CPhill

We can use something  called Pick's  Theorem  to answer  this

Area  =

Number of boundary points with integer coordinates  / 2   + Number of  Interior points with integer  coordinates    -1

Number of boubary points with integer coordinates  =   6

Number of interior points with integer coordinates  = 31

So....the area  =

6/2  +  31  -  1  =

3  +  30  =

33 units^2

Not  to  hard  to  find  this....T is  (2/3)  of  the  distance  from D to F

The coordinates  of T  can be found as

[ 1 + (2/3)(7 - 1)  ,  4 + (2/3)(1 - 4) ]  =

[ 1 + (2/3)(6) , 4 + (2/3)(-3) ]  =

[ 1 + 4.   4  - 2 ]  =

( 5 , 2)

The  octagon will  be formed  8 congruent  isosceles triangles  with  equal sides of  5  and  an included angle  of 45°

The area of the octagon  is

8 (1/2) (5^2) sin (45°)  =

100  ( sqrt (2)  / 2)   =

50 sqrt (2)   units^2  

See heureka's excellent answer, here :

https://web2.0calc.com/questions/geometry_136

If MB  =  6.5    then  BC  =  2(6.5)   =  13

Using  the Law  of  Cosines  we  have  that

AB^2  = AM^2  + MB^2  -  2(AM*MB) cos (AMB)

5^2  =  (6.5)^2  + (6.5)^2  - 2(6.5)^2 * cos (AMB)

[ 25  - 6.5^2  - 6.5^2]  / [ - 2* 6.5^2  ]  =  cos (AMB)

-59.5 / -84.5    =  119/169

Then the sin of AMB   =  sqrt [ 169^2  - 119^2]   / 169   =  120/169

Using  the Law  of Sines

sin(AMB)  /AB = sin (ABM) / 6.5

(120/169) /5  =  sin (AMB)  /6.5

6.5 ( 120/169) / 5 =  sin AMB  =  12/13

Then the area  of  ABC  =

(1/2) AB * BC  sin (AMB)  =

(1/2) (5)(13) * (12/13)  =

(1/2) 60  =

30 units^2

We need to  first  find  out  how long  it  will take A and B working together to fill the tank....this  is easy

A  fills    1/6  of the tank in one hour

B fills  1/4  of  the tank in one hour

Add these fractions   =   (1/6) + (1/4)  =  10/24   =  5/12  =  portion of tank filled by both working together in one hour

Time  for  A  and B  ti fill the tank  =  12/5  hrs

So.....C  fills   1 / (12/5)  =  5/12 of the tank in one hr

So....if all  three  work together    (5/12)  + (5/12)  =  10/12  =  5/6  of the tank is filled in one hour

Let D  =  (0,0)

Let C = (3,0)

Let G  = (2,0)

Let  F= ( 3,1)

Let B  =  (3,3)

The slope  of DF  =  [ 1 - 0] /  [ 3 - 0 ]  =  1/3

So   the equation of  the  line  containing  DF  =  (1/3)x

The slope  of BG   =  [ 3-0]  / [3 - 2]  =  3/1  =  3

So  the equation of  the line containing BG  =  3(x - 3)  +  3   =   3x - 6

We can find  the x coordinate of the intersection of  these lines can be found as

(1/3)x  =  3x - 6     rearrange  as

6  =  [ 3 - 1/3]x

6  = (8/3]  x

x  = (3 * 6 / 8]  =  18/8   =  9/4  =  x coordinate of E

The  y coordinate of E  =   (1/3)(9/4)  =  9/12   =  3/4  =  height of  triangle DGE

Area  of triangle  DGE  = (1/2)DG * height  =  (1/2) (2) * (3/4)  =  3/4

Area of triangle DFC = (1/2) DC * CF  = (1/2)(3) (1)  =  3/2

Area  of EFGC  =   area of triangle DFC  - area of triangle DGE =  (3/2)  - (3/4)  =  (3/4)cm^2  = .75 cm^2

1 mm  =  .1 cm

So  the radius  of  the wire  =  .1 cm / 2  =  .05cm

The  cross-section  of the  wire  =  pi *(.05)cm^2   =  [.0025 pi]cm^2

The  volume of the wire  =  Cross-Section * Length  =  The volume of the block of metal

So  we  have this

50cm^3   =  [ .0025]cm^2  *  length

50 / .0025   = length   =   20,000cm

Surface  area  of  cylinder   =   2pi * r  * (2r)   =  4 pi r^2

Area  of  the two bases  =   2 pi *r^2

Area of two bases  /   Surface area of cylinder   =   [ 2 pi  r^2  ]  / [ 4 pi r^2]   =    1  / 2

I'll  try

1m = 100cm

3m  = 300 cm

The  volume of water displaced  =    (300cm) (300cm) (1 cm)   =  90000 cm^3

 1 L   =  1000cm^3

So

90000 cm^3    =   90 L

And

90L  =   90  kg  =  the man's weight