A rectangle that is 15 by 10 is similar to a second rectangle with perimeter 40. Find the length and the area of the second rectangle.
Let’s begin by sketching the first rectangle that has dimensions 15 and 10 units. We are told that this is similar to a second rectangle, which means that each rectangle is an enlargement or dilation of the other. This means that each length of the first rectangle will be multiplied by the same scale factor to calculate the corresponding length of the second rectangle. If we let the dimensions of the second rectangle be 𝑥 and 𝑦, we can create two equations: 15 multiplied by our scale factor is equal to 𝑥 and 10 multiplied by our scale factor is equal to 𝑦.
We are also told in the question that the perimeter of the second rectangle is equal to 40. As the perimeter is the distance around the outside of our rectangle, the perimeter of the first rectangle is equal to 50. 10 plus 15 plus 10 plus 15 is equal to 50. As the perimeter is a length, multiplying 50 by the same scale factor will give us an answer of 40. We can write this as an equation and then divide both sides by 50. The scale factor is therefore equal to 40 over 50. Dividing the numerator and denominator by 10, this simplifies to four-fifths. The second rectangle will therefore have dimensions four-fifths of the size of the first rectangle.
Substituting this into our two initial equations gives us 15 multiplied by four-fifths is equal to 𝑥, and 10 multiplied by four-fifths is equal to 𝑦. 𝑥 is therefore equal to 12 units and 𝑦 is equal to eight units. The length of the second rectangle is therefore equal to 12. We are also asked to calculate the area of this rectangle. And we know that the area of any rectangle is equal to its length multiplied by its width. The area is therefore equal to 12 multiplied by eight. This is equal to 96 and would be in square units. If a rectangle that is 15 by 10 is similar to a second rectangle with perimeter 40, then the length and area of the second rectangle are 12 and 96, respectively.