21
If the area of the smaller rectangle is 17 cm², then its length and width can be denoted as ( l ) and ( w ), where ( l \times w = 17 ) cm². The area of the larger rectangle, whose sides are twice as long, would be ( (2l) \times (2w) = 4(l \times w) ). Therefore, the area of the larger rectangle is ( 4 \times 17 = 68 ) cm².
15 and 20 inches because these dimensions comply with Pythagoras' theorem and the area of the rectangle.
Area of rectangle = Length * Breadth 126 = 18 * B => B = 126/18 = 7 inches.
The area of a rectangle = length x width = 5 x 8 = 40 square inches.
Perimeter: 7+3+7+3 = 20 inches Area: 7 times 3 = 21 square inches
The area of rectangle is : 15.0
Two
The area of rectangle is : 35.0
The area of rectangle is : 28.0
area = base X height if rectangle is 4 inches by 8 inches it the area would be 32 inches perimeter = sum of the length of the 4 sides...so in the rectangle above you have 2 sides at 4 inches and 2 sides at 8 so the total of those 4 sides will be 24 inches
48.488m2.
If the area of the smaller rectangle is 17 cm², then its length and width can be denoted as ( l ) and ( w ), where ( l \times w = 17 ) cm². The area of the larger rectangle, whose sides are twice as long, would be ( (2l) \times (2w) = 4(l \times w) ). Therefore, the area of the larger rectangle is ( 4 \times 17 = 68 ) cm².
15 and 20 inches because these dimensions comply with Pythagoras' theorem and the area of the rectangle.
Perimeter:20 inches Area:35
Area of rectangle = Length * Breadth 126 = 18 * B => B = 126/18 = 7 inches.
the width of the rectangle is 12.....because area= lenght x width so if the lenght is 5 inches and the area is 60....u divide 60 divided by 5 = 12
A rectangle with sides of 6 inches and 10 inches would have an area of 60 square inches.