DE/AH*BC
=AK/AH * DE/BC
=DE/BC * DE/BC
=DE^2/BC^2

Is it clear to U now Virtualanimosity?

Kudos [?]: 19 [1], given: 9

Manager
Joined: 19 Aug 2009
Posts: 78

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Re: Geometry 2 [#permalink]

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02 Oct 2009, 08:38
Thanks man...crystal clear
great detailed explanation..really helpful!!

Mikko wrote:
virtualanimosity wrote:
Pls explain this step again-

My method is, area of tri(ADE)/area of tri(ABC) = DE^2 / BC^2 since tri(ADE) ~ tri(ABC)

ok...understood triangles are similar..but confusion over how u reached DE^2/ BC^2...pls elaborate this step

Attachment:
untitled.JPG

ADK & ABH are similar triangle too:

Tri ADE & ABC are similar triangle:

=AK*DE/AH*BC
=AK/AH * DE/BC
=DE/BC * DE/BC
=DE^2/BC^2

Is it clear to U now Virtualanimosity?

Kudos [?]: 284 [0], given: 46

Re: Geometry 2   [#permalink] 02 Oct 2009, 08:38
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# Geometry 2

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