Question 285614

Let the width of the 1st rectangle be W. Then its length = 4W, since its length is 4 times its width


The width and length of the 2nd rectangle are W  + 2, and 4W + 5, respectively, since the second rectangle is 2 centimeters wider and 5 centimeters longer than the 1st.


The area of the 1st rectangle is then: W*4W, or {{{4W^2}}}, and the area of the 2nd rectangle is (W + 2)(4W + 5 ), or {{{4W^2 + 13W + 10)}}}.


Since the area of the 2nd rectangle is 270 square centimeters greater than the area of the 1st rectangle, then we’ll have: 

{{{4W^2 + 270 = 4W^2 + 13W + 10}}}

{{{4W^2 - 4W^2 - 13W = 10 - 270}}}

- 13W = - 260

{{{W = (-260)/-13 = 20}}}


Now, since W, or width = 20, then the dimensions of the 1st rectangle are: W, or width = {{{highlight_green(20)}}} cm, and L, or length = {{{highlight_green(80)}}} cm ------> 4W, or 4(20)

Check:
Width of 1st rectangle = 20 cm
Length of 1st rectangle = 4(20) = 80 cm


Area of 1st rectangle = 20 * 80 = 1,600 sq centimeters


Width of 2nd rectangle = 22 cm (20 + 2)
Length of 2nd rectangle = 85 cm (4*20 + 5)


Area of 2nd rectangle = 22 * 85 = 1,870 sq centimeters


Area of 2nd rectangle (1,870 sq centimeters) is 270 sq centimeters greater than area of 1st rectangle (1,600 sq centimeters).


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