A 5-in. by 8-in. photo is enlarged by a scale factor of 1 in. : 2 in.
a. Find the perimeter and area of the enlarged postcard.
b. Suppose the length and width of the enlarged postcard is doubled. Describe how the perimeter and area are affected.
c. Suppose the width of the enlarged photograph is doubled. Describe how the perimeter and area are affected.
a. To find the perimeter and area of the enlarged postcard, we first need to determine the dimensions of the enlarged photo.
The scale factor is 1 in. : 2 in., which means that for every 1 inch in the original photo, it is enlarged to 2 inches in the enlarged photo.
Given that the original photo is 5 in. by 8 in., we can multiply each dimension by the scale factor to find the dimensions of the enlarged photo:
Length of enlarged photo = Scale factor * Length of original photo
= 2 in./1 in. * 5 in. = 10 in.
Width of enlarged photo = Scale factor * Width of original photo
= 2 in./1 in. * 8 in. = 16 in.
Now we can calculate the perimeter and area of the enlarged postcard.
Perimeter = 2 * (Length + Width)
= 2 * (10 in. + 16 in.)
= 2 * 26 in.
= 52 in.
Area = Length * Width
= 10 in. * 16 in.
= 160 in^2.
Therefore, the perimeter of the enlarged postcard is 52 inches and the area is 160 square inches.
b. If the length and width of the enlarged postcard are doubled, the new dimensions would be:
Length = 2 * Length of enlarged photo
= 2 * 10 in.
= 20 in.
Width = 2 * Width of enlarged photo
= 2 * 16 in.
= 32 in.
Comparing the new dimensions to the original dimensions, we can see that both the length and width have doubled.
The perimeter is given by:
Perimeter = 2 * (Length + Width)
= 2 * (20 in. + 32 in.)
= 2 * 52 in.
= 104 in.
The area is given by:
Area = Length * Width
= 20 in. * 32 in.
= 640 in^2.
Therefore, when the length and width of the enlarged postcard are doubled, the perimeter becomes 104 inches and the area becomes 640 square inches.
c. If only the width of the enlarged photo is doubled, the new dimensions would be:
Length of the enlarged photo remains the same: 10 in.
Width = 2 * Width of enlarged photo
= 2 * 16 in.
= 32 in.
Comparing the new dimensions to the original dimensions, we can see that only the width has doubled.
The perimeter remains the same as in part a:
Perimeter = 2 * (Length + Width)
= 2 * (10 in. + 32 in.)
= 2 * 42 in.
= 84 in.
The area is given by:
Area = Length * Width
= 10 in. * 32 in.
= 320 in^2.
Therefore, when only the width of the enlarged photograph is doubled, the perimeter becomes 84 inches and the area becomes 320 square inches.
To find the perimeter and area of the enlarged postcard, we first need to calculate the dimensions of the enlarged photo.
The scale factor given is 1 in. : 2 in. This means that for every 1 inch in the original photo, it is scaled to 2 inches in the enlarged photo.
a. To find the dimensions of the enlarged postcard, we can simply multiply the dimensions of the original photo by the scale factor.
Original photo dimensions: 5 in. by 8 in.
Enlarged photo dimensions: (5 in. x 2) by (8 in. x 2)
Simplifying, we have: 10 in. by 16 in.
To find the perimeter, we add all the sides of the enlarged postcard:
Perimeter = 2(Length + Width)
Perimeter = 2(10 in. + 16 in.)
Perimeter = 2(26 in.)
Perimeter = 52 in.
To find the area, we multiply the length and width of the enlarged postcard:
Area = Length x Width
Area = 10 in. x 16 in.
Area = 160 square inches
Therefore, the perimeter of the enlarged postcard is 52 inches and the area is 160 square inches.
b. Suppose the length and width of the enlarged postcard is doubled. If we double both the dimensions, the new dimensions would be 20 inches by 32 inches.
The perimeter is affected by doubling both the length and width. So, the new perimeter would be:
New Perimeter = 2(20 in. + 32 in.)
New Perimeter = 2(52 in.)
New Perimeter = 104 in.
The area is affected by multiplying both the length and width. So, the new area would be:
New Area = 20 in. x 32 in.
New Area = 640 square inches.
Therefore, if we double the dimensions, the new postcard will have a perimeter of 104 inches and an area of 640 square inches.
c. Suppose only the width of the enlarged photograph is doubled. In this case, the new dimensions would be 10 inches by 16 inches, with the length remaining the same.
The perimeter is affected by doubling only the width. So, the new perimeter would be:
New Perimeter = 2(10 in. + 16 in.)
New Perimeter = 2(26 in.)
New Perimeter = 52 in.
However, the area is not affected by doubling only the width, as it is still calculated by multiplying length and width:
New Area = 10 in. x 16 in.
New Area = 160 square inches.
Therefore, if only the width is doubled, the postcard will still have a perimeter of 52 inches, but the area will remain 160 square inches.
(a) each side is doubled; so the perimeter is also.
(b) perimeter is doubled again. Since area is length*width, the area is multiplied by 2^2.
(c) if only the length is doubled, the perimeter is increased, but not doubled.
2(w + 2L) = 2w + 4L. The area is doubled, since (w)(2L) = 2(wL) = 2A
(a) each side is doubled; so the perimeter is also.
(b) perimeter is doubled again. Since the area is length*width, the area is multiplied by 2^2.
(c) if only the length is doubled, the perimeter is increased, but not doubled.
2(w + 2L) = 2w + 4L. The area is doubled, since (w)(2L) = 2(wL) = 2A