Question

Given that the points (-7, 6), (3, 6), (3, -6), and (-7, -6) are vertices of a rectangle, how much longer is the length than the width?

Answer 1

Given that ABCD is a rectangle
So, AD = BC ( Width )
and AB = DC ( Length )

Solution :-

Let A ( - 7 , 6 ) , B ( 3 , 6 ) , C ( 3 , - 6 ) ,
D ( -7 , - 6 ) are the vertices of rectangle.

• Length of AD ( width ) = Distance of A to D

x1 = - 7 x2 = - 7

y 1 = 6 y2 = - 6

Distance of A to D =

$= \sqrt{ {(x2 - x1)}^{2} + {(y2 - y1)}^{2} }$

$= \sqrt{ {( - 7 + 7)}^{2} + {( - 6 - 6)}^{2} }$



$= \sqrt{0 + { ( - 12)}^{2} }$


$= \sqrt{144}$


= 12 Units

AD = BC

BC = 12 units


_________________________

• Length of AB = Distance between A To B

x1 = - 7 y1 = 6

x2 = 3 y2 = 6


Distance between A To B =
$= \sqrt{ {(3 + 7)}^{2} + {(6 - 6)}^{2} }$

$= \sqrt{ {(10)}^{2} }$


$= 10 {cm}^{2}$

• AB = DC

DC = 10 cm²


✴ Length of rectangle = 10 cm²

Width of rectangle = 12 cm²