CPhill

See the following :

We  have   30-60-90  triangle GIF   where  GI  = 3.5  and  IF =  3.5sqrt (3)

So....the  area of this triangle is  (1/2) (3.5)(3/5) sqrt (3)  = (49/4)sqrt (3)/2            (1)

And then we have   the  area  of   60°  sector   of a circle with a rdius of 7  less a triangle with sides of  7  and an included angle of 60°

So this area is    (1/2)7^2 ( pi/3)  - (1/2) ( 7)^2 sqrt (3) / 2  =  (49/2)  ( pi/3 - sqrt (3) / 2)   =

49pi/6 - (49/4) sqrt (3)     (2)

So   1/4 area  of this  shaded region  is  (1) + (2)   =    (49/6) pi - (49sqrt (3)  /8)

So....the total area is    4  [ (49/6) pi  - 49sqrt (3) / 8 '  =

(98/3) pi  - (49/2) sqrt (3)    ≈  60.19  units^2

Exactly  what Guest found   !!!

s/4    -2t    = 2    multiply through  by   4 ⇒   s  - 8t  = 8 ⇒  s  = 8 + 8t

Sub this into the second equation for s

4t  - 5(8+ 8t)    =   9   simplify

4t  - 40  - 40t  =  9

-36t =  9 + 40

-36t =  49

t = -49/36

And s  = 8 + 8(-49/36)    =  -26/9

Note....for any plate..... the word "MATH" can appear in  any one of 4 positions

Since  the  characters are distinct, I'm assuming that  no letter can be repeated

In the other 3 positions, we  can choose  any of  22 remaining letters * any of 21 remaining letters * 10 digits  and these can be arranged in 3!  = 6 ways

So.....in this scenario, we have

4 * 22 * 21 * 10 * 3    =

55440 possible plates

Excellent, Guest   !!!!

****

Call the part of  room  that  Ryan can paint in the first 20 minutes,  1/x

When Leonardo  joins in   the part that is painted between 9:20 and 9:40 (20 minutes) is   (1/x) + (2/x)   =  3/x

So  in 40 minutes  the whole job is done....so

(1/x + 3/x)  =  4/ (x)   = 1    ⇒   x  = 4

So....the part painted  between 9  and 9:30   is

(1/4)  +  ( 1/2) (3/4)   =

(1/4)  + 3/8  =

2/8  + 3/8  = 

5/8

Let A  =  (0,6)   B  = (0,0)   and C = (6,0)

AC   =    6sqrt (2)

Triangle   ADE  is similar to  triangle ABC

Call the  side of the square, S =  DE

So

AD/ DE  = AB/BC

AD /S  = 6/6

AD  = S

And triangle   CFG is similar to triangle CBA

So

FG/ CF  = BA/ CB

S / CF  = 6/6

CF = S

So

AD + DF  + FC   =  6sqrt (2)

S  + S + S    = 6sqrt (2)

3S  = 6sqrt (2)

S  = 2sqrt (2)  =  sqrt(8)  cm

Area of square  = 8  cm^2

Using the Law of Cosines

80 - 13 - 61   = -2(sqrt (793)  cos x

6 / [ -2 sqrt (793) ]   = cos x =   -3 /sqrt (793)

sin x   = sqrt  [ 1 - (-3/ sqrt (793))^2  ]   =  sqrt [ 1  - 9/793 ] =  sqrt [ 784/793  ]  = 28/ sqrt (793)

Area  =  (1/2) ( 13) (61) sin x   = 

(1/2) (793) (28/ sqrt (793) )   =

14sqrt (793)  units^2

Area of triangle ABC 

Find the semi-perimeter, S   = ( 8 + 6+ 4)  /2   =  9

Area  = sqrt  [  9 (9-8) (9-6) (9-4)  ] =    sqrt  [ 9 * 3 * 5  ] =  3sqrt (15)

The circle is an inscribed circle in a triangle....its radius, R, can be found as

Area / S  =  R

3sqrt (15)  / 9   = R   =   sqrt (15) /  3

And  the height of  triangle ABC  can be found as

3sqrt (15)  = (1/2)  BC * height

6sqrt (15)  =  4 * height

(3/2) sqrt (15)  = height  of ABC

Height of triangle AMN  =   (3/2)sqrt (15)  - 2R  =  (3/2)sqrt (15) - (2/3)sqrt (15)   = (5/6)sqrt (15)

And triangle AMN is similar to triangle ABC....so....

MN/ height of AMN  = BC /height of ABC

MN/ (5/6)sqrt (3)   = 4 / [(3/2)sqrt (15)]

MN  / (5/6)  = 4 /(3/2)

MN  = 4 (5/6) / (3/2)

MN  = (4)(2/3)(5/6)  = 40/18  =   20/9

Diagonal of square   = 2

Side of square   =  2/sqrt (2)  = sqrt (2)

Area of white area between the circle and  shaded area  =

(1/4)  (area of  circle - area of  square)  =

(1/4)[  pi (1)^2  - sqrt (2)^2 ]   =    (1/4)  [ pi - 2 ]  =  ( pi/4  - 1/2 )    (1)

Area of (1/2) circle constructed on the side of the square = (1/2) pi (sqrt (2)/2)^2   =  (1/2)pi (2/4)    = 

(1/2)pi (1/2)  =    pi/4      (2)

So.....shaded area =      (2)  - (1)   = 

pi/4  - (pi/4 - 1/2)  =

1/2  units^2

So  4 (1/2)   = 2  units^2