S3_Mock0_13_E

B
D
C


ARE THE ANS
VERY HARD MOCK
WHICH SCHOOL ARE U

2014-12-26 18:17:36 補充：
Refer to the graph,
Let radius of the four small identical circles be r,
Hence, the radius of large cirlce = [ 2 sq rt (2) r + 2 r ] / 2 = (sq rt 2+1)r
Area of four small circle = 4 pie r^2
Area of the large circle = (sq rt 2 +1)^2 x r^2 x pie
The required ans = 4 pie r^2 / (sq rt 2 +1)^2

2014-12-26 19:00:42 補充：
The required ans = 4 pie r^2 / (sq rt 2 +1)^2 x r^2 x pie x100% = 68.6%
https://www.flickr.com/photos/129477157@N07/

2014-12-26 20:26:08 補充：
Q2
Let PQ be h
PQ = QR = h (given)
QS = sq rt (h^2 + 30^2) (pyth. theorem)
WU = QS = sq rt (h^2 + 30^2) (opp.sides, //gram)
VK = PQ = h (given)
KU = sq rt ( 17^2 - h^2) (pyth. theorem)
HU = HK + KU = 22 + sq rt (17^2 - h^2)
WH = PQ = h (given)
In triangle WHU,
WH ^2 + HU ^2 = WU^2

2014-12-26 20:32:18 補充：
Q2 continue.
h^2 + [22 + sq rt (17^2 - h^2)]^2 = h^2 + 30^2 (pyth. theorem)
22 + sq rt (17^2 - h^2) = 30
17^2 - h^2 = 64
h = 15
Extend VK to RT at Z, where VZ perpendicular to RT.
So PVRZ is a rectangle.
QW = SU = 25 (opp.sides, //

2014-12-26 20:38:33 補充：
Q2 continue.
PW = sq rt (25^2 - 15^2) = 20
Area of PVRZ = (20+22) x ( 2x15) = 1260
In triangle VZT,
VK = KZ
VU = VT
KU// ZT
so KU = 1/2 (ZT)
KU = sq rt ( 17^2 - 15^2) = 8
Then ZT = 16
Area of triangle VZT = 16 x 30 x 1/2 = 240
So ans = 1260 + 240 = 1500

2014-12-26 20:47:24 補充：
Q3.
For easy observe,
(sin3)^2 + (sin6)^2 + (sin9)^2 + ... + (sin81)^2 + (sin84)^2 + (sin87)^2 + (sin90)^2
As sin(90-x) = cos x
then sin81= sin(90-9)=cos 9
sin84= sin(90-6)=cos 6
sin87= sin(90-3)=cos 3
As (sinx)^2 + (cosx)^2 = 1
We can group them together,

2014-12-26 20:52:11 補充：
i.e. (sin3)^2 + (sin87)^2 = 1
(sin6)^2 + (sin84)^2 = 1
(sin9)^2 + (sin81)^2 = 1
................................= 1
................................ =1
(sin42)^2+(sin48)^2 =1
There are total 14 pairs,
they equal to 14

2014-12-26 20:54:07 補充：
(sin 45)^2 and (sin90)^2 alone were left
(sin 45)^2 =0.5
(sin90)^2 = 1
so the ans is 14+1+0.5= 15.5