Show that the points (-3, 2), (-5, -5), (2, -3) and (4, 4) are the vertices of a rhombus. Find the area of this rhombus.
Answers
the area of rhombus ABCD = 45 sq units.
Given,
(-3, 2), (-5, -5), (2, -3) and (4, 4)
Let, A = (-3, 2), B = (-5, -5), C = (2, -3) and D = (4, 4)
Now, we need to find the distance between the points
AB = √ [ (-5+3)^2 + (-5-2)^2 ]
⇒ AB = √ [ (2)^2 + (-7)^2 ]
⇒ AB = √ ( 4+49 )
⇒ AB = √ (53 )
BC = √ [ (2+5)^2 + (-3+5)^2 ]
⇒ BC = √ [ (7)^2 + (2)^2 ]
⇒ BC = √ ( 49+4 )
⇒ BC = √ (53 )
CD = √ [ (4-2)^2 + (4+3)^2 ]
⇒ CD = √ [ (2)^2 + (7)^2 ]
⇒ CD= √ ( 4+49 )
⇒ CD = √ (53 )
AD = √ [ (4+3)^2 + (4-2)^2 ]
⇒ AD = √ [ (7)^2 + (2)^2 ]
⇒ AD = √ ( 49+4 )
⇒ AD = √ (53 )
AC = √ [ (2+3)^2 + (-3-2)^2 ]
⇒ AC = √ [ (5)^2 + (-5)^2 ]
⇒ AC= √ ( 25+25 )
⇒ AC = √ (50 )
BD = √ [ (4+5)^2 + (4+5)^2 ]
⇒ BD = √ [ (9)^2 + (9)^2 ]
⇒ BD = √ ( 81+81 )
⇒ BD = √ (162 )
AB = BC = CD = AD = √ (53 ) Sides
AC ≠ BD Diagonals
Therefore the points represent a rhombus.
Area of rhombus ABCD = 1/2 × AC × BD
= 1/2 × √50 × √162
= 1/2 × 90
Therefore, the area of rhombus ABCD = 45 sq units.