If the sum of the diagonals of a rhombus is L and the perimeter is 4P, find the area of the rhombus?
Answer & explanation
Correct answer: option 2
We know that,
(d1/2)2 + (d2/2)2 = (side)2
Area = 1/2 × d1× d2
a2 + b2 = ( a + b)2 - 2ab
Where, d1 and d2 are diagonals of the rhombus respectively
And, a and b are 1st and 2nd term respectively
We have,
Sum of the diagonals of a rhombus = L
Perimeter = 4P
According to the question,
Perimeter = 4P
= 4× Side = 4 P
= Side = P
Now, sum of the diagonals = L
And, let us assume one diagonal be d1 and another diagonal be d2
= (d1/2)2 + ( d2/2)2 = P2
= d12 + d22 = 4 P2
= ( d1 + d2)2 - 2d1d2 = 4P2 [a2 + b2 = ( a + b)2 - 2ab ]
= L2 - 2d1d2 = 4P2 [As, d1 + d2 = L]
= 2d1d2 = L2 - 4 P2
= d1d2 = ( L2 - 4P2 )/2
Area of the Rhombus = \(\frac{1}{2}\) × d1d2 = \(\frac{1}{2}\) × (L2 - 4P2)/2 = \(\frac{1}{4}\) ( L2 - 4P2)