→ ooww : 有辦法求出a和b嗎?11/24 12:00
Yes, you can.
a^2+b^2=256
a+b=22
Set a>b
=>a^2+(22-a)^2=256
=>a^-22a+114=0
=>a=11+sqrt(7), b=11-sqrt(7)
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※ 編輯: angel07 (140.115.189.46), 11/24/2014 12:09:19
※ 引述《ooww (另外一個我)》之銘言：
: 此為某國中段考填充題
: 已知 直角三角形 外接圓半徑8, 內切圓半徑3
: 求此三角形周長
: 答案是38,
: 我猜38 這答案是由兩股(3+8)+(3+8) +斜邊16 = 38而來
: 那就怪了
: 如果真是這樣 應該是等腰直角三角形, 斜邊應該是11√2,而不是8
Sol:
R=8, r=3
Due to Right-angled triangle, Hypotenuse c=2R=16
r=2A/(a+b+c)
Due to Right-angled triangle, a+b=c+2r
S=a+b+c=2c+2r=32+6=38