See the attached figure to better understand the problem
we know that
in the triangle BCD
y²+5.66²=z²-------> y²=z²-5.66²------> equation 1
in the triangle ABC
y²+5.66²=x²-------> y²=x²-5.66²------> equation 2
equals 1 and 2
z²-5.66²=x²-5.66²--------> z²=x²---------> z=x
if z=x then
angle A=45°
angle D=45°
in the triangle ABD
cos 45=z/(5.66*2)-------> z=11.32*cos 45-----> 11.32*√2/2----> 8
z=8
x=8
y²=z²-5.66²------> 8²-5.66²-----> y²=31.96------> y=5.65
the answer isthe value of x is 8the value of z is 8the value of y=5.65
Let us find the co ordinates of each vertex of the triangle .
Vertex A ( in firs second quadrant) = ( -5 ,3)
vertex B in third quadrant = ( -5, -5)
vertex C in fourth quadrant = ( 4, -2)
let us use distance formula AB^2 = ( -5 - 3)^2 + (-5 - -5 )^2 = 64 + 0
AB= 8
BC^2 = ( -2 - -5 )^2 + ( 4 - - 5)^2 = 9 + 81 = 90
BC = 9.48
AC^2 = ( -2 -3)^2 +( 4- -5)^2 = 25 + 81 = 106
AC= 10.29
Perimeter = sum of length of AB+ BC+ Ac = 8 + 10.29 + 9.48= 22.77
Y=-9+6
y=-3
The answer is B
54,702÷25=2,188.08.
The remainder is.08