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11 Sep 2021, 00:21
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
37% (02:04) correct
63%
(02:15)
wrong
based on 49
sessions
History
Date
Time
Result
Not Attempted Yet
Calculate the area of cross-section:
Topic 03.png [ 18.8 KiB | Viewed 2174 times ]
A. 0.6 \(cm^2\)
B. 0.8 \(cm^2\)
C. 1.0 \(cm^2\)
D. 1.2 \(cm^2\)
E. 1.4 \(cm^2\)
Attachment:
Topic 03.png [ 18.8 KiB | Viewed 2174 times ]
A. 0.6 \(cm^2\)
B. 0.8 \(cm^2\)
C. 1.0 \(cm^2\)
D. 1.2 \(cm^2\)
E. 1.4 \(cm^2\)
12 Sep 2021, 23:44
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ostrick5465
Let us mark the points as shown in the attached figure.
Attachment:
Topic 03.png [ 44.79 KiB | Viewed 1995 times ]
(I) Take \(\triangle ABC\) and \(\triangle BCD\)
\(\angle BCD\) is common
\(\angle CBD=\angle CAB=x\)
The third angle will also be equal, so \(\angle BDC = \angle BDA=90\)
Also, both \(\triangle AFE\) and \(\triangle BCD\) are congruent.
(II) \(\triangle ADB\), a right angled triangle
Let side BD be x, so AF=BD=x, and since AFE and ABD are similar, AF=FD=t
Area of cross section, which is a square with side t = \(t^2\)
In \(\triangle ABD\), \(AD^2+BD^2=AB^2\)
\((2t)^2+t^2=2^2..........4t^2+t^2=4........5t^2=4........t^2=\frac{4}{5}=0.8\)
Area of cross section, which is a square with side t = \(t^2\)=0.8 \(cm^2\)
B
13 Sep 2021, 05:15
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Attachment:
Topic 03 explanation.png [ 34.47 KiB | Viewed 1933 times ]
=> IMO B
