有一直角三角形的三邊長成等差數列，且三角形的面積為216，則其斜邊長為何？
回答 (3)
2020-02-17 3:28 pm
✔ 最佳答案
設△3邊長成等差數列 的 首項=a,公差=d則 △3邊長= a,a+d,a+2d
a²+(a+d)²=(a+2d)² - - - - - ∵ {這是直角△}
2a²+2ad+d²=a²+4ad+4d²
a²-2ad-3d²=0
(a-3d)(a+d)=0
a=3d 或 a= -d (不合,因斜邊a+2d一定 >其餘2邊:a,a+d)
∴ a=3d . . . . . . . . . . . . . ①
△面積=216
½a(a+d)=216
a(a+d)=432
3d(3d+d)=432 (代入①)
12d²=432
∴ d=6
代入①: a=3(6)=18
∴ 斜邊長=a+2d=18+2(6)=30
2020-02-17 3:19 pm
邊長成等差數列 → 3：4 ：5，設 r 倍 → 3r : 4r : 5r
加起來，變12r， 12r =216 ，得r =18。
斜邊是最長的（ 5r )，5 × 18 = 90 。😗😗
加起來，變12r， 12r =216 ，得r =18。
斜邊是最長的（ 5r )，5 × 18 = 90 。😗😗
2020-02-17 5:02 pm
直角邊高=h
斜長邊=h+e
底短邊=h-e
Pythagoras Theorem畢氏定理
(h-e)^2 + h^2 = (h+e)^2
h^2 -2he+e^2 + h^2 = h^2 +2he +e^2
h^2-4he = 0
h (h-4e) = 0
so,
h=0
or
h-4e=0 => h=4e
Since h cannot be 0,
Put h = 4e into Area
A = 0.5 x h x (h-e)
216 = 0.5 (h^2 – he)
432 = h^2 – he
432 = (4e)^2 – (4e)e
432 = 16e^2-4e^2
432 = 12e^2
36 = e^2
e = 6
h = 4x6 = 24
斜邊長=h+e = 24+6 = 30
斜長邊=h+e
底短邊=h-e
Pythagoras Theorem畢氏定理
(h-e)^2 + h^2 = (h+e)^2
h^2 -2he+e^2 + h^2 = h^2 +2he +e^2
h^2-4he = 0
h (h-4e) = 0
so,
h=0
or
h-4e=0 => h=4e
Since h cannot be 0,
Put h = 4e into Area
A = 0.5 x h x (h-e)
216 = 0.5 (h^2 – he)
432 = h^2 – he
432 = (4e)^2 – (4e)e
432 = 16e^2-4e^2
432 = 12e^2
36 = e^2
e = 6
h = 4x6 = 24
斜邊長=h+e = 24+6 = 30
收錄日期: 2021-04-11 23:03:34
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200217065643AAA8ck5