Question

If a rhombus has area 12 sq cm and side length 5 cm, then the length, in cm, of its longer diagonal is

• A. \(\frac{\sqrt{37}+\sqrt{13}}{2}\)
• B. \(\frac{\sqrt{13}+\sqrt{12}}{2}\)
• C. \(\sqrt{(37)} + \sqrt{(13)}\)
• D. \(\sqrt{(13)} + \sqrt{(12)}\)

Answer 1

The Correct Option is C

Let's designate the length of the longer diagonal of the rhombus as '2a' and the length of the shorter diagonal as '2b.'

The area of the rhombus is equal to 12 square centimeters, which can be expressed as:

\((\frac{1}{2}) \cdot (2a) \cdot (2b) = 12\) sq cm.

This simplifies to:
ab = 6.

The side length of the rhombus is 5 cm. Therefore,
a² + b² = 25.

Now, we can use the above equations to find the values of 'a' and 'b':
(a + b)² = a² + b² + 2ab
(a + b)² = 25 + 2(6) = 37
\(a + b = \sqrt{(37)}\)       (equation 1).

Similarly,
(a - b)² = a² + b² - 2ab
(a - b)² = 25 - 2(6) = 13
\(a - b = \sqrt{(13)}\)     (equation 2).

By solving equations 1 and 2, we can determine that the length of the long diagonal is 2a and is equal to:
\(2a=\sqrt{(37)} + \sqrt{(13)}.\)

Answer 2

Let's assume that the rhombus's longer and shorter diagonals are, respectively, "2a" and "2b" in length.
12 sq cm is the rhombus's area.
\(12 \text{ sq cm} = \frac{1}{2} \times 2a \times 2b\)
ab is equal to 6. 
The rhombus's side measures 5 cm. 
Consequently, \(a^2 + b^2 = 25.\) 
\((a + b)^2 = a^2 + b^2 + 2ab\)
\((a + b)^2 = 25 + 2(6) = 37\)
\(a + b = \sqrt{37} \quad \text{(eq. 1)}\)
\((a - b)^2 = a^2 + b^2 - 2ab\)
\((a - b)^2 = 25 - 2(6) = 13 \)
\(a - b = \sqrt{13} \quad \text{(eq. 2)}\)
Solving equations (1) and (2) yields 
Diagonal length = \(2a = \sqrt{37} + \sqrt{13}\)