Find the length of one side of a rhombus whose area is 24 cm2 and the sum of the lengths of its diagonals is 14 cm.

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  1. 4 cm
  2. cm
  3. cm
  4. cm

Answer (Detailed Solution Below)

Option 3 : 5 cm
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Detailed Solution

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Given:

Area of the rhombus = 24 cm2

Sum of the lengths of its diagonals = 14 cm

Concept used:

The diagonals of a rhombus always bisect each other at an angle of 90°.

Area of a rhombus =  Product of its diagonals

(a + b)2 = (a - b)2 + 4ab

Calculation:

Let, the length of the diagonals of the rhombus be P and Q respectively. (Where P > Q)

​According to the concept,

PQ/2 = 24

⇒ PQ = 48

​According to the question,

P + Q = 14     ---(1)

⇒ (P + Q)2 = 196

⇒ (P - Q)2 + 4PQ = 196

⇒ (P - Q)2 = 196 - 4PQ

Putting the value of PQ in the equation,

⇒ (P - Q)2 = 196 - 4 × 48 

⇒ (P - Q)2 = 196 - 192

⇒ (P - Q)2 = 4

⇒ (P - Q) = ± 2

 (P - Q) = 2     [∵  P > Q, taking +2]     ---(2)

Adding equation (1) and (2), we get,

P + Q + P - Q = 14 + 2

⇒ 2P = 16

⇒ P = 8 cm

So, Q = 8 - 2 = 6 cm

​According to the concept,

ΔCOD is a right-angled triangle at O.

So, DO = OB = 8/2 = 4 cm

AO = OC = 6/2 = 3 cm

Now, DC =  = 5 cm

∴ The length of each side of the rhombus is 5 cm.

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