解下列方程? 唔該呀

2006-12-08 6:33 am
等腰三角形ABC,周界41cm.AB=AC,它的底BC比AB長2cm,求三邊長度?

回答 (4)

2006-12-08 6:42 am
✔ 最佳答案
由於AB=AC,設x是AB和AC;BC比AB長2cm,故BC=x+2

AB+AC+BC=周界
x+x+(x+2)=41
3x=41-2
3x=39
x=13

AB=13cm
AC=13cm
BC=13+2
=15cm
2006-12-08 6:44 am
AB=AC
BC=AB+2
AB+AB+(AB+2)=41
3AB+2=41
3AB=41-2
3AB=39
AB=13
AC=13
BC=13+2=15
2006-12-08 6:44 am
Let y be the length of AB & AC, then (y + 2) be the length of BC

y + y + ( y + 2 ) = 41
3y + 2 = 41
3y = 39
y = 13

Hence, the length of AB & AC is 13 cm and the length of BC is ( 13 + 2) = 15 cm.
2006-12-08 6:39 am
AB+2=BC
2AB+(AB+2)=41
3AB=39
AB=13CM
AC=AB=13CM
BC=AB+2
=15CM
參考: ME


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