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

Let the width of the rectangle be \( w \) cm, then the length \( l = 2w \) cm.

Step2: Use area formula

The area of a rectangle is \( A = l \times w \). Given \( A = 5000 \) cm², so \( 2w \times w = 5000 \), which simplifies to \( 2w^2 = 5000 \). Then \( w^2 = \frac{5000}{2} = 2500 \), so \( w = \sqrt{2500} = 50 \) cm.

Step3: Find length

Since \( l = 2w \), then \( l = 2\times50 = 100 \) cm.

Step4: Find diagonal length

Using the Pythagorean theorem, the diagonal \( d \) of a rectangle is \( d = \sqrt{l^2 + w^2} \). Substituting \( l = 100 \) and \( w = 50 \), we get \( d = \sqrt{100^2 + 50^2} = \sqrt{10000 + 2500} = \sqrt{12500} \approx 111.8 \) cm. Since a rectangle has two diagonals of equal length, the sum of the two diagonals is \( 2d \approx 2\times111.8 = 223.6 \) cm.

Answer:

C. 223.6 cm