Q212 / 270 · NDA 2025 (Phase 2) · Mathematics
If the vertices $B$ and $D$ of a square $ABCD$ are $(2, 3)$ and $(4, 1)$ respectively, then what is the area of the square?
- A$2$ square units
- B$3$ square units
- Correct answerC$4$ square units
- D$8$ square units
Solution
The diagonal of the square is the distance between the opposite vertices $B(2, 3)$ and $D(4, 1)$.
$BD = \sqrt{(4 - 2)^2 + (1 - 3)^2}$
$BD = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8}$
Let $a$ be the side length of the square. The relation between the diagonal and the side of a square is $BD = a\sqrt{2}$.
$a\sqrt{2} = \sqrt{8}$
$2a^2 = 8$
$a^2 = 4$
The area of the square is $a^2 = 4$ square units.
Answer: $4$ square units
$BD = \sqrt{(4 - 2)^2 + (1 - 3)^2}$
$BD = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8}$
Let $a$ be the side length of the square. The relation between the diagonal and the side of a square is $BD = a\sqrt{2}$.
$a\sqrt{2} = \sqrt{8}$
$2a^2 = 8$
$a^2 = 4$
The area of the square is $a^2 = 4$ square units.
Answer: $4$ square units
Hence, the correct answer is option: 'C'.