Geometry Help

4) Three coplanar squares with sides of lengths two, four and six units, respectively, are arranged side-by-side, as shown so that one side of each square lies on line AB and a segment connects the bottom left corner of the smallest square to the upper right corner of the largest square. What is the area of the shaded quadrilateral?

Let h be  the height of the shaded  region

By similar triangles, we  have  that

6/12  =  h/ 6

6 * 6 /12 = h 

36/12  = h

3  = h

So......the  shaded area  plus  the smaller triangular area at the bottom of the left-most square has the area

(1/2)(6)(3)  =  9  units^2

And using similar triangles again....the height of this smaller triangular area is

h / 2  = 6 / 12

h = 2*6 / 12  = 1

So....the area of this triangular region  is  (1/2) (2)(1)  = 1  units^2

So....the shaded area =  9 - 1 =   8 units^2

Thanks, helps a ton 

(5)

Let  the  center of the  circle  be  O

Let  the  sides of the square  be tanglent to the  circle at  M  and N

Let  the bottom right vertex of the square  be   P

So  OMPN  forms a smaller square with sides  of 4  and  a diagonal of  4√2

So....the diagonal of the  larger square  =    4√2 + 4  =  4 ( √2 + 1)

So....the side of the  larger square is  =   (4/√2) ( √2 + 1) 

So....the area of this square is

[ (4/√2)   ( √2 + 1 )  ]^2  =

( 8)  ( 2  + 2√2 + 1 ]   =

(8) [ 3 + 2√2]  =

[24 + 16√2]  units^2

3) Circle O has diameter AE with AE = 8 cm. Point C is on the circumference of the circle such that AC and CE are congruent. Also, AC is a diameter of semicircle ABC, and CE is a diameter of semicircle CDE, as shown in the figure. In square centimeters, what is the total combined area of the shaded regions?

Believe it or not, the two shaded regions  have  the same combined area  as  that of right triangle ACE

Proof  :

Radius  of semi-circle  CDE  = (1/2)CE

So...area  of   semi-circle CDE  = (1/2)pi [(1/2)CE]*2 =  (pi / 8) CE^2

And similarly.....the area of semi-circle  ABC =  (pi/8)AC^2

Area  of  right triangle  ACE  =  (1/2)(AC)(CE)

Area  of    two  unshaded   regions  between the right triangle and  the circle  =

(1/2)pi[(1/2)AE]^2  - (1/2)(AC)(CE)  =  

(1/2) pi (1/4) [ AE]^2  - (1/2)(AC)(CE)  =    { note that AE^2  = CE^2 + AC^2 }

(1/8) pi [ CE^2 + AC^2 ]  - (1/2)(AC)(CE)

So....the  area  of  the two shaded regions  =

(pi/8) (CE^2 + AC^2]  -  [ (1/8)pi (CE^2 + AC^2 ) - (1/2)(AC)(CE) ]  =

(1/2)(AC)(CE).....which is the same  area as right triangle ACE

And  the area of  right triangle  ACE = (1/2) (CE)(AC) =   (1/2)(8/√2)(8/√2)  = 64 / 4  =  16 cm^2 = area of two  shaded regions

Chris should get paid (by this cheater not by the site manager) AND, for giving all his/her assignment to Chris to answer, the asker should get shot, or at least failed.

The asker was lucky Chris saw the questions before I did. I would have deleted them all.

Shot? I mean, thats a bit harsh, isn't it? Chris does deserve to get paid, though.

True. I don't know if there is a possible way to identify cheats, though. Some of them are extremely obvious, and others are hidden. People sometimes rephrase a problem differently to make it seem like its not a test problem.

Cheats? How do you cheat on this?

Can someone answer the other equations?