Given ,

`y = 9-|x| , y = 0`

first let us find the total area of the bounded by the curves.

so we shall proceed as follows

as given ,

`y = 9-|x| , y = 0`

=>` 9-|x|=0`

=> `|x| -9 =0`

=>` |x|=9`

so `x=+-9`

the the area of the region is = `int _-9 ^9 (9-|x| -0) dx`

=>`int _-9 ^0 (9+x -0) dx`+`int _0 ^9 (9-x -0) dx`

=>`[9x+x^2 /2]_-9 ^0 + [9x-x^2/2]_0 ^9`

=>`[0]-[-81+81/2] +[81-81/2]-[0]`

=>`81/2 +81/2 =81`

So now we have to find the horizonal line that splits the region into two regions with area 81/2

as when the line y=b intersects the curve `y=9-|x|` then the area bounded is 81,so

let us solve this as follows

first we shall find the intersecting points

as ,

`9-|x|=b`

`|x|= 9-b`

`x=+-(9-b)`

so the area bound by these curves `y=b` and `y=9-|x| ` is as follows

A= `int _-(9-b) ^(9-b) (9-|x|-b)dx = 81/2`

=>`int _-(9-b) ^0 (9+x-b)dx +int _0 ^(9-b) (9-x-b)dx =81/2`

=>`[9x+x^2/2-bx]_-(9-b) ^0 +[9x-x^2/2-bx]_0 ^(9-b) = 81/2`

=>`[0]-[9(-(9-b))+(-(9-b))^2 /2-b(-(9-b))]+`

`[9((9-b))-((9-b))^2/2-b((9-b))]-[0]=81/2`

=>` [9((9-b))-((9-b))^2/2-b((9-b))]`

`-[9(-(9-b))+(-(9-b))^2/2 -b(-(9-b))]=81/2`

let `t= 9-b`

so

=> `[9(t)-(t)^2/2 -b(t)] -[9(-t)+(-t)^2/2 -b(-t)]=81/2`

=>`[9t-t^2/2 -bt]+[9t-t^2/2 -bt]=81/2`

=>`18t-t^2-2bt =81/2`

but we know half the Area of the region between `y=9-|x|,y=0` curves =`81/2`

so now ,

`18t-t^2-2bt =81/2`

`18t-2bt-t^2 = 81/2`

=>`t(18-2b)-t^2=81/2`

=> `t^2=t(18-2b)-81/2`

=>`t^2 -t(18-2b)+81/2=0`

this is like quadratic equation

`ax^2+bx+c=0`

so `t = (-b+-sqrt(b^2-4ac))/2a`

`=(18-2b+-sqrt((-(18-2b))^2-4*(81/2)))/2`

but

`t=9-b`

so,

`9-b=(18-2b+-sqrt((-(18-2b))^2-4*(81/2)))/2`

=> `18-2b=(18-2b+-sqrt((-(18-2b))^2-4*(81/2)))`

=>`+-sqrt((-(18-2b))^2-4*(81/2))=0`

=>`sqrt((-(18-2b))^2-4*(81/2))=0`

=>`(-(18-2b))^2-4*(81/2)=0`

=>`(-(18-2b))^2=4*(81/2)`

=>`(-(18-2b))^2=2*(81)`

=>`(-(18-2b))= +- sqrt(2) *9`

=> `-18+2b=+-9sqrt(2)`

=>`2b=+-9sqrt(2)+18`

=>`b=(18+-9sqrt(2))/2`

so `b= (18+-9sqrt(2))/2`