Answer:
The measure of side MO is 8 unit
Step-by-step explanation:
Given as :
In the Triangle ΔMOP ,
∠P = 90°
∠M = 60°
The perimeter of ΔMOP = 12 + 4√3
Now, From figure , in Triangle ΔMOP
∠O = 180° - ( ∠P + ∠M )
or, ∠O = 180° - ( 90° + 60° )
or, ∠O = 180° - 150°
∴ ∠O = 30°
Now, from Triangle
Sin 90° = [tex]\dfrac{\textrm perpendicular}{\textrm hypotenuse}[/tex]
Or, Sin 90° = [tex]\dfrac{\textrm OP}{\textrm OM}[/tex]
Or, [tex]\dfrac{\textrm OP}{\textrm OM}[/tex] = 1
Again
Sin 60° = [tex]\dfrac{\textrm perpendicular}{\textrm hypotenuse}[/tex]
Or, Sin 60° = [tex]\dfrac{\textrm OP}{\textrm OM}[/tex]
Or, [tex]\dfrac{\textrm OP}{\textrm OM}[/tex] = [tex]\dfrac{\sqrt{3} }{2}[/tex]
Similarly
Sin 30° = [tex]\dfrac{\textrm base}{\textrm hypotenuse}[/tex]
Or, Sin 30° = [tex]\dfrac{\textrm PM}{\textrm OM}[/tex]
Or, [tex]\dfrac{\textrm PM}{\textrm OM}[/tex] = [tex]\frac{1}{2}[/tex]
So, The ratio of the sides as
PM : MO : OP = 1 : 2 : [tex]\sqrt{3}[/tex]
Let PM = x
MO = 2 x
OP = x[tex]\sqrt{3}[/tex]
Now, from question
The perimeter of triangle ΔMOP = 12 + 4√3
I.e, The sum of sides of triangle ΔMOP = 12 + 4√3
or, PM + MO + OP = 12 + 4√3
or, x + 2 x + x[tex]\sqrt{3}[/tex] = 12 + 4√3
or, 3 x + x[tex]\sqrt{3}[/tex] = 12 + 4√3
Or, x ( 3 +[tex]\sqrt{3}[/tex] ) = 4 ( 3 +[tex]\sqrt{3}[/tex] )
Now, equating both side we get
x = 4
So, The measure of side MO = 2 x
I.e The measure of side MO = 2 × 4 = 8 unit
Hence The measure of side MO is 8 unit Answer
