Question

On plotting the points O(0,0), A(3, – 4), B(3, 4) and C(0, 4) and joining OA, AB, BC
and CO, which of the following figure is obtained?
(a) Square (b) Rectangle (c) Trapezium (d) Rhombus

Answer 1

first of all we have to find out sides of OABC
use distance formula to find sides OA, AB,BC and CO.
OA=√{(3 - 0)² + (-4 - 0)²} = 5 unit
AB = √{(3-3)² + (4 + 4)²} = 8 unit
BC = √{(0 - 3)² + (4 - 4)²} = 3 unit
CO =√{(0 - 0)² + (4 - 0)²} = 4 unit

also find OB = √{(3 - 0)² + (4-0)²} = 5 unit
AC = √{(3 - 0)² + (4 + 4)²} = √73 unit

I think something is wrong in your question, because data's don't satisfy any options .
it is simple quadrateral . as you know, for square all sides must be equal and angle between two consecutive sides must be right angle. similarly , a rectangle , when two opposite sides be equal and angle between two consecutive side must be right angle.
for rhombus , all sides must be equal. for trapezium at least one pair of parallel sides.

hence, this question must be given wrong data's