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# ABC and AED are triangles with BC parallel to ED. Find the a

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ABC and AED are triangles with BC parallel to ED. Find the a  [#permalink]

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Updated on: 11 Oct 2013, 03:43
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Question Stats:

74% (01:27) correct 26% (01:53) wrong based on 119 sessions

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ABC and AED are triangles with BC parallel to ED. Find the area of BCDE if the area of ABC is 16.
Attachment:

Triangle_Barrons.png [ 45.21 KiB | Viewed 4666 times ]

(1) BC = 8
(2) ED = 5

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Originally posted by obs23 on 11 Oct 2013, 03:42.
Last edited by Bunuel on 11 Oct 2013, 03:43, edited 1 time in total.
Renamed the topic and edited the question.
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Re: ABC and AED are triangles with BC parallel to ED. Find the a  [#permalink]

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11 Oct 2013, 03:56
1
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ABC and AED are triangles with BC parallel to ED. Find the area of BCDE if the area of ABC is 16.

(1) BC = 8. We don't know where is ED positioned. Not sufficient.
(2) ED = 5. We don't know the other sides. Not sufficient.

(1)+(2) Notice that angles of triangles ABC and AED are equal, which means that the triangles are similar.

Property:
If two similar triangles have sides in the ratio $$\frac{x}{y}$$, then their areas are in the ratio $$\frac{x^2}{y^2}$$.
OR in another way: in two similar triangles, the ratio of their areas is the square of the ratio of their sides: $$\frac{AREA}{area}=\frac{SIDE^2}{side^2}$$.

Since BC/ED=8/5, then $$\frac{AREA_{ABC}}{area_{AED}}=\frac{8^2}{5^2}=\frac{64}{25}$$. As given that the area of ABS is 16, then we can find the area of AED and then find the area of BCDE. Sufficient.

Hope it's clear.
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Re: ABC and AED are triangles with BC parallel to ED. Find the a  [#permalink]

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14 Oct 2013, 04:41
Bunuel wrote:
ABC and AED are triangles with BC parallel to ED. Find the area of BCDE if the area of ABC is 16.

(1) BC = 8. We don't know where is ED positioned. Not sufficient.
(2) ED = 5. We don't know the other sides. Not sufficient.

(1)+(2) Notice that angles of triangles ABC and AED are equal, which means that the triangles are similar.

Property:
If two similar triangles have sides in the ratio $$\frac{x}{y}$$, then their areas are in the ratio $$\frac{x^2}{y^2}$$.
OR in another way: in two similar triangles, the ratio of their areas is the square of the ratio of their sides: $$\frac{AREA}{area}=\frac{SIDE^2}{side^2}$$.

Since BC/ED=8/5, then $$\frac{AREA_{ABC}}{area_{AED}}=\frac{8^2}{5^2}=\frac{64}{25}$$. As given that the area of ABS is 16, then we can find the area of AED and then find the area of BCDE. Sufficient.

Hope it's clear.

Wonderfully done - one question please: "the area of ABC is 16" (in the stimulus) - is there anything we can infer from this information? Or does this information help us in any way before looking at S1 and S2?
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Posts: 54493
Re: ABC and AED are triangles with BC parallel to ED. Find the a  [#permalink]

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14 Oct 2013, 04:44
1
obs23 wrote:
Bunuel wrote:
ABC and AED are triangles with BC parallel to ED. Find the area of BCDE if the area of ABC is 16.

(1) BC = 8. We don't know where is ED positioned. Not sufficient.
(2) ED = 5. We don't know the other sides. Not sufficient.

(1)+(2) Notice that angles of triangles ABC and AED are equal, which means that the triangles are similar.

Property:
If two similar triangles have sides in the ratio $$\frac{x}{y}$$, then their areas are in the ratio $$\frac{x^2}{y^2}$$.
OR in another way: in two similar triangles, the ratio of their areas is the square of the ratio of their sides: $$\frac{AREA}{area}=\frac{SIDE^2}{side^2}$$.

Since BC/ED=8/5, then $$\frac{AREA_{ABC}}{area_{AED}}=\frac{8^2}{5^2}=\frac{64}{25}$$. As given that the area of ABS is 16, then we can find the area of AED and then find the area of BCDE. Sufficient.

