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1. Triangle ABC is a right angles triangle with right-angled at B. AB=2t cm, AC = (t^2+1)cm and BC = 15cm. Find the value of AB and AC.
F.4 MATHS
2007-02-14 4:24 am
回答 (2)
2007-02-14 4:32 am
✔ 最佳答案
From the pyth. theorem , (AB) ^ 2 + (BC) ^ 2 = (AC) ^2
4t^ 2 + 225 = t^4 + 2t^ 2 + 1
t^ 4 - 2 t^2 - 224 = 0
(t ^ 2 + 14 ) ( t^2 - 16) = 0
because 2t > 0(side length > 0 )
so t ^ 2 + 14 = 0 is rejected .
Therefore ,
t^ 2 - 16 = 0
t = 4 or - 4
because 2t > 0(side length > 0 )
t = -4 is rejected
So t = 4
AB = 2 x 4 = 8 cm
AC = 4 + 1 = 5 cm
2007-02-14 4:34 am
AB^2 + BC^2 = AC^2
(2t)^2 + 15^2 = (t^2+1)^2
4t^2 + 225 = t^4 + 2t^2 + 1
t^4 - 2t^2 - 224 = 0
(t^2-16) (t^2+14) = 0
t^2 -16 =0 or t^2+14 = 0
t^2=16 or t^2 = -14
t= +4 or -4
then AB= 2 x 4 = 8
AC= (4^2+1) = 17
(2t)^2 + 15^2 = (t^2+1)^2
4t^2 + 225 = t^4 + 2t^2 + 1
t^4 - 2t^2 - 224 = 0
(t^2-16) (t^2+14) = 0
t^2 -16 =0 or t^2+14 = 0
t^2=16 or t^2 = -14
t= +4 or -4
then AB= 2 x 4 = 8
AC= (4^2+1) = 17
收錄日期: 2021-04-12 22:10:17
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