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705-805 (Hard)|   Geometry|      
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nick1816
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Line segments are drawn from the vertices of the large square to the m 10 May 2020, 06:18
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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85% (hard)

Question Stats:

41% (02:17) correct 59% (03:10) wrong based on 34 sessions

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Attachment:
6c4cOJzjJV-group-57.png
6c4cOJzjJV-group-57.png [ 13.1 KiB | Viewed 2636 times ]
Line segments are drawn from the vertices of the large square to the midpoints of the opposite sides to form a smaller, white square. If each red or blue line-segment measures 10 m long, what is the area of the smaller, white square?

A. 64
B. 75
C. 80
D. 81
E. 90
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C
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Line segments are drawn from the vertices of the large square to the m 10 May 2020, 12:00
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nick1816
Attachment:
The attachment 6c4cOJzjJV-group-57.png is no longer available
Line segments are drawn from the vertices of the large square to the midpoints of the opposite sides to form a smaller, white square. If each red or blue line-segment measures 10 m long, what is the area of the smaller, white square?

A. 64
B. 75
C. 80
D. 81
E. 90
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correct answer C
follow the attachment. solution is in details.
Attachments

geometry.jpg
geometry.jpg [ 1.82 MiB | Viewed 2555 times ]

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Line segments are drawn from the vertices of the large square to the m 10 May 2020, 23:01
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Attachment:
6c4cOJzjJV-group-57.png
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Triangle ABC is similar to triangle ADE

Hence, \(\frac{AC}{CE} = \frac{AB}{BD}\)

AC=CE = 2x

Also,
\(\frac{AB}{AD} = \frac{BC}{ED}\)

We get BC = x = EF

AC+CE+EF = AF

\(5x = \sqrt{20^2+10^2}\)

\(2x = 4\sqrt{5}\)

\(CE^2=(2x)^2 = 16*5=80\)
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Line segments are drawn from the vertices of the large square to the m 19 May 2020, 09:38
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Attachment:
6c4cOJzjJV-group-57.png
Triangle ABC is similar to triangle ADE

Hence, \(\frac{AC}{CE} = \frac{AB}{BD}\)

AC=CE = 2x

Also,
\(\frac{AB}{AD} = \frac{BC}{ED}\)

We get BC = x = EF

AC+CE+EF = AF

\(5x = \sqrt{20^2+10^2}\)

\(2x = 4\sqrt{5}\)

\(CE^2=(2x)^2 = 16*5=80\)
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When you say triangles ABC & ADE are similar, shouldn't the proportionality be AB/AD = AC/AE and not AB/BD = AC/CE;

The reason is BD and CE are not sides of a triangle from your figure. Can you or someone please clarify ??

Thank you.
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Line segments are drawn from the vertices of the large square to the m 19 May 2020, 10:55
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\(\frac{AD}{AB}= \frac{AE}{AC} =k\)

\(\frac{BD}{CE} = \frac{AD-AB}{AE-AC} = \frac{AB*k-AB}{ AC*k - AC} = \frac{AB(k-1)}{AC(k-1)} = \frac{AB}{AC }\)

\(\frac{BD}{CE}= \frac{AB}{AC }\)

\(\frac{AB}{BD} = \frac{AC}{CE}\)

OR we could apply Dividendo (Ratios property) to prove it.

OR The Triangle Proportionality Theorem states that if a line parallel to one side of a triangle intersects the other two sides, then it divides those sides proportionally.





SiddharthR


When you say triangles ABC & ADE are similar, shouldn't the proportionality be AB/AD = AC/AE and not AB/BD = AC/CE;

The reason is BD and CE are not sides of a triangle from your figure. Can you or someone please clarify ??

Thank you.
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Line segments are drawn from the vertices of the large square to the m 21 May 2020, 20:42
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\(\frac{AD}{AB}= \frac{AE}{AC} =k\)

\(\frac{BD}{CE} = \frac{AD-AB}{AE-AC} = \frac{AB*k-AB}{ AC*k - AC} = \frac{AB(k-1)}{AC(k-1)} = \frac{AB}{AC }\)

\(\frac{BD}{CE}= \frac{AB}{AC }\)

\(\frac{AB}{BD} = \frac{AC}{CE}\)

OR we could apply Dividendo (Ratios property) to prove it.

OR The Triangle Proportionality Theorem states that if a line parallel to one side of a triangle intersects the other two sides, then it divides those sides proportionally.





SiddharthR


When you say triangles ABC & ADE are similar, shouldn't the proportionality be AB/AD = AC/AE and not AB/BD = AC/CE;

The reason is BD and CE are not sides of a triangle from your figure. Can you or someone please clarify ??

Thank you.
Show more



That's interesting. Thank you for that insight. Could you please also tell me how you arrived at length "2x" for AC & CE ?

I got the following - AB/AD = 10/20 = 1/2;
Let BC = x. Then BC/DE = 1/2; From this we get DE = 2x

From here, I am unable to understand how you reached 2x distance for AC & CE.
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Line segments are drawn from the vertices of the large square to the m 29 Jul 2024, 13:40
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