Question

Find the perimeter of a rhombus whose one diagonal is 16 cm long and area is 240 cm\(^2\).

• A. \( 68 \text{ cm} \)
• B. \( 30 \text{ cm} \)
• C. \( 24 \text{ cm} \)
• D. \( 36 \text{ cm} \)

Hint:

\textbf{Rhombus Properties.} Remember that the diagonals of a rhombus bisect each other at right angles, and all four sides of a rhombus are equal in length. The area of a rhombus can be calculated using its diagonals, and the side length can be found using the Pythagorean theorem with half the lengths of the diagonals.

Answer 1

The Correct Option is A

Let the diagonals of the rhombus be \(d_1\) and \(d_2\). Given, one diagonal \(d_1 = 16\) cm and the area of the rhombus \(A = 240\) cm\(^2\). The area of a rhombus is given by the formula: $$ A = \frac{1}{2} d_1 d_2 $$ Substituting the given values: $$ 240 = \frac{1}{2} \times 16 \times d_2 $$ $$ 240 = 8 d_2 $$ $$ d_2 = \frac{240}{8} = 30 \text{ cm} $$ So, the lengths of the diagonals are 16 cm and 30 cm. The diagonals of a rhombus bisect each other at right angles. Let the sides of the rhombus be \(a\). Half the lengths of the diagonals are \(\frac{d_1}{2} = \frac{16}{2} = 8\) cm and \(\frac{d_2}{2} = \frac{30}{2} = 15\) cm. By the Pythagorean theorem, the side of the rhombus \(a\) can be found using the right-angled triangle formed by half of each diagonal: $$ a^2 = \left( \frac{d_1}{2} \right)^2 + \left( \frac{d_2}{2} \right)^2 $$ $$ a^2 = (8)^2 + (15)^2 $$ $$ a^2 = 64 + 225 $$ $$ a^2 = 289 $$ $$ a = \sqrt{289} = 17 \text{ cm} $$ The length of each side of the rhombus is 17 cm. The perimeter of a rhombus is given by \(P = 4a\). $$ P = 4 \times 17 = 68 \text{ cm} $$ Therefore, the perimeter of the rhombus is 68 cm.