Question

A circle with centre P is inscribed in the $\Delta {\rm{ABC}}$. Side AB, side BC and side AC
touch the circle at points L, M and N respectively. Radius of the circle is $r$
Prove that:${\rm{Area}}\;{\rm{of}}\;\Delta {\rm{ABC}}\;{\rm{ = }}\dfrac{1}{2} \times r\left( {{\rm{AB
+ BC + CA}}} \right)$

 

Answer 1

Hint: In this problem it is given that a circle with centre P is inscribed in the $\Delta
{\rm{ABC}}$. Side AB, side BC and side AC touch the circle at points L, M and N respectively.
Radius of the circle is $r$. We have to prove that${\rm{Area}}\;{\rm{of}}\;\Delta {\rm{ABC}}\;{\rm{ = }}\dfrac{1}{2} \times r\left( {{\rm{AB
+ BC + CA}}} \right)$
. Firstly we draw the lines between PA, PB and
PC and use tangent radius theorem. Then we prove the problem.

It is given that a circle with centre P is inscribed in the $\Delta {\rm{ABC}}$. Side AB, side BC
and side AC touch the circle at points L, M and N respectively. Radius of the circle is $r$.
We have to prove that ${\rm{Area}}\;{\rm{of}}\;\Delta {\rm{ABC}}\;{\rm{ = }}\dfrac{1}{2} \times r\left( {{\rm{AB
+ BC + CA}}} \right)$
Let,
Draw the lines between PA, PB and PC
AB is the tangent on the circle, touch the circle at the point L, and PL is the radius.
Now, we use the tangent radius theorem.
According to the tangent radius theorem we can easily say that ${\rm{PL}} \bot {\rm{AB}}$
Also, we can prove that ${\rm{PN}} \bot {\rm{AC}}$ and ${\rm{PM}} \bot {\rm{BC}}$.
It is known that the radius of the circle $r = {\rm{PL = PM = PN}}$.
And
$\Delta {\rm{APL,}}\Delta {\rm{APN,}}\Delta {\rm{BPL,}}\Delta {\rm{BPM,}}\Delta
{\rm{CPM}}$ and $\Delta {\rm{CPN}}$ are right angled triangles.
Thus,
Area of $\Delta {\rm{ABC}}\;{\rm{ = }}$ Area of $\Delta {\rm{APL}}$$ + $ Area of $\Delta
{\rm{APN}}$ $ + $ Area of $\Delta {\rm{BPL}}$$ + $ Area of $\Delta {\rm{BPM}}$$ + $
Area of$\Delta {\rm{CPM}}$ $ + $ Area of $\Delta {\rm{CPN}}$

$\begin{array}{l} = \dfrac{1}{2} \times {\rm{PL}} \times {\rm{AL + }}\dfrac{1}{2} \times
{\rm{PN}} \times {\rm{AN + }}\dfrac{1}{2} \times {\rm{PL}} \times {\rm{BL +
}}\dfrac{1}{2} \times {\rm{PM}} \times {\rm{BM + }}\dfrac{1}{2} \times {\rm{PM}} \times
{\rm{CM + }}\dfrac{1}{2} \times {\rm{PN}} \times {\rm{CN}}\\\end{array}$
Substitute $r = {\rm{PL = PM = PN}}$.in the above equation.
${\rm{Area}}\;{\rm{of}}\;\Delta {\rm{ABC}}\;{\rm{ = }}\dfrac{1}{2} \times r\left( {{\rm{AL
+ AN + BL + BM + CM + CN}}} \right)$
Simplify further.
${\rm{Area}}\;{\rm{of}}\;\Delta {\rm{ABC}}\;{\rm{ = }}\dfrac{1}{2} \times r\left( {{\rm{AL
+ AN + BL + BM + CM + CN}}} \right)$
${\rm{Area}}\;{\rm{of}}\;\Delta {\rm{ABC}}\;{\rm{ = }}\dfrac{1}{2} \times r\left( {{\rm{AL
+ BL + BM + CM + CN + AN}}} \right)$
From the figure ${\rm{AB = AL + BL,}}\;{\rm{BC = BM + CM,}}\;{\rm{AC = AN + CN}}$
Substitute ${\rm{AB = AL + BL,}}\;{\rm{BC = BM + CM,}}\;{\rm{AC = AN + CN}}$ in the
above equation.
${\rm{Area}}\;{\rm{of}}\;\Delta {\rm{ABC}}\;{\rm{ = }}\dfrac{1}{2} \times r\left( {{\rm{AB
+ BC + CA}}} \right)$
Hence proved.
Note: Here it is given $\Delta {\rm{ABC}}$ is a triangle with centre is ${\rm{P}}$ and side
AB, side BC and side AC touch the circle at points L, M and N respectively.We apply tangent
radius theory and formula of area triangle. Then simplify the result and we get the answer.