Answer
64.8k+ views
Hint: Use theorem tangent to the circle to find sides of the triangle and Heron’s formula to compute the area of the triangle.
![](https://www.vedantu.com/question-sets/f56d2401-de35-4012-9d6d-a8d185ecf97b2662106727949385828.png)
Let ABC be the triangle with incircle inside as given in the problem.
Let the incircle touch the sides of the triangle ABC at points D,E and F respectively .
It is given that $\alpha $,$\beta $,$\gamma $ are the distance of the vertices of a triangle of $ABC$ from the corresponding point of contact with the incircle.
Therefore, we assume,
$
AD = \alpha \\
BD = \beta \\
CE = \gamma \\
$
Theorem tangent to the circle states that a point external to the circle subtends equal tangents to the circle.
Using this theorem in triangle $ABC$,we get
$
AE = \alpha \\
BF = \beta \\
CF = \gamma \\
$
We need to find the area of the triangle $ABC$ using Heron’s formula.
Heron’s formula states that area of the triangle is given by,
\[\Delta = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} {\text{ (1)}}\]
In the above equation, $s$ is the semi-perimeter of the triangle and $a,b,c$ are the sides of the triangle.
Sides of the triangle $ABC$ are given by
$\left(
a = AB = \alpha + \beta \\
b = BC = \beta + \gamma \\
c = AC = \alpha + \gamma \\
\right){\text{ (2)}}$
Semi-perimeter of triangle $ABC$ is given by,
$
s = \dfrac{{a + b + c}}{2} \\
\Rightarrow s = \dfrac{{\alpha + \beta + \beta + \gamma + \alpha + \gamma }}{2} = \alpha + \beta + \gamma {\text{ (3)}} \\
$
Using equation $(2)$and $(3)$ in $(1)$,we get
\[
\Delta = \sqrt {\left( {\alpha + \beta + \gamma } \right)\left( {\alpha + \beta + \gamma - \alpha - \beta } \right)\left( {\alpha + \beta + \gamma - \beta - \gamma } \right)\left( {\alpha + \beta + \gamma - \alpha - \gamma } \right)} \\
\Rightarrow \Delta = \sqrt {\left( {\alpha + \beta + \gamma } \right)\alpha \beta \gamma } {\text{ }} \\
\]
Squaring both sides of the above equation, we get
\[{\Delta ^2} = \left( {\alpha + \beta + \gamma } \right)\alpha \beta \gamma {\text{ }}\]
Hence option $(C)$. ${\Delta ^2} = \left( {\alpha + \beta + \gamma } \right)\alpha \beta \gamma $ is the correct answer.
Note: Always try to draw the diagram in problems related to incircle. Try to remember the properties of incircle. Geometrical theorems should also be kept in mind while solving similar problems.
![](https://www.vedantu.com/question-sets/f56d2401-de35-4012-9d6d-a8d185ecf97b2662106727949385828.png)
Let ABC be the triangle with incircle inside as given in the problem.
Let the incircle touch the sides of the triangle ABC at points D,E and F respectively .
It is given that $\alpha $,$\beta $,$\gamma $ are the distance of the vertices of a triangle of $ABC$ from the corresponding point of contact with the incircle.
Therefore, we assume,
$
AD = \alpha \\
BD = \beta \\
CE = \gamma \\
$
Theorem tangent to the circle states that a point external to the circle subtends equal tangents to the circle.
Using this theorem in triangle $ABC$,we get
$
AE = \alpha \\
BF = \beta \\
CF = \gamma \\
$
We need to find the area of the triangle $ABC$ using Heron’s formula.
Heron’s formula states that area of the triangle is given by,
\[\Delta = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} {\text{ (1)}}\]
In the above equation, $s$ is the semi-perimeter of the triangle and $a,b,c$ are the sides of the triangle.
Sides of the triangle $ABC$ are given by
$\left(
a = AB = \alpha + \beta \\
b = BC = \beta + \gamma \\
c = AC = \alpha + \gamma \\
\right){\text{ (2)}}$
Semi-perimeter of triangle $ABC$ is given by,
$
s = \dfrac{{a + b + c}}{2} \\
\Rightarrow s = \dfrac{{\alpha + \beta + \beta + \gamma + \alpha + \gamma }}{2} = \alpha + \beta + \gamma {\text{ (3)}} \\
$
Using equation $(2)$and $(3)$ in $(1)$,we get
\[
\Delta = \sqrt {\left( {\alpha + \beta + \gamma } \right)\left( {\alpha + \beta + \gamma - \alpha - \beta } \right)\left( {\alpha + \beta + \gamma - \beta - \gamma } \right)\left( {\alpha + \beta + \gamma - \alpha - \gamma } \right)} \\
\Rightarrow \Delta = \sqrt {\left( {\alpha + \beta + \gamma } \right)\alpha \beta \gamma } {\text{ }} \\
\]
Squaring both sides of the above equation, we get
\[{\Delta ^2} = \left( {\alpha + \beta + \gamma } \right)\alpha \beta \gamma {\text{ }}\]
Hence option $(C)$. ${\Delta ^2} = \left( {\alpha + \beta + \gamma } \right)\alpha \beta \gamma $ is the correct answer.
Note: Always try to draw the diagram in problems related to incircle. Try to remember the properties of incircle. Geometrical theorems should also be kept in mind while solving similar problems.
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