中六數學題

In right-angled triangle ABC，angle ACB =90degree，AC=3 and BC=4
A moving line L cuts AB and BC at M and N respectively. It is given that
area of triangle BMN = 1/2 (area of triangle ABC )，Find the minimum value of MN
A. 2
B. 3
C. 4
D. 5
Sol
△ABC=3*4/2=6
△BMN=6/2=3
SinB=3/5，CosB=4/5
BMN=(1/2)*BM*BN*SinB=(1/2)*BM*BN*(3/5)=3
BM*BN=10
CosB=(BM^2+BN^2－MN^2)/(2*BM*BN)
=(BM^2+BN^2－MN^2)/20=4/5
BM^2+BN^2－MN^2=16
MN^2+16=BM^2+BN^2>=2*BM*BN
MN^2+16>=20
MN^2>=4
MN>=2
(A)