2014 DSE Maths 練習卷2 MC

21. Let radius of sector = r
So area of triangle OAC = (r^2 sin 60)/2 = (r^2 sqrt3)/4 = 9 sqrt 3
r^2 = 36, r = 6.
Arc length AC = 2 pi r x 60/360 = 2 pi x 6 x 60/360 = 2 pi (C).
16.
Hypotenus of the "white" triangle = 12 - 2 - 2 = 8
The " white " triangle is a right - angled triangle, let its side = x
So by Pythagoras thm., x^2 + x^2 = 8^2
2x^2 = 64
x = sqrt 32
So area of each " white" triangle = x^2/2 = 16
So area of the shaded region = area of square - 4 x area of each " white " triangle = 144 - 4 x 16 = 144 - 64 = 80 cm^2 (C)
22.
Let angle A = z
so angle DCB = 180 - z (opp. angle of cyclic quad.)
so angle YCX = 180 - z (vert. opp. angle)
There is a very frequently used result for this figure :
x + y + z = 180 - z
so z = 90 - (x + y)/2
angle DCY = z ( ext. angle of cyclic quad.) = 90 - (x + y)/2 (D).