题目内容
(2011•青浦区一模)已知:线段a、b、c,且
=
=
.
(1)求
的值.
(2)如线段a、b、c满足a+b+c=27.求a、b、c的值.
| a |
| 2 |
| b |
| 3 |
| c |
| 4 |
(1)求
| a+b |
| b |
(2)如线段a、b、c满足a+b+c=27.求a、b、c的值.
分析:(1)根据比例的性质得出
=
,即可得出
的值;
(2)首先设
=
=
=k,则a=2k,b=3k,c=4k,利用a+b+c=27求出k的值即可得出答案.
| a |
| b |
| 2 |
| 3 |
| a+b |
| b |
(2)首先设
| a |
| 2 |
| b |
| 3 |
| c |
| 4 |
解答:解:(1)∵
=
,
∴
=
,
∴
=
,
(2)设
=
=
=k,
则a=2k,b=3k,c=4k,
∵a+b+c=27,
∴2k+3k+4k=27,
∴k=3,
∴a=6,b=9,c=12.
| a |
| 2 |
| b |
| 3 |
∴
| a |
| b |
| 2 |
| 3 |
∴
| a+b |
| b |
| 5 |
| 3 |
(2)设
| a |
| 2 |
| b |
| 3 |
| c |
| 4 |
则a=2k,b=3k,c=4k,
∵a+b+c=27,
∴2k+3k+4k=27,
∴k=3,
∴a=6,b=9,c=12.
点评:此题主要考查了比例的性质,根据已知得出a=2k,b=3k,c=4k进而得出k的值是解题关键.
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