Question

Assertion (A): Quadrilateral formed by vertices A(0, 0, 0), B(3, 4, 5), C(8, 8, 8) and D(5, 4, 3) is a rhombus.
Reason (R): ABCD is a rhombus if $AB = BC = CD = DA$, and $AC \ne BD$.

• A. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
• B. Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of the Assertion (A)
• C. Assertion (A) is true, but Reason (R) is false.
• D. Assertion (A) is false, but Reason (R) is true.

Hint:

For a rhombus in 3D, all sides equal and diagonals unequal ensures it’s not a square.

Answer 1

The Correct Option is A

Step 1: Find the lengths of all sides. \[ AB = \sqrt{ (3-0)^2 + (4-0)^2 + (5-0)^2 } = \sqrt{9 + 16 + 25} = \sqrt{50}. \] \[ BC = \sqrt{ (8-3)^2 + (8-4)^2 + (8-5)^2 } = \sqrt{25 + 16 + 9} = \sqrt{50}. \] \[ CD = \sqrt{ (5-8)^2 + (4-8)^2 + (3-8)^2 } = \sqrt{9 + 16 + 25} = \sqrt{50}. \] \[ DA = \sqrt{ (0-5)^2 + (0-4)^2 + (0-3)^2 } = \sqrt{25 + 16 + 9} = \sqrt{50}. \] So, all four sides are equal. Step 2: Find the lengths of diagonals. \[ AC = \sqrt{ (8-0)^2 + (8-0)^2 + (8-0)^2 } = \sqrt{64 + 64 + 64} = \sqrt{192} = 8\sqrt{3}. \] \[ BD = \sqrt{ (5-3)^2 + (4-4)^2 + (3-5)^2 } = \sqrt{4 + 0 + 4} = \sqrt{8} = 2\sqrt{2}. \] So, \[ AC \neq BD. \] Rhombus condition:
A quadrilateral is a rhombus if:
1. All sides are equal
2. Opposite sides are parallel (which holds by construction if diagonals are not equal and both diagonals bisect each other)
Since all sides equal and diagonals are not equal ⇒ it’s a rhombus.
So the Assertion is True.
The Reason says:
“A quadrilateral is a rhombus if $AB = BC = CD = DA$ and $AC \neq BD$.” This is True: equal sides and diagonals not equal ensures it’s not a square but a rhombus.