Question

In a triangle the sum of two sides is x and the product of the same two sides is y. If \[{{x}^{2}}-{{c}^{2}}=y\], where c is the third side of the triangle, then the ratio of the in-radius to the circum-radius of the triangle is
a. \[\dfrac{3y}{2x\left( x+c \right)}\]
b. \[\dfrac{3y}{2c\left( x+c \right)}\]
c. \[\dfrac{3y}{4x\left( x+c \right)}\]
d. \[\dfrac{3y}{4c\left( x+c \right)}\]

Answer 1

Hint: Consider 2 sides as a and b. Thus substitute the values in formulas of in-radius and circumradius. Find the ratio and substitute the value of the area and simplify it to get the required expression.

Complete step-by-step answer:

It is said that the sum of 2 sides of a triangle = x.
The product of the 2 sides of the triangle = y.
Let ‘c’ be the third side of the triangle.
It has been given that, \[{{x}^{2}}-{{c}^{2}}=y\].

We need to find the ratio of the in-radius to the circum-radius of the triangle.
The circumradius of a triangle is the radius of the triangle’s circumcircle or the line segment from the circumcenter to any of the vertices of the triangle.
In radius is the radius of the circle which is inscribed inside the triangle.

In-radius of triangle = (\[2\times \] Area)/ Sum of all sides
Circumradius of triangle = Product of sides / (\[4\times \] area)

Let us consider the 3 sides of the triangle as a, b and c.
Let us consider the area as a triangle.

The sum of all sides \[=s=\dfrac{a+b+c}{2}\], which is semi – perimeter.
\[\therefore \] In radius \[=\dfrac{2\times \Delta }{\dfrac{\left( a+b+c \right)}{2}}=\dfrac{\Delta }{s}-(1)\]
Circum – radius = product of sides / \[4\times \] area \[=\dfrac{abc}{4\Delta }-(2)\]

Let us take the ratio of (1) and (2).
\[\therefore \left( 1 \right)/\left( 2 \right)=\dfrac{\dfrac{\Delta }{s}}{\dfrac{abc}{4\Delta }}=\dfrac{4{{\Delta }^{2}}}{abcs}-(3)\]

We have been given that the product of 2 numbers is y.
i.e. ab = y.
The sum of 2 numbers is, a + b = x.

We know sum \[=\dfrac{a+b+c}{2}=\dfrac{x+c}{2}=s\]
Substitute the values in equation (3).
\[\therefore \] Ratio \[=\dfrac{4{{\Delta }^{2}}}{yc\dfrac{\left( x+c \right)}{2}}=\dfrac{8{{\Delta }^{2}}}{yc\left( x+c \right)}\]
We know that area of a triangle \[=\dfrac{1}{2}\times \] base \[\times \] height

Consider the figure drawn, to get the area , we know \[=\dfrac{1}{2}\times \] base \[\times \] height.

Area of \[\Delta ABC=\dfrac{1}{2}ah\]
But, \[\sin C=\dfrac{h}{b}\Rightarrow h=b\sin C\]
\[\therefore \] Area becomes \[=\dfrac{1}{2}ab\sin C\].
\[\therefore \] Area of the triangle, \[\Delta =\dfrac{1}{2}ab\sin C\].

We have been given, \[{{x}^{2}}-{{c}^{2}}=y\].
We know, a + b = x and y = ab.
\[\begin{align}
  & \therefore {{\left( a+b \right)}^{2}}-{{c}^{2}}=ab \\
 & {{a}^{2}}+{{b}^{2}}+2ab-{{c}^{2}}=ab\Rightarrow {{a}^{2}}+{{b}^{2}}-{{c}^{2}}=-2ab \\
 & {{a}^{2}}+{{b}^{2}}-{{c}^{2}}=-ab \\
\end{align}\]

Divide both sides by \[2ab\Rightarrow \dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}=\dfrac{-ab}{2ab}\]

Take cosine rule formula,
Thus, \[\cos C=\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}=\dfrac{{{x}^{2}}-2y-{{c}^{2}}}{2y}=\dfrac{-1}{2}\]
\[\begin{align}
  & \therefore \cos C=\dfrac{-1}{2} \\
 & \therefore C={{\cos }^{-1}}\left( \dfrac{-1}{2} \right) \\
\end{align}\]

From the trigonometric formula, \[{{\cos }^{-1}}\left( \dfrac{-1}{2} \right)={{120}^{\circ }}\].
\[\therefore \] We got, \[C={{120}^{\circ }}\].
Thus, \[\sin C=\sin {{120}^{\circ }}=\dfrac{\sqrt{3}}{2}\] [From trigonometric table]
\[\begin{align}
  & \therefore \Delta =\dfrac{1}{2}ab\sin C \\
 & \Delta =\dfrac{1}{2}ab\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{3}}{4}y \\
\end{align}\]

Thus, Ratio \[=\dfrac{8{{\Delta }^{2}}}{yc\left( x+c \right)}=\dfrac{8{{\left( \dfrac{\sqrt{3}}{4}y \right)}^{2}}}{yc\left( x+c \right)}\]
                                             \[=\dfrac{8\times 3{{y}^{2}}}{16yc\left( x+c \right)}=\dfrac{3y}{2c\left( x+c \right)}\]
\[\therefore \] We got the ratio as, \[\dfrac{3y}{2c\left( x+c \right)}\].
\[\therefore \] Option (b) is the correct answer.

Note: The area of the triangle can be expressed using the lengths of two sides and the sine of the included angle. You may see that this is referred to as the SAS formula for the area of a triangle. With this formula we don’t have to rely on finding the altitude of a triangle to get its area.