(解答) |
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ECとAH,FH,FI,GIの交点をそれぞれ、P,Q,R,Sとすると、 |
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ABとFH,GIは、それぞれ互いに平行であり、CI:CH:CB=1:2:3であるから、 |
SI:QH:EB=1:2:3 |
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△PAEと△PHQは相似であるから、AP:PH=1:2 |
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△ABHは、平行四辺形ABCDの |
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よって、△AEPの面積は、平行四辺形ABCDの |
1 |
× |
1 |
× |
1 |
= |
1 |
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6 |
1+3 |
1+2 |
72 |
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△HFAも、平行四辺形ABCDの |
1 |
であるから、△HQPの面積は、 |
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6 |
|
1 |
× |
2 |
× |
2 |
= |
1 |
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6 |
2+1 |
2+2 |
18 |
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△FQRは、△HQPと合同であるから、その面積は、平行四辺形ABCDの |
1 |
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18 |
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△ISRは、△AEPと合同であるから、その面積は、平行四辺形ABCDの |
1 |
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72 |
|
△GICも、平行四辺形ABCDの |
1 |
であるから、△GSCの面積は、 |
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6 |
|
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よって、黒くぬった部分の面積は、平行四辺形ABCDの面積の |
1 |
×2 |
+ |
1 |
×2+ |
1 |
= |
19 |
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72 |
18 |
8 |
72 |
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(答え) |
19 |
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72 |
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