CPhill

Using the triangle inequality which says that the sum of any two sides of a triangle is greater than the remaining side, we have two triangles to consider

Top triangle....

22 + 18   = 40

So the greatest integer value that we can have for the dashed line is 39

18 + x  must be greater than 22

So the least integer value that we can have for the dashed line is 5

So the interval  that  solves  the dashed line in the top riangle is  [ 5, 39]     (1)

Bottom triangle

21 + 25 =  46

So the greatest integer value that we can have for the dashed line is  45

21 + x  = must be greater than 25

So the least integer value that we can have for the dashed line is 5

So the interval that solves the dashed line in the bottom triangle is  [5, 45 ]    (2)

Taking the intersection of (1)  and (2)  we have  that the   interval that solves the problem is

[ 5, 39 ]

So....the number of whole numbers that satisfies  the dashed line length  is

39  - 5  +  1   =

34  +  1  =

35

Since DI =3, BD = 4  and BI = 5, then triangle IDB  is right with angle IDB = ADB = 90°

And since  ID = 3  and BI = 5.....then sin (DBI)  = 3/5

And cos DBI = 4/5

But since BE is an angle bisector, then sin (ABC)  = sin (2 * DBI)  =  2 sin(DBI)cos(DBI) = 2 ( 3/5) (4/5) = 24/25

And ADB is also a right triangle with BD  = 4

And since sin ABC = 24/25  then  triangle ADB  is a 7 - 24- 25  Pythagorean right triangle

Then sin ABD = 24/25

So sin of DAB  = 7/25

Using the Law of Sines

AB/ sin 90 = BD / sin (DAB)

AB/ 1  = 4 / (7/25)  =  100/7

And

AD  / sin (ABC)  = BD / sin (DAB)

AD = sin (ABC) * BD / sin (DAB)

AD = (24/25) * 4  / ( 7/25)  = 96/7

And since  angle ADB is right, then so is angle ADC

And since AD is an  angle bisector then  sin (BAD)  = sin (DAC)  = 7/25

So sin (ACD)   = 24/25

So we can find DC using the Law of Sines

DC / sin (DAC)  = AD / sin (ACD)

DC = sin (DAC) * AD / sin (ACD)

DC = (7/25) * ( 96/7) / (24/25)

DC = (7/25) * (96/7) * (25/24)

DC = (96/24)  = 4

So  BC =  BD + DC  = 4 + 4  = 8

And  the area of triangle ABC =  (1/2) (BC) ( AB) * sin ( ABC)  = (1/2) ( 8) ( 100/7) ( 24/25)  = 

(4) ( 4) (24/7)  = 

384/7  units^2

See the figure below to  get a feel  for this :

This seems trickier  than it really is

Look  at the following figure  :

Note  that   since the sides of the  square = 1....then DB = sqrt 2 = AC

E is the  center of the  square so  BE  = CE  =  sqrt (2)/2

And angle CEB = 90°

And the circle has a radius of 1/2

So....if we take  the area of 1/2 of this  circle  and  subtract  the  area of triangle CEB we will have the same area  as  the shaded region

Area of half-circle  with radius of 1/2  =  pi (1/2)^2  / 2 =  pi/8

Area of triangle CEB  =  area of an isosceles triangle  with equal sides of sqrt (2)/2  and an included angle of 90° =

(1/2) [ sqrt (2) / 2 ] ^2 *sin 90° =  (1/2) ( 2/4) * 1  =  2/8  = 1/4

So....the shaded area =   pi/8  - 1/4  =    [ pi - 2 ] / 8

The interior angles of a pentagon sum to 540°

So.....let  X  be  the measure of angle M , angle T  and angle H......and we  have that

X + X + X  +   58  =  540

3X  +  58  =  540      subtract  58  from both sides

3X   =  482   divide both sides  by  3

X =  482/3  =   ( 160 + 2/3  )°  =  measure of angle  H

(1,11)  and  (4,5)

Slope  =   [ 5 - 11] / [ 4 -1] =   -6/3  =  -2

Equation  of line

y = -2 (x -1)  + 11     simplify

y = -2x + 2 + 11

y = -2x + 13       add 2x to both sides

2x + y =  13        divide through  by 13

(2/13) x  +  (1/13) y =  1        note that we can write this as

  x               y

____    +  _____     =       1

(13/2)         13

This implies  that  a =   (13/2)

Thanks, Alan  !!!!!

If  I understand what  you're  asking

Look at the  graph here :  https://www.desmos.com/calculator/mqxce8lgk4

This graph  will  approach  0 as x increases from 3   (but never quite gets there)...so.....the lowest integer value  for  f(x)  = y = 1

And note that when  y = 1,  then x has the real number value  (5)

Note  that  when  x = -4  and y = 4   the equation will be  true

And when x = 4  and y = -4  the equation will also be true

So.....the points  ( -4,4)  and ( 4,-4) will be  on the graph

The  slope is  [ -4-4 ] / [4 - -4 ]  = -8 / 8  =   -1

Algebraically....we can see that

x/4  + y/4  =  0

[4x + 4y] / [xy] = 0     multiply through by xy

4x + 4y  = 0     divide through by 4

x + y =  0   rearrange as

y =  -x

So...this confirms that the slope is -1

Since    XZ/XY  = 1/2     and  ZY/XY  = 1/2......we can see that  Z is  the midpoint  of XY

To find this point  we  have that

x value of Z  =  [ x yalue  of X  +  x value of Y ] / 2

-1  =  [ x value of X +  1 ] / 2   multipy both sides by 2

-2  = x value of X + 1      suvtract 1 from  both sides

-3 = x value of  X

Also

y value of of Z =  [ y value of X  + y value of Y ] / 2

- 7 =  [ y value  of X  +   7]  /  2    multiply both sides by 2

-14 =  y value of X  + 7    subtract 7 from both sides

-21 = y value of X

So  X =  (-3, -21)     and the sum of the its coordinates =   -24

Here's one way of solving this.....

Note that from a point ourside a circle, two tangents  drawn  to a  circle have the same length

We have two circles

One  has the equation x^2  + (y-8)^2   =  64    [ this is the inscribed circle ]

The  other  has the equation   (x - 10)^2  + y^2  = 100

Solving these two equations for their intersections  we  have  that   (x,y)  = (320/41, 400/41) = point C  in the figure below

We have two points  (10,0)  and (320/41, 400/41)

The slope  of this tangent line  to the inscribed circle   =  (400/41 - 0) / ( 320/41 -10)  = (400/41) / ( -90/41)  =

-400/90 =  -40/9

So.....the equation of this tangent line through these rwo points is

y = (-40/9) ( x - 10)

y =( -40/9)x  + 400/9

Since the  inscribed circle touches the top of the trapezoid, the line will pass through  y= 16

So....to find the x value at this point  we can solve this

16 =  (-40/9)x  + 400/9

16 - 400/9  = (-40/9)

-256/9 =  *(-40/9)

-256/-40  =  6.4

Double this to find the  upper base length  =  12.8   [ your answer is  correct  !!!  ] 

See the following figure  :