Question

select the correct answer. a garden is in the shape of a rhombus, with two diagonal walking paths. what is the length of \\(\overline{bd}\\)? a. 28.3 m b. 56.6 m c. 20 m d. 40 m

Explanation:

Step1: Recall rhombus properties

In a rhombus, diagonals bisect each other at right angles. So, triangle \( AEB \) is a right triangle with \( AE = 20 \, \text{m} \), \( AB = 28.3 \, \text{m} \), and \( BE \) is half of \( BD \) (let \( BD = 2x \), so \( BE = x \)).

Step2: Apply Pythagorean theorem

For right triangle \( AEB \), \( AB^2 = AE^2 + BE^2 \). Substitute \( AB = 28.3 \), \( AE = 20 \):
\( 28.3^2 = 20^2 + x^2 \)
\( x^2 = 28.3^2 - 20^2 \)
\( x^2 = 800.89 - 400 = 400.89 \)
\( x \approx 20 \) (since \( 20^2 = 400 \), close to \( 400.89 \)).
Then \( BD = 2x \approx 40 \, \text{m} \)? Wait, no—wait, maybe miscalculation. Wait, \( AB = 28.3 \), \( AE = 20 \). Wait, \( 28.3 \) is approx \( 20\sqrt{2} \) (since \( 20\sqrt{2} \approx 28.28 \)). So if \( AE = 20 \), \( BE = 20 \) (since \( (20\sqrt{2})^2 = 20^2 + 20^2 \)). So \( BE = 20 \), so \( BD = 2 \times 20 = 40 \)? Wait, no, the options: D is 40m. Wait, but let's recheck. Wait, the diagonals bisect each other, so \( AE = EC = 20 \), \( BE = ED \). In triangle \( AEB \), \( AB = 28.3 \approx 20\sqrt{2} \), so it's an isosceles right triangle, so \( BE = AE = 20 \), so \( BD = BE + ED = 20 + 20 = 40 \, \text{m} \).

Answer:

D. 40m