The rectangle below has an area of 4 (x+3) [4 is width, x+3 is length] square units. When the dimensions of the rectangle are doubled, the area of the new rectangle in terms of x is 8 (2x+6). [ 8 being the width and 2x+6 being the length.] Will the ratio of the area of the original rectangle to the larger rectangle be the same for any positive value of x?

Question

Mathematics

Zander Sampson

6 March, 14:15

The rectangle below has an area of 4 (x+3) [4 is width, x+3 is length] square units. When the dimensions of the rectangle are doubled, the area of the new rectangle in terms of x is 8 (2x+6). [ 8 being the width and 2x+6 being the length.] Will the ratio of the area of the original rectangle to the larger rectangle be the same for any positive value of x?

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Answers (1)

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Jace Henson · Answer 1

Yes it will be. The ratio will always be

4 (x+3) / 8 (2x+6)

4x+12/16x+48

ratio = 1/4

You can test this out by substituting some values of x.

For example,

x=1

Orig=16 New=64

16/64=1/4

x=2

Orig=20 New=80

20/80=1/4

and so on ...