Question

In a right angled isosceles triangle the ratio of the circumradius and $r = \dfrac{{{\text{area}}}}{{{\text{semiperimeter}}}}$is:-
A.$2\left( {\sqrt 2 + 1} \right):1$
B.$\left( {\sqrt 2 + 1} \right):1$
C.$2:1$
D.$\sqrt 2 :1$

Answer 1

Hint: First find the hypotenuse of the given isosceles triangle using the Pythagoras theorem and then use the formula for finding the circumradius(R) and in-radius(r).
$R = \dfrac{{abc}}{{4\left( {{\text{Area of the triangle}}} \right)}}$
Then find the ratio which is required in the problem.

Complete step by step answer:
We have given a right-angled isosceles triangle which means it contains one right angle and other two angles are equal to each other which means the height and base are also equal to each other.
We have to find the ratio of the circumradius and in-radius. That is, $R:r$.

First we will write the properties of the isosceles triangle.
In this type of triangle, two sides are equal.
Let the length of sides be
Then the other side will also be $a$ as the given triangle is an equilateral triangle.
Now, we can find out the hypotenuse with the help of Pythagoras theorem, which is:
${\left( {{\text{hypotenuse}}} \right)^{\text{2}}}{\text{ = }}{\left( {{\text{height}}} \right)^{\text{2}}}{\text{ + }}{\left( {{\text{base}}} \right)^{\text{2}}}$
Now, substitute the assumed values of height and base.
$
  {\text{hypotenuse}} = \sqrt {{a^2} + {a^2}} \\
   = \sqrt {2{a^2}} \\
   = a\sqrt 2 \\
$
So, the hypotenuse of the triangle is $a\sqrt 2 $
Now, we know that the sides of the triangle are $a$,$a$ and $2\sqrt a $ .
Now, we will find the circumradius which is equal to the product of sides divided by four times the area.
Let, circumradius be $R$ , then the circumradius is given as:
$R = \dfrac{{a \times a \times 2\sqrt a }}{{4 \times {\text{area}}}}$
We know that base of the given triangle is $\left( a \right)$ and the height of the triangle is $\left( a \right)$ so, the area of the triangle is given as:
Area of the triangle$ = \dfrac{1}{2}\left( {{\text{base}}} \right)\left( {{\text{height}}} \right)$
Substitute the values of the base and height:
Area of the triangle$ = \dfrac{1}{2}\left( a \right)\left( a \right) = \dfrac{{{a^2}}}{2}$
Substitute the value of the area of the triangle to find the circumradius of the circle.
$R = \dfrac{{a \times a \times a\sqrt 2 }}{{4 \times \dfrac{{{a^2}}}{2}}}$
$R = \dfrac{{a\sqrt 2 }}{2}$
$R = \dfrac{a}{{\sqrt 2 }}$
Therefore, the circumradius of the given triangle is $\dfrac{a}{{\sqrt 2 }}$.
Now, we will calculate the in-radius which is equal to the ratio of area of triangle and semi perimeter of triangle.
We already know that the area of the triangle is:
Area of the triangle$ = \dfrac{1}{2}\left( a \right)\left( a \right) = \dfrac{{{a^2}}}{2}$
We find the semi-perimeter of the triangle.
$s = \dfrac{{{\text{Sum of sides}}}}{2}$
Substitute the value of the side lengths of the triangle.
$s = \dfrac{{a + a + a\sqrt 2 }}{2}$
$s = \dfrac{{a\left( {2 + \sqrt 2 } \right)}}{2}$
$s = a + \dfrac{a}{{\sqrt 2 }}$
So, the in-radius of the triangle is given as:
$r = \dfrac{{{\text{area}}}}{{{\text{semiperimeter}}}}$
So, in-radius will be:-
$r = \dfrac{{\dfrac{{{a^2}}}{2}}}{{a + \dfrac{a}{{\sqrt 2 }}}}$
$r = \dfrac{{{a^2}}}{{a\left( {\sqrt 2 + 2} \right)}}$
$r = \dfrac{a}{{\left( {\sqrt 2 + 2} \right)}}$
So, we get that the in-radius is equal to$\dfrac{a}{{\left( {\sqrt 2 + 2} \right)}}$ .
Now, we will calculate the ratio of circumradius to the in-radius.
Which is given as:$\dfrac{R}{r}$
Now, put the obtained values of $R$ which is circumradius and $r$ which is in-radius.
$R:r = \dfrac{{\dfrac{a}{{\sqrt 2 }}}}{{\dfrac{a}{{\left( {\sqrt 2 + 2} \right)}}}}$
$R:r = \dfrac{{\sqrt 2 + 2}}{{\sqrt 2 }}$
$R:r = \dfrac{{\sqrt 2 + 1}}{1}$
So, the radius of circumradius to the in-radius is $\left( {\sqrt 2 + 1} \right):1$.
Hence, option B is the correct answer.


Note:
 To calculate the circumradius and in-radius first we need to calculate the area and semiperimeter of the triangle, and we know the area of right angled isosceles triangle is $\dfrac{1}{2} \times a \times a$ and the semi perimeter of triangle is $\dfrac{{{\text{perimeter}}}}{2}$ .