Question

If the length of each side of a rhombus is 36 cm, and the area of the rhombus is 396 cm\(^2\), then what is the absolute value of the difference between the lengths of its diagonals?

Hint:

For rhombus problems, the two key formulas relating area, side, and diagonals are essential. Combining them with algebraic identities like \((x \pm y)^2\) is a common solution pattern.

Answer 1

Correct Answer: 60

Approach: A rhombus's diagonals are perpendicular and bisect each other, which links them to both the area and the side. Get the product and the sum of squares of the diagonals, then use \((d_1 - d_2)^2 = (d_1^2 + d_2^2) - 2d_1 d_2\).

Step 1: Given side \(s = 36\), area \(= 396\). Area \(= \tfrac12 d_1 d_2\), so \[ d_1 d_2 = 2 \times 396 = 792. \]

Step 2: Each side is the hypotenuse of a right triangle with legs \(d_1/2\) and \(d_2/2\), giving \(d_1^2 + d_2^2 = 4s^2\): \[ d_1^2 + d_2^2 = 4 \times 36^2 = 4 \times 1296 = 5184. \]

Step 3: Now \[ (d_1 - d_2)^2 = (d_1^2 + d_2^2) - 2 d_1 d_2 = 5184 - 2(792) = 5184 - 1584 = 3600. \]

Step 4: Take the square root: \(|d_1 - d_2| = \sqrt{3600} = 60\).

Final Answer: The diagonals differ by \(\boxed{60}\) cm.

Answer 2

Approach: Use the two standard rhombus identities, one from the Pythagorean relation between the side and half-diagonals, one from the area formula, and combine them using the algebraic identity for \((d_1-d_2)^2\).

Let the diagonals be \(d_1,d_2\). Since the diagonals of a rhombus bisect each other at right angles, each side is the hypotenuse of a right triangle with legs \(\tfrac{d_1}2,\tfrac{d_2}2\): \[ 36^2=\left(\frac{d_1}2\right)^2+\left(\frac{d_2}2\right)^2 \implies d_1^2+d_2^2=4(36^2)=5184. \]
The area formula gives \(\tfrac12 d_1 d_2=396 \implies d_1d_2=792\).

Now use \((d_1-d_2)^2=d_1^2+d_2^2-2d_1d_2\): \[ (d_1-d_2)^2=5184-2(792)=5184-1584=3600. \]
So \[ |d_1-d_2|=\sqrt{3600}=\boxed{60 \text{ cm}} \]