Question

Show that pts A(-4,-7) , B(-1,2) ,C(8,5) D[5,-4)] are vertices of a rhombus ABCD , I HAVE FOUND THE SIDES MEASURES AND ONLY WANT THE DIAGONALS MEASURE , pls let me know only both the DIAGONALS , the sides r equal so it is √90 of all four sides , I want diagonal in root answer ... genius expert pls dooo ittt , will be added to the brainliested , so no unnecessary message , Use *distance formula*

Answer 1

Given vertices are A(-4,-7), B(-1,2),C(8,5) and D(5,-4).

We know that $Distance : \sqrt{(x2 - x1)^2 + (y2 - y1)^2}$

Now,

$AB = \sqrt{(-1 + 4)^2 + (2 + 7)^2}$

$= \ \textgreater \ \sqrt{3^2 + 9^2}$

$= \ \textgreater \ \sqrt{9 + 81}$

$= \ \textgreater \ \sqrt{90}$

$= \ \textgreater \ 3 \sqrt{10}$



$BC = \sqrt{(8 + 1)^2 + (5 - 2)^2 }$

$= \ \textgreater \ \sqrt{(9)^2 + (3)^2}$

$= \ \textgreater \ \sqrt{81 + 9}$

$= \ \textgreater \ \sqrt{90}$

$= \ \textgreater \ 3 \sqrt{10}$



$CD = \sqrt{(5 - 8)^2 + (-4 - 5)^2}$

$= \ \textgreater \ \sqrt{(-3)^2 + (-9)^2}$

$= \ \textgreater \ \sqrt{9 + 81}$

$= \ \textgreater \ \sqrt{90}$

$= \ \textgreater \ 3 \sqrt{10}$




$AD = \sqrt{(5 + 4)^2 + (-4 + 7)^2}$

$= \ \textgreater \ \sqrt{9^2 + 3^2}$

$= \ \textgreater \ \sqrt{81 + 9}$

$= \ \textgreater \ \sqrt{90}$

$= \ \textgreater \ 3\sqrt{10}$


Now,

$Diagonals AC = \sqrt{(8 + 4)^2 + (5 + 7)^2}$

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

$= \ \textgreater \ \sqrt{144 + 144}$

$= \ \textgreater \ \sqrt{288}$

$= \ \textgreater \ 12 \sqrt{2}$




$Diagonals BD = \sqrt{(5 + 1)^2 + (-4 - 2)^2}$

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

$= \ \textgreater \ \sqrt{36 + 36}$

$= \ \textgreater \ \sqrt{72}$

$= \ \textgreater \ 6 \sqrt{2}$


Now,

Given that sides are equal but diagonals are not equal. Thus ABCD is a rhombus.


Hope this helps!

Answer 2

Hope this helps u..
Any doubts...can ask me