There are three diagrams:
Diagram 1:
Comparing the corresponding sides
#x/(3sqrt55)=(3sqrt55)/55#
#55x=(3sqrt55)(3sqrt55)#
#55x=(9)(55)#
Solving for x,
#x=9#
Diagram 2:
The other side in big triangle is obtained by using the pythagoras theorem as
#=sqrt(25^2-x^2)#
Comparing the corresponding sides
#x/25=16/sqrt(25^2-x^2)#
Squaring both the sides
#x^2/625=256/(625-x^2)#
Cross multiplying
#x^2(625-x^2)=(256)(625)#
Let #t=x^2#
#t(625-t)=160000#
#625t-t^2=160000#
#t^2-625t-160000=0#
If#ax^2+bx+c=0#, the roots are given by
#x=(-b=-sqrt(b^2-4ac))/(2a#
Here,
#a=1, b=-625, c=-160000, x=t#
Substituting
#t=(-(-625)=-sqrt((-625)^2-4(1)(-160000)))/(2(1)#
#=(625+-sqrt(390625+4(160000)))/2#
#t=(625+-sqrt(390625+640000))/2#
#t=(625+-sqrt(1030625))/2#
#sqrt1030625=1015.197#
#t=(625+-1015.197)/2#
Let
#t_1=(625+1015.197)/2=1640.197/2=820.0985=820# approximate
#t_2=(625-1015.197)/2=-390.197/2=-195.0985=-195# approximate
Considering positive value,
#t=820#
#t=^2#
#820=x^2#
#=28.637#
#x=sqrt820=28.637#
Thus,
#x=28.637#
Diagram 3:
#3+9=12#
The other side in big triangle is obtained by using the pythagoras theorem as
#=sqrt(12^2-x^2)#
Comparing the corresponding sides
#x/12=3/sqrt(12^2-x^2)#
Squaring both the sides
#x^2/144=9/(144-x^2)#
Cross multiplying
#x^2(144-x^2)=(9)(144)#
Let #t=x^2#
#t(144-t)=1296#
#144t-t^2=1296#
#t^2-144t-1296=0#
If#ax^2+bx+c=0#, the roots are given by
#x=(-b=-sqrt(b^2-4ac))/(2a#
Here,
#a=1, b=-144, c=-1296, x=t#
Substituting
#t=(-(-144)=-sqrt((-144)^2-4(1)(-1296)))/(2(1)#
#=(144+-sqrt(20736+4(1296)))/2#
#t=(144+-sqrt(20736+5184))/2#
#t=(144+-sqrt(25920))/2#
#sqrt25920=160.997#
#t=(144+-160.997)/2#
Let
#t_1=(144+160.997)/2=304.197/2=152.5# approximate
#t_2=(144-160.997)/2=-16.997/2=-8.498=-8.5# approximate
Considering positive value,
#t=152.5#
#t=x^2#
#152.5=x^2#
#=12.349#
#x=sqrt820=12.349#
Thus,
#x=12.349#
