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

Sagot :

Hrishii

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]

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