Question 151018
We need to remember the following,
1st rectangle (original) {{{R[1]}}} is designated below,
{{{highlight(A[1])rea}}}, {{{highlight(L[1])ength}}}, & {{{highlight(W[1])idth}}}
2nd rectangle, {{{R[2]}}}
{{{highlight(A[2])rea}}}, {{{highlight(L[2])ength}}}, & {{{highlight(W[2])idth}}}
In {{{R[1]}}}: {{{L[1]=4W[1]}}}    ---> length 4 times its width
So, {{{A[1]=L[1]*W[1]=4W[1]*W[1]}}} ---> {{{A[1]=4W[1]^2}}} ----------> eqn 1
.
In {{{R[2]}}}: {{{L[2]=L[1]+5cm}}} ---> 5 cm longer
Also, {{{W[2]=W[1]+2}}} ----------> 2 cm wider
So, {{{A[2]=(L[1]+5)(W[1]+2)}}} -------------------------------------------->A
And we have to remember, {{{A[2]=A[1]+270cm^2}}} ---> greater than 270sqcm-->B
Plug in eqn 1 in this condition: {{{A[2]=4W[1]^2+270cm^2}}} ---------------->C
Therefore, equating A & C,
{{{(L[1]+5)(W[1]+2)=4W[1]^2+270}}}
Remember in {{{R[1]}}} that {{{L[1]=4W[1]}}} & substitute:
{{{(4W[1]+5)(W[1]+2)=4W[1]^2+270}}}
{{{cross(4W[1]^2)+13W[1]+10=cross(4W[1]^2)+270}}}
{{{13W[1]=270-10}}} ------>{{{W[1]=260/13}}} ---> {{{W[1]=20cm}}} ---> ORIG. WIDTH
We know {{{L[1]=4W[1]=4*20cm}}} ------------> {{{L[1]=80cm}}} ----> ORIG LENGTH
.
For checking & verification,
{{{A[1]=L[1]*W[1]=80*20=1600cm}}}
{{{A[2]=(L[1]+5)(W[1]+2)=(80+5)(20+2)=1870cm}}} 
This satisfy condition B above -----> {{{A[2]}}} is 270cm greater than {{{A[1]}}} 
1870-1600=270cm
</pre><font size=4="indigo"<b>Thank you,
Jojo</pre>