CPhill

Here's the first one....there might be an easier way to do it, however  !!!

The area of  ABC  is   (1/2) (16)^2 (√3/2)  =  64√3  cm

So...the area  of  equilateral triangle   PQR is   (1/4)  of this  = 16√3 cm^2

And we can find the   length of  PR  by using the area formula for PQR

(1/2) (PR)^2 (√3 /2 )  =  16√3 

(1/2) PR^2 (1/2)  = 16

PR^2  / 4  = 16

PR^2  = 64

PR  = 8  cm

And note that we have trapezoid BPRC  which has the same area as the small equilateral triangle

So   area  of trapezoid  BPRC  = area of triangle PQR

 also

(1/2) (height) ( sum of the bases)  = 16√3

height ( 16 + 8)  = 32√3

height (24)  = 32√3

height  =  32√3 / 24   =   (4/3)√3  cm

So...note  that if we draw a perpendicular  from P  to BC  this  will be the height of the trapezoid and will  also form one leg of a right triangle  with BP  as the hypotenuse

And the other leg length will be  ( BC  - PR)  / 2   = (16 - 8) /2  =  8/2  = 4

So....using the Pythagorean Theorem...we can find  BP  as

BP  =   √ [ 4^2  +  [ 4/3 * √3 ]^2  ]  =  √ [ 16  +  16/3  ]  =   4 √ [ 1 + 1/3] = 

4 √ [ 4/3 ]  = 8√ [ 1/3 ]  =  8 / √3  cm   =  8√3 / 3  cm

Sorry, AWESOMEEE...I had the wrong value for the radius...see the corrected answer...!!!

Here's the second one

MIR  is an inscribed angle

So....since it subtends minor arc  MR, the  angular measure of this arc  is twice 18° = 36°

So......the length  of MR  is given by :

2 pi  *  radius *  (36 / 360)  =

2pi * 80  * (1  /10) =

(160 / 10) pi  =

16 pi   cm

Draw AR

Since arc AS  is 58°....then angle  ARS  also  = 58°

Similarly...draw  RB

And since arc BT is  37°, then angle BRT  also  = 37°

So...we have quadrilateral  PARB

Since   angle SRT  = 180°  ....then angle ARB  =

180  - angle ARS  - angle BRT  =  

180  - 58  - 37  =  85°

And  angles  RAP  and RBP  = 90°

And the sum of the interior angles of a quadrilateral  = 360°

So

angle ARB + angle RAP  + angle RBP  + angle APB  = 360

85  + 90  + 90 + angle APB  = 360

265  + APB  = 360

APB  = 95°

We can find half of the area of the shaded region as follows

Area of a sector of 1/6 of a circle with a radius of 12  - Area of a triangle with sides of 12 and an included angle of 60°   =

(1/2) 12^2 ( pi/3)  - (1/2) 12^2 sin (60)  =

(1/2) 12^2  (  pi /3  - √3/2)

So...the  shaded area  is twice this  =

12^2 ( pi /3  - √3/2) units^2  ≈  26.1  units^2

Here's 3

The total  volume  of the  first  prism is  ( 4 * 7 * 5)  = 140 cm^3

So....the price per cubic centimeter  is   4000 cents/140 cm^3   ≈  28.6 cents / cm^3

And the volume  of the second prism is  (12 * 42 * 45)  = 22680 cm^3

So...the price of the second prism (block)  is just

22680 cm^3  *  28.6 cents / cm^3 =  648648 cents  ≈   $6486.48

If we look at the  small  right triangle on the bottom left of the  kite, it has a hypotenuse of 13 and a leg of 12....so...the height of the triangle is the other leg  and  is found as

√ [ 13^2  - 12^2 ] = √ [169 - 144]   = √25  = 5

So...by symmetry, we have two  congruent triangles comprising the top and bottom of the figure

The height of both triangles is 5 and the base is  12 + 8  = 20

So....the area of  one of the triangles is  (1/2) (5) (20)  =   50

And doubling this gives  the total area  of  100 units^2 

Thanks, Daniel....The graph doesn't  clearly show it, but 0  should also be added to  the range....I have amended my answer

We have that

(1/a  + 1/b)                1

________    =         __   

   a +  b                   25

[ a + b ]

______                    1

 [  ab   ]                   ___

______    =             25

  a + b

  1                    1

___       =     ____   

 ab                 25

This implies that   ab  = 25

Since a,b  appear to  be distinct  integers, the  result would be that either  1*25  = 25 or

25 * 1    = 25

Either way, their sum is 26

Look  at the graph here :

https://www.desmos.com/calculator/mdadvpqeu9

Note the graph  is "periodic"  with a period  of  1

It appears that in each "period,"  there are 6 values in the range   (0, 0.2, 0.4, 0.6, 0.8  and  1  )

Thanks to Daniel for spotting that  "0"  should be added to the range  !!!