Question

a rectangle with area 5000 square cm is 2 times as long as it is wide. what is the sum of the diagonals? round to the nearest tenth. a. 55.9 cm b. 111.8 cm c. 223.6 cm d. 300.0 cm

Explanation:

Step1: Define variables for width/height

Let width = $w$, length = $2w$

Step2: Set up area equation

$$w \times 2w = 5000$$

Step3: Solve for $w$

$$2w^2 = 5000 \implies w^2 = 2500 \implies w = 50$$
Length = $2 \times 50 = 100$

Step4: Calculate one diagonal length

Use Pythagorean theorem:
$$\text{Diagonal} = \sqrt{w^2 + (2w)^2} = \sqrt{50^2 + 100^2} = \sqrt{2500 + 10000} = \sqrt{12500}$$

Step5: Compute sum of two diagonals

$$2 \times \sqrt{12500} = 2 \times 111.803 \approx 223.6$$

Answer:

C. 223.6 cm