Questions on Word Problems: Geometry answered by real tutors!

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Question 151018: A RECTANGLE IS 4 TIMES AS LONG AS IT IS WIDE A SECOND RECTANGLE IS 5 CENTIMETERS LONGER AND 2 CENTIMETERS WIDER THAN THE FIRST. THE AREA OF THE SECOND RECTANGLE IS 270 SQUARE CENTIMETERS GREATER THAN THE FIRST. WHAT ARE THE DIMENSIONS OF THE ORIGINAL RECTANGLE: A RECTANGLE IS 4 TIMES AS LONG AS IT IS WIDE A SECOND RECTANGLE IS 5 CENTIMETERS LONGER AND 2 CENTIMETERS WIDER THAN THE FIRST. THE AREA OF THE SECOND RECTANGLE IS 270 SQUARE CENTIMETERS GREATER THAN THE FIRST. WHAT ARE THE DIMENSIONS OF THE ORIGINAL RECTANGLE
Answer by jojo14344(282) About Me  (Show Source):
You can put this solution on YOUR website!
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
Thank you,
Jojo