The diagonals of the rhombus bisect each other at right angle. The expression in terms of [tex]n[/tex] can be used to represent AD in rhombus ABCD is,
[tex]AD=\dfrac{1}{2} \sqrt{(2n+5) ^2+(4n-3) ]^2\\}[/tex]
What is property of diagonals of rhombus?
The diagonals of the rhombus bisect each other at right angle.
Given information-
The length of the one diagonal of the rhombus is,
[tex]AC=(2n+5)[/tex]
The length of the another diagonal of the rhombus is,
[tex]BD=(4n-3)[/tex]
Let the center point of the rhombus is O as shown in the figure.
In [tex]\Delta AOD[/tex] the [tex]\angle AOD[/tex] is the right angle triangle. Thus by the Pythagoras theorem,
[tex]AD^2=AO^2+DO^2[/tex]
Suppose the above equation as equation number 1.
As the diagonals of the rhombus bisect each other thus,
[tex]AO=\dfrac{1}{2} AC\\DO=\dfrac{1}{2} BD[/tex]
Put the values in equation 1,
[tex]AD^2=(\dfrac{1}{2}AC) ^2+(\dfrac{1}{2}BD) ^2[/tex]
Put the values,
[tex]AD^2=[\dfrac{1}{2}(2n+5)] ^2+[\dfrac{1}{2}(4n-3) ]^2\\AD^2=\dfrac{1}{4}[(2n+5) ^2+(4n-3) ]^2\\AD=\sqrt{\dfrac{1}{4}[(2n+5) ^2+(4n-3) ]^2}\\AD=\dfrac{1}{2} \sqrt{(2n+5) ^2+(4n-3) ]^2\\}[/tex]
Hence the expression in terms of [tex]n[/tex] can be used to represent AD in rhombus ABCD is,
[tex]AD=\dfrac{1}{2} \sqrt{(2n+5) ^2+(4n-3) ]^2\\}[/tex]
Learn more about the properties of the rhombus here;
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