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# In the figure above, square ABCE has an area of 81, and points G and F

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Math Expert
Joined: 02 Sep 2009
Posts: 60727
In the figure above, square ABCE has an area of 81, and points G and F  [#permalink]

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26 Nov 2019, 01:59
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Difficulty:

75% (hard)

Question Stats:

40% (02:37) correct 60% (02:43) wrong based on 25 sessions

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In the figure above, square ABCE has an area of 81, and points G and F trisect AB. If the area of $$\triangle{FGD}$$ is 24, what is the length of HE ?

A. 3

B. $$\frac{42}{16}$$

C. 4

D. $$\frac{14}{3}$$

E. $$\frac{31}{7}$$

Are You Up For the Challenge: 700 Level Questions

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Trapezoid2.jpg [ 15.8 KiB | Viewed 422 times ]

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Joined: 19 Oct 2018
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Location: India
Re: In the figure above, square ABCE has an area of 81, and points G and F  [#permalink]

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26 Nov 2019, 02:39
2
square ABCE has an area of 81; hence AB=BC=CE=AE=9

GF= 1/3*9=3

AD= $$\frac{24*2}{3}=16$$

DE=16-9=7

DHE is similar to DGA

$$\frac{HE}{GA}=\frac{DE}{AD}$$

HE=$$\frac{7*6}{16}=\frac{42}{16}$$

Bunuel wrote:

In the figure above, square ABCE has an area of 81, and points G and F trisect AB. If the area of $$\triangle{FGD}$$ is 24, what is the length of HE ?

A. 3

B. $$\frac{42}{16}$$

C. 4

D. $$\frac{14}{3}$$

E. $$\frac{31}{7}$$

Are You Up For the Challenge: 700 Level Questions

Attachment:
Trapezoid2.jpg
Intern
Joined: 24 Jun 2019
Posts: 13
Re: In the figure above, square ABCE has an area of 81, and points G and F  [#permalink]

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16 Dec 2019, 12:24
nick1816 wrote:
square ABCE has an area of 81; hence AB=BC=CE=AE=9

GF= 1/3*9=3

AD= $$\frac{24*2}{3}=16$$

DE=16-9=7

DHE is similar to DGA

$$\frac{HE}{GA}=\frac{DE}{AD}$$

HE=$$\frac{7*6}{16}=\frac{42}{16}$$

Bunuel wrote:

In the figure above, square ABCE has an area of 81, and points G and F trisect AB. If the area of $$\triangle{FGD}$$ is 24, what is the length of HE ?

A. 3

B. $$\frac{42}{16}$$

C. 4

D. $$\frac{14}{3}$$

E. $$\frac{31}{7}$$

Are You Up For the Challenge: 700 Level Questions

Attachment:
Trapezoid2.jpg

Hi,

Thanks
GMAT Tutor
Joined: 17 Sep 2014
Posts: 336
Location: United States
GMAT 1: 780 Q51 V45
GRE 1: Q170 V167
Re: In the figure above, square ABCE has an area of 81, and points G and F  [#permalink]

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16 Dec 2019, 13:27
Hello Krish728 !
The height of triangle GFD here is AD not FD, the height is always perpendicular to the base. Triangle GFD and FAD both have the same height (AD) for example.

Krish728 wrote:
nick1816 wrote:
square ABCE has an area of 81; hence AB=BC=CE=AE=9

GF= 1/3*9=3

AD= $$\frac{24*2}{3}=16$$

DE=16-9=7

DHE is similar to DGA

$$\frac{HE}{GA}=\frac{DE}{AD}$$

HE=$$\frac{7*6}{16}=\frac{42}{16}$$

Bunuel wrote:

In the figure above, square ABCE has an area of 81, and points G and F trisect AB. If the area of $$\triangle{FGD}$$ is 24, what is the length of HE ?

A. 3

B. $$\frac{42}{16}$$

C. 4

D. $$\frac{14}{3}$$

E. $$\frac{31}{7}$$

Are You Up For the Challenge: 700 Level Questions

Attachment:
Trapezoid2.jpg

Hi,

Thanks

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Re: In the figure above, square ABCE has an area of 81, and points G and F   [#permalink] 16 Dec 2019, 13:27
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