Final answer:
The bulletin board has a length of 60 inches, and its area is indeed four times the area of Aleyah's message board; therefore, statements A and C are true. Statements B and D are false.
Step-by-step explanation:
To determine the truth of the statements concerning the bulletin board outside Aleyah's room, let's analyze each one based on the dimensions given for Aleyah's message board. The message board has a length of 30 inches and a perimeter of 108 inches. The formula for the perimeter of a rectangle is P = 2l + 2w, where l is the length and w is the width. Given the perimeter, we can calculate the width of Aleyah's message board as w = (P/2) - l = (108/2) - 30 = 54 - 30 = 24 inches.
The bulletin board is said to be twice as long and twice as wide. Therefore, the bulletin board has a length of 60 inches (30 inches * 2) which makes statement A true. Statement B suggests that the bulletin board has a perimeter of 180 inches, but this is incorrect because if we double the length and width for the perimeter calculation, we get P = 2(60) + 2(48) = 120 + 96 = 216 inches.
For statement C, the area of a rectangle is found by multiplying its length by its width. The area of Aleyah's message board is 30 * 24 = 720 square inches. The bulletin board, being twice as long and twice as wide, will have an area of 60 * 48 = 2880 square inches. Since 2880 is four times 720, statement C is true regarding the area being four times greater than the message board.
Statement D asserts that the width of the bulletin board is 15 inches, which is not correct as we have already determined it to be 48 inches. Of the given statements, only A and C are true.