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/length

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

Step2: Set area equation

$w \times 2w = 5000$
$2w^2 = 5000$

Step3: Solve for width $w$

$w^2 = \frac{5000}{2} = 2500$
$w = \sqrt{2500} = 50$ cm
Length = $2 \times 50 = 100$ cm

Step4: Calculate one diagonal length

Use Pythagorean theorem: $d = \sqrt{w^2 + l^2}$
$d = \sqrt{50^2 + 100^2} = \sqrt{2500 + 10000} = \sqrt{12500}$
$\sqrt{12500} = 50\sqrt{5} \approx 111.80$ cm

Step5: Sum of two diagonals

A rectangle has 2 equal diagonals, so sum = $2 \times 111.80$

Answer:

C. 223.6 cm