Hope it's clear.

Wonderfully done - one question please: "the area of ABC is 16" (in the stimulus) - is there anything we can infer from this information? Or does this information help us in any way before looking at S1 and S2?

We just know that the area is 16, we can get nothing more from it.
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Re: ABC and AED are triangles with BC parallel to ED. Find the a  [#permalink]

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09 Dec 2014, 10:11
Hi Bunuel,

Without combining (1) and (2), can't we say both the triangles are similar? i am asking this because i did not understand the reason for insufficiency "We don't know where is ED positioned. Not sufficient."

I marked the answer D because i thought two triangles are similar. Hence, knowing BC=8 will give me the height of the triangle ABC because the area is given. As we know, for similar triangles their ratios of heights and bases are equal i will be able to find the area of triangle AED.

Thanks
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Re: ABC and AED are triangles with BC parallel to ED. Find the a  [#permalink]

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09 Dec 2014, 10:36
HKD1710 wrote:
Hi Bunuel,

Without combining (1) and (2), can't we say both the triangles are similar? i am asking this because i did not understand the reason for insufficiency "We don't know where is ED positioned. Not sufficient."

I marked the answer D because i thought two triangles are similar. Hence, knowing BC=8 will give me the height of the triangle ABC because the area is given. As we know, for similar triangles their ratios of heights and bases are equal i will be able to find the area of triangle AED.

Thanks

Yes, we know that they are similar from the stem but this does not help. Check the image below:
Attachment:

Untitled.png [ 49.29 KiB | Viewed 3081 times ]

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Re: ABC and AED are triangles with BC parallel to ED. Find the a  [#permalink]

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09 Dec 2014, 11:12
Bunuel wrote:
HKD1710 wrote:
Hi Bunuel,

Without combining (1) and (2), can't we say both the triangles are similar? i am asking this because i did not understand the reason for insufficiency "We don't know where is ED positioned. Not sufficient."

I marked the answer D because i thought two triangles are similar. Hence, knowing BC=8 will give me the height of the triangle ABC because the area is given. As we know, for similar triangles their ratios of heights and bases are equal i will be able to find the area of triangle AED.

Thanks

Yes, we know that they are similar from the stem but this does not help. Check the image below:
Attachment:
Untitled.png

By looking at the image attached, i understood what you meant by "We don't know where is ED positioned. Not sufficient."

I conclude that in case of similar triangles when the figure is not drawn to scale and value of base for both the triangles is not known then knowing the value of base for only one triangle will always lead to this situation and will be insufficient. Please confirm!

Thank you
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Re: ABC and AED are triangles with BC parallel to ED. Find the a  [#permalink]

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06 Sep 2017, 23:57
@Bunuel- just making sure- we cannot actually infer that both triangles are 90 necessarily right triangles right? In other words- we know it's C because of the property you explained not because we can make the assumption that both triangles are right triangles in which case it also makes sense that the answer is C? The fact that I did not think both triangles were right triangles was the reason I second guessed and picked E
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Re: ABC and AED are triangles with BC parallel to ED. Find the a  [#permalink]

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07 Sep 2017, 00:02
Nunuboy1994 wrote:
@Bunuel- just making sure- we cannot actually infer that both triangles are 90 necessarily right triangles right? In other words- we know it's C because of the property you explained not because we can make the assumption that both triangles are right triangles in which case it also makes sense that the answer is C? The fact that I did not think both triangles were right triangles was the reason I second guessed and picked E

Yes, we don't know that the triangles are right.
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Re: ABC and AED are triangles with BC parallel to ED. Find the a  [#permalink]

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08 Sep 2017, 18:18
obs23 wrote:
ABC and AED are triangles with BC parallel to ED. Find the area of BCDE if the area of ABC is 16.
Attachment:
Triangle_Barrons.png

(1) BC = 8
(2) ED = 5

The proportion of the areas would simply be

8^2/5^2

C
Re: ABC and AED are triangles with BC parallel to ED. Find the a   [#permalink] 08 Sep 2017, 18:18
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