CPhill

surface area =  2 (LW + LH + WH)  =  24    ⇒    LW + LH + WH  =  12

sum of  edge lengths = 4 [ L + W + H)  = 24   ⇒  L + W + H =   6

Using a little logic and assuming integer values for  each dimensinon, each edge measures 2

Proof

L + W + H = 6                 LW + LH + WH  =12

2 + 2 + 2  =  6                 2*2 + 2*2 + 2*2 =  12

The volume =  LWH =  2*2*2  =   8 cubic units

          P

     5           8

Q         M           R

     5.5       5.5

Law of Cosines  (twice)

[QP^2 - QR^2 - PR^2] /[-2* QR * PR] =  cos PRQ

[ 25 - 121 - 64] / [-2 * 11 *8 ] = cos PRQ  = 10/11

PM^2 = MR^2 + PR^2  - 2 (MR * PR) *cos (PRQ)

PM^2  = 30.25 + 64 - 2[5.5 * 8]* (10/11)

PM ^2  =14.25

PM =sqrt [14.25] ≈ 3.77

At 60 mph, it takes him 20 minutes to drive 20 miles

After 15 min , he has traveled 15 miles

The time it akes the taxi to arrive is irrelevant

So....he has 85  miles left to travel at 50mph

The total time in the taxi is    85 m / 50 m/h =  1.7 hrs 

2

11

Two primes

Either  the segment connecting these two points is an edge or it is the  diagonal of a square

If it  is an edge  the  area =  (12 + 98)^2 + (56 -34)^2  = 12584 units^2

If it is a diagonal the area =  (diagonal length) /sqrt (2) ) ^2   =  [ sqrt [ 12584] / sqrt 2]^2  = 6292 units^2

Sum of all possible area values =  12584 + 6292 = 18,876 units^2

Powers of 4

4^1 =  4

4^2 = 16

4^3 = 64

4^4 = 256

In general   4^(2n)  ends in 6

So...the  units digit of 1^1 +  2^2 + 3^4 + 4^8  =

1 + 4 + 1 + 6  =   2

Assuming a right triangle

Let the legs =  AC =  sqrt 7    and  BC =   sqrt 3

The hypotenuse = 

sqrt [ (sqrt .7)^2  + (sqrt .3)^2]   =  sqrt [ 7 + 3]  = sqrt 10  = AB

So....noting that cos B  = sinA

cos^2 B + cos^2 A  =

sin^2 A + cos^2A  = 1

(sqrt (3/10)^2  +(sqrt (7/10)^2  =  1

3/10  +  7/10  =  1

Then  angle C  is  opposite the hypotenuse  so its measure = 90°

And cos C =   cos (90°)  =  0

Angle DAB  =90

So angle DAE  = 90 / 3 =  30

And DAE  is a 30- 60 -90 right triangle

DE is  opposite the 30 degree angle

And AD is opposite the 60 degree angle

So AD =  6sqrt 3  =  BC

So

BF  = BC - FC   =   6sqrt 3  - 1  =   5sqrt 3

semi-circle on top  right with radius =  4  =  16pi/2  = 8pi  km^2

area of rectangle on left  side of  semi-circle = 8 * 5  = 40 km^2

area of small rectangle in the middle  = 2 * 3 = 6 km^2

area of rectangle at bottom = 5 * 8  = 40 km^2

Total area  = [  8pi + 40 + 6 + 40 ] km^2  =

[ 86 + 8pi] km^2 ≈  111.13 km^2

                        A

                F         

             3         

          D                            E

       x                                    3

  B                                            C

AF / AE  = AD / AC    ⇒  AE/AC = AF/ AD

AD/ AE = AB / AC   ⇒    AE / AC  = AD/ AB

AF / AD  = AD / AB

AD^2 = AF * AB

We need to know AF  to solve this rather than EC

If BAC  = 60

Then angle BAD = 30   and angle CAD  = 30

So CAD can't equal 45

arcos (sqrt (.7)) + arcos (sqrt(.3)) = 90°

cos C  =  180 - 90  =  cos (90°) = 0