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The figure above consists of triangle ADF and rectangle ABCD. BE = CG

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The figure above consists of triangle ADF and rectangle ABCD. BE = CG  [#permalink]

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New post 11 Oct 2018, 01:51
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The figure above consists of triangle ADF and rectangle ABCD. BE = CG = 1.5 and EG = 2. If the area of triangle EFG = x, what is the area of rectangle ABCD?


A. \(5x\)

B. \(\frac{5x}{2}\)

C. \(15x\)

D. \(\frac{15x}{2}\)

E. \(\frac{25x}{2}\)


Attachment:
figure.jpg
figure.jpg [ 10.38 KiB | Viewed 913 times ]

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Re: The figure above consists of triangle ADF and rectangle ABCD. BE = CG  [#permalink]

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New post 11 Oct 2018, 21:00
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Bunuel wrote:
Image
The figure above consists of triangle ADF and rectangle ABCD. BE = CG = 1.5 and EG = 2. If the area of triangle EFG = x, what is the area of rectangle ABCD?


A. \(5x\)

B. \(\frac{5x}{2}\)

C. \(15x\)

D. \(\frac{15x}{2}\)

E. \(\frac{25x}{2}\)


Attachment:
figure.jpg



BC = 1.5 + 2 + 1.5 = 5 = AD

Since ABCD is a rectangle, BC is parallel to AD. So triangle EFG is similar to triangle AFD.
\(Sides Ratio = \frac{EG}{AD} = \frac{2}{5}\)
Ratio of altitudes will be 2/5 too

Area of EFG = x = (1/2)*Altitude * 2
Altitude of triangle EFG = x

So altitude of triangle AFD \(= \frac{5x}{2}\)

\(CD = \frac{5x}{2} - x = \frac{3x}{2}\)

Area of rectangle ABCD \(= AD * CD = 5 * \frac{3x}{2} = \frac{15x}{2}\)
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Re: The figure above consists of triangle ADF and rectangle ABCD. BE = CG  [#permalink]

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New post 11 Oct 2018, 03:16
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Bunuel wrote:
Image
The figure above consists of triangle ADF and rectangle ABCD. BE = CG = 1.5 and EG = 2. If the area of triangle EFG = x, what is the area of rectangle ABCD?


A. \(5x\)

B. \(\frac{5x}{2}\)

C. \(15x\)

D. \(\frac{15x}{2}\)

E. \(\frac{25x}{2}\)



Attachment:
figure.jpg



Please check the video solution using similar triangle property here...



Answer: Option D
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Re: The figure above consists of triangle ADF and rectangle ABCD. BE = CG  [#permalink]

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New post 11 Oct 2018, 20:22
I think something is wrong with the framing. If you choose the triangles ABE similar to FGE. The resulting area is 15x/4 and not 15x/2. Here is my reasoning.

In triangle FGE , let the altitude be h then h = x. (Area = 1/2* 2 * h).

By similar triangle property
BE/GE= AB/x (ratio of heights) -----> AB= 3x/4
AB*5 = area required = 15x/4

:0

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Re: The figure above consists of triangle ADF and rectangle ABCD. BE = CG  [#permalink]

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New post 11 Oct 2018, 23:10
1
Hi,

Already this question has been clearly explained above.

Just adding only one point, instead of “x” as the area of triangle, we can use some values for faster calculation (only for those who are very comfortable with numbers than variables).

BC = AD = 5(Given)

Area of rectangle = length * breadth = 5 * AB = 5 * AB

We need to find the value of AB.

Triangles EGF and ADF are similar triangles (EF and AB are parallel to each other).

In similar triangles, angles are equal and sides are proportional.

EG / AD = 2/5

So, height of these two triangles are also similar,

So, if we consider, height of the triangle EGF as 4,

Then given of the triangle EGF = x = ½ * 2 * 4 = 4

So, x = 4

According to the similar triangle rule,

Heights should also be in the ratio of 2/5 = 4/ 10

Breadth of the rectangle ABCD = Height of the triangle ADF – Height of the triangle EGF = 10 – 4 = 6

So the area of rectangle = 5 * 6 = 30.

Just substitute x = 4 in the answer choices, and check where the answer is 30.

A. 5x - 20

B. 5x/2 - 10

C. 15x - 60

D. 15x/2 - 30

E. 25x/2 – 25

So answer is D.

Hope this helps.
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Re: The figure above consists of triangle ADF and rectangle ABCD. BE = CG  [#permalink]

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New post 11 Oct 2018, 23:25
Hi VeritasKarishma

I am still not clear with this. why can't we consider triangles ABE and EFG as similar and then use ratio side proportion ? Where am I going wrong ?

