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virtualanimosity
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Geometry 2 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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Geometry 2 26 Sep 2009, 12:33
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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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deepakraam
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Geometry 2 27 Sep 2009, 01:48
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Certainly it shud be 9:16
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virtualanimosity
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Geometry 2 28 Sep 2009, 14:32
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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
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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arkadiyua
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Geometry 2 28 Sep 2009, 15:49
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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(ADE) = 3x/2
S(BCDE) = (3+5)/2 *y = 4y
S(ABC) = 5*(x+y)/2

Since S(ABC)=S(ADE)+S(BCDE)
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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Geometry 2 28 Sep 2009, 15:52
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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(Triangle) = Area(ADE)
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
triangle.JPG [ 10.83 KiB | Viewed 9572 times ]

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Mikko
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Geometry 2 28 Sep 2009, 22:00
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[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
untitled.JPG [ 9.57 KiB | Viewed 9492 times ]


ADK & ABH are similar triangle too:
AD/AB=AK/AH

Tri ADE & ABC are similar triangle:
AD/AB=AE/AC=DE/BC

SADE/SABC=(AK*DE/2)/(AH*BC/2)
=AK*DE/AH*BC
=AK/AH * DE/BC
=AD/AB * DE/BC
=DE/BC * DE/BC
=DE^2/BC^2

Is it clear to U now Virtualanimosity?
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virtualanimosity
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Geometry 2 02 Oct 2009, 08:38
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Thanks man...crystal clear
great detailed explanation..really helpful!!

Mikko
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


ADK & ABH are similar triangle too:
AD/AB=AK/AH

Tri ADE & ABC are similar triangle:
AD/AB=AE/AC=DE/BC

SADE/SABC=(AK*DE/2)/(AH*BC/2)
=AK*DE/AH*BC
=AK/AH * DE/BC
=AD/AB * DE/BC
=DE/BC * DE/BC
=DE^2/BC^2

Is it clear to U now Virtualanimosity?
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