One diagonal of a kite is twice as long as the other diagonal. If the area of the kite is 240 square inches (UAE metrify: Change inches to centimeters), what are the lengths of the diagonals?

To solve the problem, let's denote the lengths of the diagonals of the kite as $d_1$ and $d_2$, where $d_1$ is the longer diagonal and $d_2$ is the shorter diagonal. According to the problem, $d_1 = 2d_2$.

The area $A$ of a kite can be calculated using the formula: $A = \frac{1}{2} \times d_1 \times d_2$

Given that the area of the kite is 240 square inches, we can set up the equation: $240 = \frac{1}{2} \times d_1 \times d_2$

Substitute $d_1 = 2d_2$ into the equation: $240 = \frac{1}{2} \times 2d_2 \times d_2$ $240 = d_2^2$

Solve for $d_2$: $d_2^2 = 240$ $d_2 = \sqrt{240}$ $d_2 = \sqrt{16 \times 15}$ $d_2 = 4\sqrt{15}$

Now, find $d_1$: $d_1 = 2d_2$ $d_1 = 2 \times 4\sqrt{15}$ $d_1 = 8\sqrt{15}$

Next, convert the lengths from inches to centimeters. Since 1 inch is equal to 2.54 centimeters: $d_2 \text{ in cm} = 4\sqrt{15} \times 2.54$ $d_2 \text{ in cm} = 10.16\sqrt{15}$

$d_1 \text{ in cm} = 8\sqrt{15} \times 2.54$ $d_1 \text{ in cm} = 20.32\sqrt{15}$

Therefore, the lengths of the diagonals in centimeters are: $d_2 \approx 39.27 \text{ cm}$ $d_1 \approx 78.54 \text{ cm}$