.........................

Answer:

[tex]2[/tex]

Step-by-step explanation:

Let [tex][n][/tex] denote the area of some polygon [tex]n[/tex].

The sum of the area of quadrilaterals BCDE and MNDE is equal to [tex][ABC]-([MBE]+[NDC]+AMN])[/tex] (big triangle subtracting the three smaller triangles that aren't part of BCDE or MNDE).

Let the area of triangle ABC be [tex]\alpha[/tex].

Triangle AMN is similar to triangle ABC. Because their side lengths are in ratio 1:2 respectively, the ratio of the area of triangle AMN to triangle ABC is [tex]1:2^2=1:4[/tex].

Therefore, [tex]\displaystyle [AMN]=\frac{1}{4}[ABC]=\frac{1}{4}\alpha[/tex].

Now for triangle NDC, NDC is similar to triangle AMC. Since M is the midpoint of AB, AMC is half the area of ABC. The ratio of the side lengths between triangle NDC and triangle AMC is [tex]1:2[/tex] because the D is the midpoint of MC, so it follows the ratio of their areas must be [tex]1:4[/tex], respectively.

Therefore, [tex]\displaystyle[NDC]=\frac{1}{4}[AMC]=\frac{1}{8}[ABC]=\frac{1}{8}\alpha[/tex].

We can use the same concept for triangle MEB. Triangle MEB is similar to triangle ANB, and the ratio of their sides is also [tex]1:2[/tex], since E is the midpoint of BN. Their areas are in the ratio [tex]1:2^2=1:4[/tex] and we have [tex]\displaystyle [MBE]=\frac{1}{4}[ANB]=\frac{1}{8}\alpha[/tex].

Thus,

[tex]\displaystyle \frac{[ABC]}{[BCDE]+[MNDE]}=\frac{\alpha}{\alpha-\frac{1}{4}\alpha-\frac{1}{8}\alpha-\frac{1}{8}\alpha}=\frac{\alpha}{0.5\alpha}=\frac{1}{\frac{1}{2}}=\boxed{2}[/tex]

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protection
factor
0
30
Thickness Count on rate-meter (counts per minute)
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Kk
2
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(cm)
0
446 444 449
1
231 238
240
113
111
110
3
65
63
65
4
34
31
35
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1
2
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5
2.
10
15
20
448
456
449
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1
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3
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1
Average
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10
70
75
73
20
19
19
21
20
90
96
94
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33
31
31
40
30
49
44
46
95
98
99
50
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