Answer:
Sum of diagonals = 20 units
Step-by-step explanation:
[tex]A(rectangle)=25\sqrt 3\: units^2\\w(rectangle) =5\: units[/tex]
To find: Sum of diagonals
[tex]A(rectangle)=l(rectangle)\times w(rectangle) [/tex]
[tex]\implies l(rectangle)=\frac{A(rectangle)}{w(rectangle)}[/tex]
[tex]\implies l(rectangle)=\frac{25\sqrt 3}{5}[/tex]
[tex]\implies l(rectangle)={5\sqrt 3}\: units[/tex]
Let the diagonals of the rectangle be [tex]d_1\:\&\:d_2\: units[/tex]
Diagonals of a rectangle are equal in measure.
[tex]\implies d_1=d_2=\sqrt{l(rectangle)^2+w(rectangle)^2}[/tex]
(By Pythagoras Theorem)
[tex]\implies d_1=d_2=\sqrt{(5\sqrt 3)^2+(5)^2}[/tex]
[tex]\implies d_1=d_2=\sqrt{75+25}[/tex]
[tex]\implies d_1=d_2=\sqrt{100}[/tex]
[tex]\implies d_1=d_2=10\: units[/tex]
[tex]Sum \:of \:diagonals = d_1+d_2[/tex]
[tex]\implies Sum \:of \:diagonals = 10+10[/tex]
[tex]\implies Sum \:of \:diagonals = 20\: units[/tex]
