If the vertices of square ABCD are A(1, 3), B | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) To identify the quadrilateral,we calculate the lengths of the sides using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.Side $A

Vedclass · Accepted Answer

rectangle

Vedclass · Answer

(C) To identify the quadrilateral,we calculate the lengths of the sides using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.Side $AB = \sqrt{(4 - 1)^2 + (3 - 3)^2} = \sqrt{3^2 + 0^2} = 3$.Side $BC = \sqrt{(4 - 4)^2 + (5 - 3)^2} = \sqrt{0^2 + 2^2} = 2$.Side $CD = \sqrt{(1 - 4)^2 + (5 - 5)^2} = \sqrt{(-3)^2 + 0^2} = 3$.Side $DA = \sqrt{(1 - 1)^2 + (3 - 5)^2} = \sqrt{0^2 + (-2)^2} = 2$.Since opposite sides are equal ($AB = CD = 3$ and $BC = DA = 2$) and adjacent sides are not equal,it is a parallelogram.Now,check the diagonals: $AC = \sqrt{(4 - 1)^2 + (5 - 3)^2} = \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}$.$BD = \sqrt{(1 - 4)^2 + (5 - 3)^2} = \sqrt{(-3)^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}$.Since the opposite sides are equal and the diagonals are equal,$\square ABCD$ is a rectangle.

If the vertices of $\square ABCD$ are $A(1, 3), B(4, 3), C(4, 5)$ and $D(1, 5)$,then $\square ABCD$ is a $\ldots$

Similar Questions

$AOBC$ is a rectangle whose three vertices are $A (0, 3)$,$O (0, 0)$,and $B (5, 0)$. The length of its diagonal is

$P(x_{1}, y_{1})$ and $Q(x_{2}, y_{2})$ are the given points. If the line segment $\overline{PQ}$ is parallel to the $Y$-axis,then which of the following is true?

Find the circumcentre and the area of the triangle with vertices $(7,9), (10,8)$ and $(12,10)$.

State whether the following statement is true or false. Justify your answer.
Points $A(3, 1)$,$B(12, -2)$,and $C(0, 2)$ cannot be the vertices of a triangle.

The area of a triangle having vertices $A(3, 0)$,$B(0, 3)$,and $C(3, 3)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module