Solutions to the nine rectangle exercises and problems:
Exercise 1: Find the area and perimeter of a rectangle with a length of 10 cm and a width of 6 cm.
Area (A): A = Length (L) × Width (W) A = 10 cm × 6 cm A = 60 square cm
Perimeter (P): P = 2(L + W) P = 2(10 cm + 6 cm) P = 2(16 cm) P = 32 cm
So, the area of the rectangle is 60 square cm, and its perimeter is 32 cm.
Exercise 2: Given the diagonal of a rectangle is 13 inches, and one of its sides is 5 inches, find the length of the other side.
Let the length of the rectangle be L and the width be W.
We can use the Pythagorean Theorem to find L:
D² = L² + W² (13 inches)² = L² + (5 inches)² 169 = L² + 25 L² = 169 - 25 L² = 144 L = √144 L = 12 inches
So, the length of the other side of the rectangle is 12 inches.
Exercise 3: The perimeter of a rectangle is 36 meters, and its length is three times its width. Find the dimensions (length and width) of the rectangle.
Let the width of the rectangle be W. Then the length is 3W.
Perimeter (P) = 2(L + W) 36 meters = 2(3W + W)
Now, solve for W: 36 meters = 2(4W) 36 meters = 8W W = 36 meters / 8 W = 4.5 meters
So, the width is 4.5 meters, and the length is 3 times the width, which is 3 × 4.5 meters = 13.5 meters.
Exercise 4: Find the length of a rectangle with an area of 72 square feet and a width of 6 feet.
Area (A) = Length (L) × Width (W) 72 square feet = L × 6 feet
Now, solve for L: L = 72 square feet / 6 feet L = 12 feet
The length of the rectangle is 12 feet.
Exercise 5: The ratio of the length to the width of a rectangle is 3:2, and its perimeter is 40 units. Find the length and width of the rectangle.
Let the common ratio between the length and width be k.
Length (L) = 3k Width (W) = 2k
Perimeter (P) = 2(L + W) 40 units = 2(3k + 2k)
Now, solve for k: 40 units = 2(5k) 40 units = 10k k = 40 units / 10 k = 4 units
So, the length is 3k = 3 × 4 units = 12 units, and the width is 2k = 2 × 4 units = 8 units.
Exercise 6: Calculate the area of a square with a side length of 9 meters, and then compare it to the area of a rectangle with a length of 12 meters and a width of 6 meters.
Area of the square = Side × Side = 9 meters × 9 meters = 81 square meters
Area of the rectangle = Length × Width = 12 meters × 6 meters = 72 square meters
So, the square has a larger area (81 square meters) compared to the rectangle (72 square meters).
Exercise 7: A rectangular garden is 18 meters long, and its width is unknown. If the area of the garden is 90 square meters, find the width of the garden.
Area (A) = Length (L) × Width (W) 90 square meters = 18 meters × W
Now, solve for W: W = 90 square meters / 18 meters W = 5 meters
The width of the garden is 5 meters.
Exercise 8: Find the diagonal of a square with an area of 64 square inches.
First, find the side length of the square using the square's area:
Area = Side × Side 64 square inches = Side × Side
Now, solve for the side length: Side = √(64 square inches) = 8 inches
Now that we know the side length, we can find the diagonal using the Pythagorean Theorem:
Diagonal² = Side² + Side² Diagonal² = (8 inches)² + (8 inches)² Diagonal² = 64 square inches + 64 square inches Diagonal² = 128 square inches
Diagonal = √(128 square inches) ≈ 11.31 inches (rounded to two decimal places)
So, the diagonal of the square is approximately 11.31 inches.
Exercise 9: A rectangle has a perimeter of 28 cm, and one of its sides is 8 cm. Find the length of the other side.
Let the length of the rectangle be L, and the given side is 8 cm.
Perimeter (P) = 2(L + W) 28 cm = 2(L + 8 cm)
Now, solve for L: 28 cm = 2(L + 8 cm) 14 cm = L + 8 cm
Subtract 8 cm from both sides: 14 cm - 8 cm = L 6 cm = L
So, the length of the other side of the rectangle is 6 cm.