Question

the area of a rectangle whose vertices are (-2,0),(5,0),(5,4) and (-2,4) taken in order is​

Answer 1

$\underline{\textsf{Given:}}$

$\textsf{The vertices of a recangle taken inorder}$

$\textsf{are (-2,0),(5,0),(5,4) and (-2,4)}$

$\underline{\textsf{To find:}}$

$\textsf{Area of the rectangle formed by the given vertices}$

$\underline{\textsf{Solution:}}$

$\textsf{Let the vertices be}$

$\textsf{A(-2,0),B(5,0),C(5,4) and D(-2,4)}$

$\textsf{First we find length and breadh of the rectangle ABCD}$

$\textsf{Length AB}$

$\mathsf{=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}}$

$\mathsf{=\sqrt{(-2-5)^2+(0-0)^2}}$

$\mathsf{=\sqrt{49}=7}$

$\textsf{Breadth BC}$

$\mathsf{=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}}$

$\mathsf{=\sqrt{(5-5)^2+(0-4)^2}}$

$\mathsf{=\sqrt{16}=4}$

$\textsf{Area of the rectangle ABCD}$

$\mathsf{=AB{\times}BC}$

$\mathsf{=7{\times}4}$

$\textsf{=28 square units}$

$\underline{\textsf{Answer:}}$

$\textsf{Area of the given rectangle formed by the given vertices is}$

$\textsf{28 square units}$