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

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

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26 Nov 2019, 00:59
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65% (hard)

Question Stats:

48% (02:45) correct 52% (02:43) wrong based on 29 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 450 times ]

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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, 01:39
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1
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, 11:24
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: 16 Sep 2014
Posts: 391
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, 12:27
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, 12:27
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