Please solve question 31 -

ABCD is a rectangle inscribed in a semi circle. If the length and the breadth of a rectangle are in the ratio 2:1 , what is the ratio of the perimeter of the rectangle to the diameter of the semi circle.

 

Dear Student,

Please find below the solution to the asked query:

We have our diagram , As :

Here we join OD , and OD  =  Radius of given semicircle =  r

Let ratio coefficient =  x  , As given :  If the length and the breadth of a rectangle are in the ratio 2 : 1 , So

Length of rectangle  = AB =  CD  =  2 x 

And

Breadth of rectangle  = BC =  DA  =  x 

Here we can see that ABCD is a largest possible rectangle that can inscribed in the given semicircle , So ' O ' is mid point of AB , So

OA =  OB =  x 

Now we apply Pythagoras theorem in triangle OAD and get :

OA²  + DA²  =  OD²  , Substitute values we get :

x² + x² = r² , 

2 x² = r² , 

r  = $\sqrt{2}$ x ,

So,

Diameter = d  = 2 $\sqrt{2}$ x

And

Perimeter of given rectangle =  AB +  BC +  CD + DA  = 2 x  + x  + 2 x  + x  = 6 x 

Then,

$$\begin{gathered} \frac{\mathrm{Perimeter} \mathrm{of} \mathrm{the} \mathrm{rectangle}}{\mathrm{Diameter} \mathrm{of} \mathrm{the} \mathrm{semi} \mathrm{circle}} = \frac{6 \mathrm{x}}{2\sqrt{2} \mathrm{x}} \\ \Rightarrow \frac{{\displaystyle \mathrm{Perimeter} \mathrm{of} \mathrm{the} \mathrm{rectangle}}}{{\displaystyle \mathrm{Diameter} \mathrm{of} \mathrm{the} \mathrm{semi} \mathrm{circle}}} = \frac{{\displaystyle 3}}{{\displaystyle \sqrt{2}}} \\ \mathbf{\Rightarrow }\mathbf{Perimeter}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{the}\mathbf{ }\mathbf{rectangle}\mathbf{ }\mathbf{:}\mathbf{ }\mathbf{Diameter}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{the}\mathbf{ }\mathbf{semi}\mathbf{ }\mathbf{circle}\mathbf{ }\mathbf{=}\mathbf{3}\mathbf{ }\mathbf{:}\mathbf{ }\sqrt{\mathbf{2}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathbf{Ans}\mathbf{ }\mathbf{)} \end{gathered}$$
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