Question

What is the radius of the circle inscribed in a triangle having side lengths\[35cm\], \[44cm\] and \[75cm\].
a) $3cm$
b) $4cm$
c) $5cm$
d) $6cm$

Answer 1

Hint:
As per the question, we know that the question is based on the concept of incenter, where the circle is inscribed in a triangle and the radius of the circle is the perpendicular bisector of the side of the triangle.

Complete step by step solution:
Given: Sides of the triangle are\[35cm\], \[44cm\] and \[75cm\].
Let the radius of the circle $ = r$
Step by step solution:
As the all sides of the triangle are different, this is known as the scalene triangle, so we use heron’s formula to find out the area of the triangle. Which is as below:
Area of $\Delta ABC = \sqrt {S(S - AB)(S - BC)(S - CA)} ......(1)$
Before putting values in heron’s formula we have to find out$S$, mentioned in heron’s formula.
We can find out the value of$S$, with the help of this formula
$S = \dfrac{{AB + BC + CA}}{2}$(We will put the values of triangle sides in this formula)
$S = \dfrac{{35 + 44 + 75}}{2} = 77$
Now we will put the values of $S$and triangle sides in heron’s formula, equation $(1)$
Area of $\Delta ABC = \sqrt {77(77 - 35)(77 - 44)(77 - 75)} $
$ = \sqrt {77 \times 42 \times 33 \times 2} = \sqrt {7 \times 11 \times 7 \times 2 \times 3 \times 3 \times 11 \times 2} $
$ = 7 \times 11 \times 2 \times 3 = 462c{m^2}$
Here we get the area of triangle $ABC$
We know that the area of triangle $ABC$is equal to the sum area of individual triangles $AOB$, $BOC$ and $COA$
Area of$\Delta ABC = \Delta AOB + \Delta BOC + \Delta COA......(2)$
For this we need to find out the area of all these three triangles individually.
To find out the area of a triangle whose perpendicular height is given, with the help of a relationship, where area of a triangle is equal to the half of the product of base and perpendicular height. Which we can write in equation as written below:
Area of$\Delta AOB = \dfrac{1}{2} \times AB \times r$
$ = \dfrac{1}{2} \times 35 \times r = 17.5r$
Here, we get area of triangle\[AOB\]in the form of$r$
Area of$\Delta BOC = \dfrac{1}{2} \times BC \times r$
$ = \dfrac{1}{2} \times 44 \times r = 22r$
Here, we get area of triangle\[BOC\]in the form of$r$
Area of$\Delta COA = \dfrac{1}{2} \times AC \times r$
$ = \dfrac{1}{2} \times 75 \times r = 37.5r$
Here, we get area of triangle\[COA\]in the form of$r$
Now, we will add area of all triangles and equate it with triangle $ABC$by putting values in equation $(2)$
Area of $\Delta ABC = 17.5r + 22r + 37.5r = 77r$
$462 = 77r$
We want to find out$r$, we will take $r$on left side remaining on right side
$r = \dfrac{{432}}{{77}} = 6cm$

Hence, we get the radius of the circle is$6cm$, means option d is the right answer.

Note:
We should keep in mind that the perpendicular of the triangle is the radius of the circle, as o is the incenter of the circle. In the incentre, the circle is inscribed in a triangle where the radius of the circle is the perpendicular bisector of the triangle’s side.