Question
Download Solution PDFFind 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.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
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 = \(\frac {1}{2} \times\) 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 = \(\sqrt {4^2 + 3^2} \) = 5 cm
∴ The length of each side of the rhombus is 5 cm.
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