Two rectangles have an area of 81 square inches. Name two possible perimeters for the rectangles
9 * 9 = 81, therefore 9 + 9 + 9 + 9 = ?
3 * 27 = 81, therefore 3 + 3 + 27 + 27 = ?
To find the possible perimeters of the rectangles, we first need to determine their dimensions. Let's assume the dimensions of the first rectangle are length (L) and width (W), and the dimensions of the second rectangle are length (L') and width (W').
Given that the area of both rectangles is 81 square inches, we have:
Rectangle 1: L * W = 81
Rectangle 2: L' * W' = 81
To find the possible values of L and W, we can consider different factors of 81 that can multiply together to give the area. These factors are (1, 81), (3, 27), and (9, 9). Therefore, the possible dimensions of the rectangles are:
Case 1: (L = 1, W = 81) and (L' = 1, W' = 81)
In this case, the perimeters of both rectangles would be:
Rectangle 1: P1 = 2(L + W) = 2(1 + 81) = 2 * 82 = 164 inches
Rectangle 2: P2 = 2(L' + W') = 2(1 + 81) = 164 inches
Case 2: (L = 3, W = 27) and (L' = 3, W' = 27)
In this case, the perimeters of both rectangles would be:
Rectangle 1: P1 = 2(L + W) = 2(3 + 27) = 2 * 30 = 60 inches
Rectangle 2: P2 = 2(L' + W') = 2(3 + 27) = 2 * 30 = 60 inches
Case 3: (L = 9, W = 9) and (L' = 9, W' = 9)
In this case, the perimeters of both rectangles would be:
Rectangle 1: P1 = 2(L + W) = 2(9 + 9) = 2 * 18 = 36 inches
Rectangle 2: P2 = 2(L' + W') = 2(9 + 9) = 2 * 18 = 36 inches
Therefore, the two possible perimeters for the rectangles are:
164 inches and 60 inches, or 36 inches for each rectangle.