s.2 maths

圖片參考：http://imgcld.yimg.com/8/n/HA00479229/o/701106040040813873455610.jpg

H係個交叉點(sorry)
angleAHG = a(alt. angles, AB//GH)
angleGHI = b-a
angleBIJ=b-a(alt. angles, GH//IJ)
angleJIF=c-(b-a)
angleJIF=c-b+a

a+c = b+d (given)
d = c-b+a
d =angle JIF

Therefore, IJ//EF(alt. anges eq.)
Therefore,AB//EF

2.(3a-b-5c)²

=[(3a-b)-5c]²

=(3a-b)²+25c²-10c(3a-b)

=9a²+b²-6ab+25c²-30ac-10bc

=9a²+b²+25c²-6ab-10bc-30ac

3.q = 2/rp - 1/p
q = 2-p/rp...........通分
rpq+p = 2
p(rq+1) = 2
p=2/rp+1

4.你問咩呀﹖﹖


5.用cosine formula搵三角形既斜邊
斜邊^2=50^2 + 50^2-2(50)(50)(cos135度)
斜邊 = 92.388 (corr to 5 sig fig)

用畢氏定理

半徑^2 + 半徑^2 = 92.388^2
半徑 (octagon)= 65.328 (corr to 5sig fig)
半徑(circle) = 65.328+3 = 68.328

area
=兀(68.328)^2-area of octagon
圖片參考：http://imgcld.yimg.com/8/n/HA00479229/o/701106040040813873455611.jpg