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The figure above consists of triangle ADF and rectangle ABCD. BE = CG  [#permalink]

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New post 14 Oct 2018, 10:32
ShankSouljaBoi wrote:
Hi VeritasKarishma

I am still not clear with this. why can't we consider triangles ABE and EFG as similar and then use ratio side proportion ? Where am I going wrong ?

Posted from my mobile device


Even if it's not drawn to scale , there's no data to prove whether ABE and EFG are similar
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Re: The figure above consists of triangle ADF and rectangle ABCD. BE = CG  [#permalink]

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New post 15 Oct 2018, 02:21
ShankSouljaBoi wrote:
Hi VeritasKarishma

I am still not clear with this. why can't we consider triangles ABE and EFG as similar and then use ratio side proportion ? Where am I going wrong ?

Posted from my mobile device


Triangles ABE and EFG are not similar to each other.

Only one pair of angles is equal.
angle BEA = angle FEG

There is no second pair of equal angles. Triangles are similar by AA i.e. two angles need to be equal to two angles of the other triangle.
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Re: The figure above consists of triangle ADF and rectangle ABCD. BE = CG  [#permalink]

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New post 15 Oct 2018, 02:56
Hi VeritasKarishma
Thank you for your response. I took the triangles similar as SAS
1.Angle as vertically opposites angles (angle BEA and angle FEG).
2.BE and BG are a part of the same straight line.
3.FE and EA are also a part of the same straight line and so they will have some sort of ratio
and thus by SAS both triangles are similar.

I think I m totally wrong here. Really need you expert advice on picking two similar triangles. I have already read both veritas articles on similarity, but this remains a common issue.

Posted from my mobile device
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The figure above consists of triangle ADF and rectangle ABCD. BE = CG  [#permalink]

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New post Updated on: 15 Oct 2018, 04:13
pandeyashwin wrote:
ShankSouljaBoi wrote:
Hi VeritasKarishma

I am still not clear with this. why can't we consider triangles ABE and EFG as similar and then use ratio side proportion ? Where am I going wrong ?

Posted from my mobile device


Even if it's not drawn to scale , there's no data to prove whether ABE and EFG are similar



pandeyashwin

Who says, ABE and EFG are similar??? Of course they are NOT similar

I mentioned in my solution that AFD and EFG are similar triangle because they share one common angle at point F and since, Ab and EG are parallel therefore, angle A and angle E also will be similar in respective similar triangles hence we can always derive that the triangles AFD and EFG are similar.

For detailed solution you may check the video solution of this question here.


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Originally posted by GMATinsight on 15 Oct 2018, 03:59.
Last edited by GMATinsight on 15 Oct 2018, 04:13, edited 2 times in total.
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Re: The figure above consists of triangle ADF and rectangle ABCD. BE = CG  [#permalink]

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New post 15 Oct 2018, 04:10
ShankSouljaBoi wrote:
Hi VeritasKarishma
Thank you for your response. I took the triangles similar as SAS
1.Angle as vertically opposites angles (angle BEA and angle FEG).
2.BE and BG are a part of the same straight line.
3.FE and EA are also a part of the same straight line and so they will have some sort of ratio
and thus by SAS both triangles are similar.

I think I m totally wrong here. Really need you expert advice on picking two similar triangles. I have already read both veritas articles on similarity, but this remains a common issue.

Posted from my mobile device


ShankSouljaBoi

Please check the Video and see if it helps in understanding how to use similar triangle property.


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Re: The figure above consists of triangle ADF and rectangle ABCD. BE = CG  [#permalink]

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New post 15 Oct 2018, 23:09
2
ShankSouljaBoi wrote:
Hi VeritasKarishma
Thank you for your response. I took the triangles similar as SAS
1.Angle as vertically opposites angles (angle BEA and angle FEG).
2.BE and BG are a part of the same straight line.
3.FE and EA are also a part of the same straight line and so they will have some sort of ratio
and thus by SAS both triangles are similar.

I think I m totally wrong here. Really need you expert advice on picking two similar triangles. I have already read both veritas articles on similarity, but this remains a common issue.

Posted from my mobile device


Being a part of the same straight line does not imply that they are in the same ratio.
For SAS, we need
BE/EG = FE/EA

Even by just looking, we see that BE is much smaller than EG while FE and EA look similar. But we should not reply on what seems.
Triangle BEA is right angled triangle whereas EFG is certainly not right angled. So they cannot be similar. Even if you use one property to establish similarity, all other properties must be applicable too.
On the other hand, EFG and AFD are certainly similar triangles. There are some figures that should remind you of similar triangles immediately. Some of them are shown here: https://www.veritasprep.com/blog/2014/0 ... -the-gmat/
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Re: The figure above consists of triangle ADF and rectangle ABCD. BE = CG &nbs [#permalink] 15 Oct 2018, 23:09
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