Question

The perimeter of the original rectangle on the left is 30 meters. The perimeter of the reduced rectangle on the right is 24 meters.

A small rectangle has a length of 8 meters. A larger rectangle has a width of x.


What is x, the width of the original rectangle on the left? Round to the nearest hundredth if necessary.

5 meters

8 meters

10 meters

12 meters

Answer 1

5

Explanation:

Find the scale factor

we know that

If two figures are similar, then the ratio of its perimeters is equal to the scale factor

Let

z -----> the scale factor

P1 -----> the perimeter of the reduced rectangle on the right

P2 ----> the perimeter of the original rectangle on the left

substitute

step 2

Find the width of the reduced rectangle on the right

substitute the given values.

Find the width of the original rectangle on the left.

To find the width of the original rectangle on the left, divide the width of the reduced rectangle on the right by the scale factor.

Answer 2

5 meters

Explanation:

If the smallest rectangle has a length of 8 meters, and a perimeter of 24 meters then its width is given by:

`24 = 2*8+2*W_1\\W_1= 4\ meters`

We can conclude that the width is half of the length for both rectangles. Therefore, the width of the larger rectangle, with a perimeter of 30 meters, is:

`30 = 2*W_2+2*L_2\\30 = 2*W_2+4*W_2\\W_2=5\ meters`

The width of the original rectangle is 5 meters.