Question

Find the area of a rhombus whose perimeter is 4a and the length of one of the diagonals is x.

Answer 1

Given, perimeter of rhombus = 4a
D1=x
Therefore side of rhombus = a
We know that diagonal of rhombus are perpendicular bisector
We have to find D2
By pyth. Theorem
a^2=(x/2)^2+(D2/2)^2
D2/2= (2a-x)/2
D2=2a-x
Area of rhombus=1/2xD1xD2
=1/2xXx(2a-x)
Area of rhombus=ax-x^2