Answer:
42m
Step-by-step explanation:
Given: Area of rectangle= 108[tex]m^{2}[/tex]
Diagonal= 15m
Lets assume the value of width of rectangle be `w` and length be `l`.
We know that Area of rectangle= [tex]w\times l[/tex]
∴ [tex]108= w\times l[/tex]
cross multiplying both side
⇒ [tex]w= \frac{108}{l}[/tex]
∴ w= [tex]\frac{108}{l}[/tex]
Now solving to get value for l and w.
Remember, [tex]Diagonal^{2} =l^{2} +w^{2}[/tex]
⇒ [tex]15^{2} = l^{2} +w^{2}[/tex]
Next substituting the value of w in the equation.
⇒ [tex]225= l^{2} +(\frac{108}{l}) ^{2}[/tex]
Opening parenthesis
⇒[tex]225= l^{2} + \frac{11664}{l^{2} }[/tex]
Now taking LCD and cross multiplying both side.
⇒ [tex]225l^{2} = l^{4} + 11664[/tex]
Use quadratic equation to solve.
∴ [tex]l^{4} -225l^{2} +11664= 0[/tex]
⇒ [tex]l^{4} -144l^{2} -81l^{2} -11664=0[/tex]
Solving it we get two value of l, which is 9 and 12.
We can use any these value of l in substituting.
∴ w= [tex]\frac{108}{9} = 12m[/tex]
l= 9m and w= 12m
Now, solving to get perimeter of the rectangle.
perimeter= [tex]2(w+l)[/tex]
⇒ Perimeter= [tex]2\times (12+9)= 2\times 21[/tex]
∴ Perimeter of the rectangle is 42m.
