Question

Rectangle ABCD is dilated to create rectangle A'B'C'D' . The width of rectangle ABCD is 8 feet. The width of rectangle A'B'C'D' is 6 feet. The area of rectangle ABCD is 96 square feet.
What is the area of rectangle A'B'C'D' ? I NEED HELP PLEASE!!

Answer 1

Answer:

54 square feet

Step-by-step explanation:

The width of the rectangle ABCD is 8 feet and the area of the same rectangle is 96 square feet.

Hence, the length of the rectangle is (96 ÷ 8) = 12 feet.

Now, if the rectangle ABCD is dilated to rectangle A'B'C'D', then the length and the width of the new rectangle will be changed in the same ratio.

So, $\frac{AB}{A'B'} = \frac{BC}{B'C'}$  {Where AB is the length and BC is the width}

⇒ $\frac{12}{A'B'} = \frac{8}{6}$

⇒ A'B' = 9 feet

Therefore, the area of the dilated rectangle A'B'C'D' will be A'B' × B'C' = 9 × 6 = 54 square feet. (Answer)