CPhill

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

x^2 + 5x + 3  =   0

By Vieta

The sum of the roots =  -5/1   = -5  =   p +  q

So

(p + q)^2 = (-5)^2

p^2 + 2pq + q^2 = 25      (1)

And the product of the roots =  3/1  = 3

So

pq = 3

2pq = 6   (2)

Sub (2) into (1)  and we get that

p^2 + 6  + q^2  = 25

p^2 + q^2   =19

So

P^2 + q^2   =  -b / 1

19 = -b

-19  =  b

And, again, by Vieta :

p^2*q^2  = c / 1

(pq)^2  = c

(3)^2  = c

So

b + c =     

-19  + 9  =

-10

Call R  the radius of the  larger quarter-circle and  r  be  the  radius  of the smaller circle

Let P  be  the  center of the  smaller circle

And we can  show that the distance  from O to P =  √2r

Then

R =  r  + √2 r

R = r ( 1 + √2)

So

r =    R  / ( 1 + √2)

3) A, B, C, and D are points on a circle, and segments AC and BD intersect at P, such that AP = 8, PC = 1, and BD = 6. Find BP, given that BP < DP.

I believe  that we can use the intersecting chord theorem here.........we have that

AP * PC  = BP * DP

8 * 1  =   BP * DP

    8 =   BP * DP

We  know that   BD  = 6

So.....let   BP = x  and DP  =  6 - x

So

8 =  x * (6 - x)      simplify

6 = 6x  - x^2      rearrange as

x^2  - 6x  +  8   = 0      factor

(x - 4) ( x - 2)  = 0

Setting each factor to 0  and  solving for  x we get that  x  =  4 or  x  = 2

But BP < DP....so.....x =  2  must be correct ....and this equals  BP

(2)

The interior  angles of the pentagon each measure 108°

So....  base angles PBA  and PAB are both supplemental to 108°  = 72°

So  angle  BPA =   180 - 2(72)  =   180  -144   =  36°

1)  NM  is  the  midline  of the trapezoid

Its  length  is the  average  of the two base lengths....so....

(AB + DC)   / 2     = 11

AB + DC  =   22

5 + DC  = 22

DC   =  17

Thanks, jfan   !!!!

Here's another way to  find (3)

Call the side of the larger hexagon  = S

Using the Law of Cosines  we can find  the side of the smaller hexagon,s,  as

s^2  =  [(1/2)S]^2 + [(1/2)S]^2  - 2 [ (1/2)S (1/2)S ]  cos (120°)

s^2  =   (1/2)S^2  -(1/2) S^2 * ( -1/2)

s^2  = (3/4)S^2

s = (√3/2) S

Because  the  hexagons  are  similar, then  the  scale  factor  of the  smaller  hexagon to the larger hexagon =

    (√3/2) S

    _______ =     √3/2

         S

Then   the  area of  the  smaller  hexagon to the larger =  square  of the  scale factor  =  3/4

See here :

https://web2.0calc.com/questions/help_37192

If 10%  evaporates  each day.....then  90%  of the previous day's amount remain

So

(.90)^3   = .729  = 72.9%    of the original amount remains after 3 days

We can use  the tangent here....let H  be the  height of the cloud

tan  (37° 55')  =   H  / 1100     multiply both sides  by 1100

1100 * tan (37° 55')  =  1100 * tan ( 37 + 11/12)°   =   H ≈  856.8  ft