Question 1 of 8
Rectangle is divided into four rows of five equal squares.
Julia drew this rectangle on grid paper. Which rectangle has the same area but a greater perimeter than Julia's rectangle?
A.A square is divided into four rows of four equal squares.
B.A rectangle is divided into 2 rows of 10 equal squares.
C.A rectangle is divided into 3 rows of 6 equal squares.
D.A rectangle is divided into 3 rows of 5 equal squares.
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C. A rectangle is divided into 3 rows of 6 equal squares.
In order to answer this question, we need to compare the area and perimeter of Julia's rectangle with the given options.
1. Start by finding the area of Julia's rectangle. Since it is divided into four rows of five equal squares, we know that there are a total of 4 x 5 = 20 squares.
2. The area of Julia's rectangle is therefore 20.
3. Now, let's compare the perimeter of Julia's rectangle with the options given. The perimeter is the sum of all the sides of the rectangle.
4. Julia's rectangle has 4 sides, each with a length of 5 squares, so its perimeter is 4 x 5 = 20 squares.
5. To compare, let's calculate the area and perimeter for each of the given options:
- A square divided into four rows of four equal squares has an area of 4 x 4 = 16 and a perimeter of 4 x 4 = 16.
- A rectangle divided into 2 rows of 10 equal squares has an area of 2 x 10 = 20 and a perimeter of 2(2+10) = 24.
- A rectangle divided into 3 rows of 6 equal squares has an area of 3 x 6 = 18 and a perimeter of 3(3+6) = 27.
- A rectangle divided into 3 rows of 5 equal squares has an area of 3 x 5 = 15 and a perimeter of 3(3+5) = 24.
6. Comparing the options, we find that the rectangle with the same area but a greater perimeter than Julia's rectangle is option C, which is a rectangle divided into 3 rows of 6 equal squares, with an area of 18 and a perimeter of 27.
Therefore, the correct answer is option C.
To find a rectangle with the same area but a greater perimeter than Julia's rectangle, we can compare the dimensions of each option.
Julia's rectangle has dimensions 4 rows x 5 squares, so the length is 5 and the width is 4.
Let's calculate the perimeter for each option:
- Option A: A square divided into four rows of four equal squares.
Since it's a square, the length and width are the same. Each side of the square is 4 squares long. Therefore, the perimeter would be 4 + 4 + 4 + 4 = 16.
- Option B: A rectangle divided into 2 rows of 10 equal squares.
The length is 10 and the width is 2. The perimeter for this rectangle would be 10 + 10 + 2 + 2 = 24.
- Option C: A rectangle divided into 3 rows of 6 equal squares.
The length is 6 and the width is 3. The perimeter for this rectangle would be 6 + 6 + 3 + 3 = 18.
- Option D: A rectangle divided into 3 rows of 5 equal squares.
The length is 5 and the width is 3. The perimeter for this rectangle would be 5 + 5 + 3 + 3 = 16.
Comparing the perimeters of each option, we can see that Option B, a rectangle divided into 2 rows of 10 equal squares, has the same area but a greater perimeter than Julia's rectangle.
Therefore, the answer is B. A rectangle divided into 2 rows of 10 equal squares.