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Question

Answers

(A) $\sqrt 3 $

(B) $\dfrac{{\sqrt 3 }}{{2\sqrt 2 }}$

(C) $\sqrt {\dfrac{1}{2}} $

(D) $\sqrt {\dfrac{1}{3}} $

Answer
Verified

In fig AB=b, AC=c, BC=a , AD=$\sqrt {\dfrac{2}{3}} $

According to the question it is given that

b = c = 1,

So, the nature of the triangle is isosceles.

The length of the altitude AD = C,

Radius of the circle = R

Now we find the area of triangle ABC where BC is the base and AD is the height of the triangle

$\Delta = \dfrac{1}{2} \times b \times h$

$\Delta = \dfrac{1}{2} \times BC \times AD$

Now put the values BC=a , AD=$\sqrt {\dfrac{2}{3}} $, we get,

$\Delta = \dfrac{1}{2} \times a \times \sqrt {\dfrac{2}{3}} $

By solving we get value of a,

\[a = 2 \times \Delta \times \sqrt {\dfrac{3}{2}} \]

Now we find the circumradius of the circle in which triangle is inscribed

$R = \dfrac{{a \times b \times c}}{{4 \times \Delta }}$ where $\Delta$ is the area of the triangle.

We put the values of \[a = 2 \times \Delta \times \sqrt {\dfrac{3}{2}} \],b=c=1 in the formula we get,

$ \Rightarrow R = \dfrac{{(\dfrac{{2\Delta \sqrt 3 }}{{\sqrt 2 }}) \times 1 \times 1}}{{4 \times \Delta }}$

Now $\sqrt 2 $ comes in the denominator,

$ \Rightarrow R = \dfrac{{2 \times \Delta \times \sqrt 3 }}{{4 \times \sqrt 2 \times \Delta }}$

Now, We cancel out the similar terms

$ \Rightarrow R = \dfrac{{\sqrt 2 \times \sqrt 2 \times \Delta \times \sqrt 3 }}{{2 \times 2 \times \sqrt 2 \times \Delta }}$

We get,

$ \Rightarrow R = \dfrac{{\sqrt 3 }}{{2\sqrt 2 }}$

So, the circumradius of the circle$R = \dfrac{{\sqrt 3 }}{{2\sqrt 2 }}$.