In a square, one pair of opposite sides was made 50% longer and the other pair of opposite sides was made 50% shorter.
What percent of the square’s area is the area of the new rectangle?
What percentage of the square’s perimeter is the perimeter of the new rectangle?
Let S1 = S2 = S3 = S4 = 10 units.
A1 = L*W = 10 * 10 = 100 sq. units.
S1 = S3 = 1.5*10 = 15.
S2 = S4 = 0.5*10 = 5.
A2 = L*W = 15 * 5 = 75 sq. units.
A2/A1 = 75/100 = 0.75 = 75%.
P2/P1 = (2L+2W)/(4*10) = (2*15+2*5) = 40/40 = 1.00 = 100%.
To find the percentage of the square's area that is the area of the new rectangle, we need to compare the areas of the two shapes.
1. Let's assume the side length of the original square is "s" units.
2. After increasing one pair of opposite sides by 50%, the new length of those sides becomes 1.5s.
3. After decreasing the other pair of opposite sides by 50%, the new length of those sides becomes 0.5s.
4. The area of the original square is given by the formula A = s^2.
5. The area of the new rectangle is given by the formula A' = (1.5s)(0.5s) = 0.75s^2.
To find the percentage of the square's area that is the area of the new rectangle, we can use the following formula:
Percentage area = (Area of new rectangle / Area of square) * 100
Now let's calculate:
Percentage area = (0.75s^2 / s^2) * 100
Percentage area = 0.75 * 100
Percentage area = 75%
Therefore, the area of the new rectangle is 75% of the original square's area.
To find the percentage of the square's perimeter that is the perimeter of the new rectangle, we need to compare the perimeters of the two shapes.
1. The perimeter of the original square is given by the formula P = 4s.
2. The perimeter of the new rectangle is given by the formula P' = 2(1.5s) + 2(0.5s) = 3s + s = 4s.
To find the percentage of the square's perimeter that is the perimeter of the new rectangle, we can use the following formula:
Percentage perimeter = (Perimeter of new rectangle / Perimeter of square) * 100
Now let's calculate:
Percentage perimeter = (4s / 4s) * 100
Percentage perimeter = 1 * 100
Percentage perimeter = 100%
Therefore, the perimeter of the new rectangle is 100% of the original square's perimeter.
In conclusion:
- The area of the new rectangle is 75% of the original square's area.
- The perimeter of the new rectangle is 100% of the original square's perimeter.
To solve this problem, we need to understand the concepts of area and perimeter.
Let's start with the first part of the question: What percent of the square’s area is the area of the new rectangle?
1. Assume the side length of the square is "S".
2. The area of the square is given by A = S * S.
3. One pair of opposite sides of the square was made 50% longer, so their new length is 1.5 times the original length. Therefore, their length is 1.5S.
4. The other pair of opposite sides was made 50% shorter, so their new length is 0.5 times the original length. Therefore, their length is 0.5S.
5. The new rectangle formed by these side lengths has an area given by A_new = (1.5S) * (0.5S).
Now, let's calculate the ratio of the new rectangle's area to the square's area:
( A_new / A ) * 100 = ( (1.5S) * (0.5S) / (S*S) ) * 100 = ( 0.75S^2 / S^2 ) * 100 = 0.75 * 100 = 75%.
Therefore, the area of the new rectangle is 75% of the square's area.
Now let's move on to the second part of the question: What percentage of the square’s perimeter is the perimeter of the new rectangle?
1. The perimeter of the square is given by P = 4S, where S is the side length.
2. Since one pair of opposite sides was made 50% longer and the other pair was made 50% shorter, the new rectangle formed will have sides measuring 1.5S and 0.5S.
3. The perimeter of the new rectangle is given by P_new = 2(1.5S) + 2(0.5S) = 3S + S = 4S.
Now, let's calculate the ratio of the new rectangle's perimeter to the square's perimeter:
( P_new / P ) * 100 = ( 4S / 4S ) * 100 = 100%.
Therefore, the perimeter of the new rectangle is 100% of the square's perimeter.
In summary, the area of the new rectangle is 75% of the square’s area, and the perimeter of the new rectangle is 100% of the square’s perimeter.