G12_MATH_P2_13 好難啊! 救命啊!

1.
AB=AD=3, then AE=1 and EB=2
Note that triangle ADE ~ triangle BEH
So area of triangle ADE:area of triangle BEH = (AD/BE)^2=(3/2)^2=9/4
Area of triangle ADE=1/2 x 3 x 1=3/2
Area of triangle BEH=3/2 x 4/9=2/3
Area of EHDC = 3x3-(3/2)-(2/3)=41/6
ED=sq rt.(1^2+3^2)=sq rt(10)
So area of square EFDG = 10
Area of shaded region = 10-41/6 = 19/6

B

2015-01-06 11:02:59 補充：
2.
當三角錐計=1/3 x (area of BCD) x FC
Ans:1/3 x (x^2/2) x (x) = 1/6 x^3
Ans:C

2015-01-06 11:03:34 補充：
2.
當三角錐計=1/3 x (area of BCD) x FC
Ans:1/3 x (x^2/2) x (x) = 1/6 x^3
Ans:C

2015-01-06 14:41:06 補充：
For Q5) https://www.flickr.com/photos/129477157@N07/16026164947/

2015-01-06 14:47:05 補充：
For Q4) https://www.flickr.com/photos/129477157@N07/16024665250/

2015-01-06 19:38:47 補充：
Q3.) https://www.flickr.com/photos/129477157@N07/15593611213/
不用理我在圖的兩條線

2015-01-06 19:50:18 補充：
For Q1,
This is a typical pair of similar triangle
Angle EAD = Angle HBE = 90 (prop. of square)

Ange AED = 180-angle DEF-angle BEH = 180-90-angle BEH=90-angle BEH
(adj. angles on st. line)

Angle BHE = 180-angle HBE-angle BEH = 180-90-angle BEH=90-angle BEH
(angle sum of triangle)

2015-01-06 19:52:51 補充：
Therefore, AED = BHE

angle ADE = angle BEH (angle sum of triangle , or u may say thrid angle of triangle )

so triangleADE~triangleBEH (A.A.A)

You should recognize this relationship quickly,
this type of question is typical!!

2015-01-06 20:06:02 補充：
AED=BHE
should be
angle AED = angle BHE