If a rhombus has area 12 sq cm and side length 5 cm, then the length, in cm, of its longer diagonal is
\(\frac{\sqrt{37}+\sqrt{13}}{2}\)
\(\frac{\sqrt{13}+\sqrt{12}}{2}\)
\(\sqrt{(37)} + \sqrt{(13)}\)
\(\sqrt{(13)} + \sqrt{(12)}\)
The Correct Option is C
Approach Solution - 1
Given:
- Area of rhombus = 12 cm²
- Side of the rhombus = 5 cm
Step 1: Let the diagonals be:
Longer diagonal = \( 2a \),
Shorter diagonal = \( 2b \)
Step 2: Use the area formula for a rhombus
\[ \text{Area} = \frac{1}{2} \cdot d_1 \cdot d_2 = \frac{1}{2} \cdot 2a \cdot 2b = 2ab = 12 \] \[ \Rightarrow ab = 6 \tag{1} \]
Step 3: Use Pythagoras Theorem in half-diagonals
Each side of rhombus forms a right triangle with half-diagonals: \[ \text{Side} = \sqrt{a^2 + b^2} = 5 \Rightarrow a^2 + b^2 = 25 \tag{2} \]
Step 4: Use identities to find \( a \) and \( b \)
Sum and difference formulas:
\[ (a + b)^2 = a^2 + b^2 + 2ab = 25 + 2 \cdot 6 = 37 \Rightarrow a + b = \sqrt{37} \tag{3} \] \[ (a - b)^2 = a^2 + b^2 - 2ab = 25 - 12 = 13 \Rightarrow a - b = \sqrt{13} \tag{4} \]
Step 5: Add equations to find \( a \)
Add Eq. (3) and (4): \[ a + b + a - b = 2a = \sqrt{37} + \sqrt{13} \Rightarrow a = \frac{\sqrt{37} + \sqrt{13}}{2} \] So, the longer diagonal: \[ 2a = \sqrt{37} + \sqrt{13} \]
✅ Final Answer: \( \boxed{2a = \sqrt{37} + \sqrt{13}} \)
Approach Solution -2
Let the rhombus have:
- Longer diagonal: \( 2a \)
- Shorter diagonal: \( 2b \)
- Area given as 12 cm²
- Each side of the rhombus is 5 cm
Step 1: Use the area formula of a rhombus
\[ \text{Area} = \frac{1}{2} \cdot d_1 \cdot d_2 = \frac{1}{2} \cdot 2a \cdot 2b = 2ab = 12 \] \[ \Rightarrow ab = 6 \tag{1} \]
Step 2: Use Pythagoras' Theorem for the triangle formed by diagonals
Since the diagonals bisect each other at right angles: \[ \left(\frac{2a}{2}\right)^2 + \left(\frac{2b}{2}\right)^2 = (\text{side})^2 \Rightarrow a^2 + b^2 = 25 \tag{2} \]
Step 3: Use sum and difference identities
\[ (a + b)^2 = a^2 + b^2 + 2ab = 25 + 2 \cdot 6 = 37 \Rightarrow a + b = \sqrt{37} \tag{3} \]
\[ (a - b)^2 = a^2 + b^2 - 2ab = 25 - 12 = 13 \Rightarrow a - b = \sqrt{13} \tag{4} \]
Step 4: Solve for a
Adding equations (3) and (4): \[ (a + b) + (a - b) = 2a = \sqrt{37} + \sqrt{13} \] \[ \Rightarrow \boxed{\text{Longer diagonal} = 2a = \sqrt{37} + \sqrt{13}} \]
✅ Final Answer: \( \boxed{2a = \sqrt{37} + \sqrt{13}} \) cm
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