Question

Show that the points A(3, 0), B(4, 5), C(–1, 4) and D(–2, –1) are the vertices of a rhombus. Find its area.

Answer 1

The given points are  A(3,0), B(4,5), C(-1,4) and D(-2,-1) 

`AB = sqrt((3-4)^2 + (0-5)^2 ) = sqrt((-1)^2 +(-5)^2)`

`= sqrt(1+25) = sqrt(26)`

`BC = sqrt((4+1)^2 +(5-4)^2) = sqrt((5)^2 +(1)^2)`

`= sqrt(25+1) = sqrt(26)`

`CD = sqrt((-1+2)^2 +(4+1)^2) = sqrt((1)^2 +(5)^2)`

`= sqrt(1+25) = sqrt(26)`

`AD = sqrt((3+2)^2 +(0+1)^2) = sqrt((5)^2 +(1)^2)`

`= sqrt(25+1) = sqrt(26)`

`AC = sqrt((3+1)^2 + (0-4)^2) = sqrt((4)^2+(-4)^2)`

`= sqrt(16+16) =4sqrt(2)`

`BD = sqrt((4+2)^2 +(5+1)^2 ) = sqrt((6)^2+(6)^2)`

`= sqrt((36+36)) = 6 sqrt(2) `

`∵ AB = BC =CD =AD = 6 sqrt(2) and AC ≠ BD `

Therefore, the given points are the vertices of a rhombus

Area (Δ ABCD ) =`1/2 xx AC xxBD`

`= 1/2 xx 4 sqrt(2) xx 6 sqrt(2 ) = 24 ` sq. units

Hence, the area of the rhombus is 24 sq. units.