Question

What is the sum of the diagonals in a rectangle whose area is 25v3 and whose width is 5?

Answer 1

Sum of diagonals = 20 units

Explanation:

`A(rectangle)=25\sqrt 3\: units^2\\w(rectangle) =5\: units`

To find: Sum of diagonals

`A(rectangle)=l(rectangle)* w(rectangle)`

`\implies l(rectangle)=(A(rectangle))/(w(rectangle))`

`\implies l(rectangle)=(25\sqrt 3)/(5)`

`\implies l(rectangle)={5\sqrt 3}\: units`

Let the diagonals of the rectangle be
`d_1\:\&\:d_2\: units`

Diagonals of a rectangle are equal in measure.

`\implies d_1=d_2=√(l(rectangle)^2+w(rectangle)^2)`

(By Pythagoras Theorem)

`\implies d_1=d_2=√((5\sqrt 3)^2+(5)^2)`

`\implies d_1=d_2=√(75+25)`

`\implies d_1=d_2=√(100)`

`\implies d_1=d_2=10\: units`

`Sum \:of \:diagonals = d_1+d_2`

`\implies Sum \:of \:diagonals = 10+10`

`\implies Sum \:of \:diagonals = 20\: units`