We want to find the perimeter and area of the new image that was made by enlarging $1.5$ times the dimensions of the original image. To do so, we will start by finding the perimeter and area of the original image. Let's start by recalling the formulas for the perimeter and area of a rectangle with length $ℓ$ and width $w .$

Perimeter of a Rectangle	$2 ℓ + 2 w$
Area of a Rectangle	$ℓ w$

In the case of the original image, $ℓ = 7$ and $w = 5 .$ We can substitute these values into each formula to calculate the perimeter and area of the original image. Let's do it!

Perimeter of the Original Image, $in .$	$2 (7) + 2 (5) = 14 + 10 = 24$
Area of the Original Image, $in^{2}$	$7 (5) = 35$

We got that the perimeter of the original image is $24$ inches and the area is $35$ square inches. Next, we will recall the relationships between perimeters and areas of similar figures.

Perimeter of Similar Figures	If figure $B$ is similar to figure $A$ by a scale factor, then the perimeter of $B$ is equal to the perimeter of $A$ times the scale factor.
Area of Similar Figures	If figure $B$ is similar to figure $A$ by a scale factor, then the area of $B$ is equal to the area of $A$ times the square of the scale factor.

Note that the new image is similar to the original image by a scale factor of $1.5 .$ Since we know the perimeter and area of the original image, we can now calculate the perimeter and area of the new image.

Perimeter of the New Image, $in .$	$24 (1.5) = 36$
Area of the New Image, $in^{2}$	$35 (1.5)^{2} = 35 (2.25) = 78.75$

Therefore, the perimeter of the new image is $36$ inches and the area is $78.75$ square inches.

Exercise