Question 724841
It is not clear to me how the 5 rectangles would be arranged.
If 5 congruent rectangles were attached to one another by their long sides to form a similar, larger rectangle, it would look like this:
{{{drawing(300,150,-0.5,5.5,-0.5,2.5,
rectangle(0,0,5,2.236),
line(1,0,1,2.236),line(2,0,2,2.236),
line(3,0,3,2.236),line(4,0,4,2.236),
locate(0.4,2.235,a),locate(0.05,1.2,b),
locate(1.4,2.235,a),locate(2.4,2.235,a),
locate(3.4,2.235,a),locate(4.4,2.235,a)
)}}} with {{{a}}} and {{{b}}} being the small rectangles' sides' measures in cm.
Since the large rectangle is similar to the small ones,
{{{5a/b=b/a}}} --> {{{5a^2=b^2}}} --> {{{sqrt(5)*a=b}}}
 
So the area of the large rectangle is
{{{(5a)*b=240}}} --> {{{(5a)*(sqrt(5)*a)=240}}} --> {{{5a^2*sqrt(5)=240}}}
Multiplying both sides of the equal sign times {{{sqrt(5)}}} we get
{{{5a^2*sqrt(5)*sqrt(5)=240*sqrt(5)}}} --> {{{25a^2=240*sqrt(5)}}} --> {{{a^2=240*sqrt(5)/25}}} --> {{{a=sqrt(240)*sqrt(sqrt(5))/sqrt(25)}}} --> {{{a=sqrt(16*15)*root(4,5)/5}}} --> {{{a=sqrt(16)*sqrt(15)*root(4,5)/5}}} --> {{{a=4sqrt(15)*root(4,5)/5}}}
Then,
{{{b=sqrt(5)*4sqrt(15)*root(4,5)/5}}} --> {{{b=4sqrt(5*15)*root(4,5)/5}}} --> {{{b=4sqrt(25*3)*root(4,5)/5}}} --> {{{b=4sqrt(25)*sqrt(3)*root(4,5)/5}}} --> {{{b=4*5*sqrt(3)*root(4,5)/5}}} --> {{{b=4sqrt(3)*root(4,5)}}}
and {{{5a=4sqrt(15)*root(4,5)}}}
 
The perimeter would be
{{{2(length+width)=2(5a+b)}}}
That is
{{{2(4sqrt(15)*root(4,5)+4sqrt(3)*root(4,5))=2*4*root(4,5)(sqrt(15)+sqrt(3))=highlight(8root(4,5)(sqrt(15)+sqrt(3))=8root(4,5)sqrt(3)(sqrt(5)+1))}}}