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655-705 (Hard)|   Geometry|      
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Bunuel
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In the figure above, square ABCE has an area of 81, and points G and F 26 Nov 2019, 01:59
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65% (hard)

Question Stats:

64% (03:08) correct 36% (02:31) wrong based on 94 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. \(\frac{31}{7}\)


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In the figure above, square ABCE has an area of 81, and points G and F 26 Nov 2019, 02:39
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square ABCE has an area of 81; hence AB=BC=CE=AE=9

GF= 1/3*9=3

Area(FGD)= 1/2*GF*AD=24

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

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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Krish728
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In the figure above, square ABCE has an area of 81, and points G and F 16 Dec 2019, 12:24
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nick1816
square ABCE has an area of 81; hence AB=BC=CE=AE=9

GF= 1/3*9=3

Area(FGD)= 1/2*GF*AD=24

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

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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Hi,

Can you please explain why did you use AD instead of FD?

Thanks
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In the figure above, square ABCE has an area of 81, and points G and F 16 Dec 2019, 13:27
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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
nick1816
square ABCE has an area of 81; hence AB=BC=CE=AE=9

GF= 1/3*9=3

Area(FGD)= 1/2*GF*AD=24

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

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


Show Spoiler
Attachment:
Trapezoid2.jpg

Hi,

Can you please explain why did you use AD instead of FD?

Thanks
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In the figure above, square ABCE has an area of 81, and points G and F 06 May 2025, 07:04
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