Incircle of a Triangle: Construction and Inradius

The three angle bisectors of any triangle always cross through the incircle of a triangle. Assume we have a large dining table with a triangle-shaped top surface. And we want to keep a water jug or a fruit tray in the centre of the table so that it is easily and equally accessible to people from all three sides. What should the table’s position be?

To have equal access to the jug or any other item from all three sides, place it on the table, so it is equidistant from all three. We can place it at or near the triangle’s inception point. The circle inscribed in a triangle is called the incircle of a triangle. The centre of the circle, which touches all the sides of a triangle, is called the incenter of the triangle. The radius of the incircle is called inradius.

This article is about the definition of the incircle of a triangle, its construction and the formula to calculate the radius of the incircle of a triangle.

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Define Incircle of a Triangle

A circle is drawn inside a triangle such that it touches all three sides of the triangle is called the incircle of a triangle.

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The sides of the triangle which touches the circle are tangents to the circle. Hence, the centre of the circle is situated at the intersection of the triangle’s internal angle bisectors. This point is known as the incentre of the triangle and it is always equidistant from the sides of the triangle. The length of the perpendicular is called the inradius.

A circle can be inscribed in any triangle, whether it is isosceles, scalene, an equilateral triangle, an acute-angled triangle, an obtuse-angled triangle or a right triangle. And incentre of a triangle always lies inside the triangle.

Construction of Incircle of a Triangle

To construct an incircle, we require a Ruler and a Compass.

Let us construct incircle by using the following example.  

Step 1: Construct the incircle of the triangle \(△ABC\) with \(AB = 7\,{\rm{cm,}}\) \(\angle B = {50^{\rm{o}}}\) and \(BC = 6\,{\rm{cm}}.\)

Step 2: Draw the angle bisectors of any two angles (\(A\) and \(B\)) of the triangle and let these bisectors meet at point \(I.\)

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Step 3: From the point \(I\) drop a perpendicular \(ID\) on \(AB.\)

Step 4: With \(I\) as centre and \(ID\) as radius, draw the circle. This circle will touch all three sides of the triangle.

Note: The point where the bisectors of the angles of a triangle meet, shown above as \(I,\) is called the incentre. The length of the perpendicular, here \(ID\) is called the inradius.

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Radius of an Incircle (Inradius) of a Triangle

The radius of an incircle of a triangle is called its inradius. The inradius can be calculated by finding the length of perpendicular to the sides of the triangle. Inradius can also be calculated as the ratio of the area of the triangle and the semi-perimeter of the triangle.

Given \(△ABC\) with incentre at \(O,\) we observe that

\({\rm{Area}}(\Delta ABC) = {\rm{Area}}(\Delta AOB) + {\rm{Area}}(\Delta BOC) + {\mathop{\rm Area}\nolimits} (\Delta COA)\)

\({\rm{Area}}(\Delta ABC) = \frac{1}{2}(AB \times r) + \frac{1}{2}(BC \times r) + \frac{1}{2}(CA \times r)\)

\({\rm{Area}}(\Delta ABC) = \frac{r}{2}(AB + BC + CA)\)

\({\rm{Area}}(\Delta ABC) = \frac{r}{2} \times {\rm{Perimeter}}\,{\rm{of}}\,\Delta ABC\)

Hence, \(r = \frac{{2 \times {\rm{ Area }}(\Delta ABC)}}{{{\mathop{\rm Perimeter}\nolimits} \,(\Delta ABC)}}\)

Radius of Incircle (Inradius) of an Equilateral Triangle

We know that all the sides of an equilateral triangle are equal.

Let the side of an equilateral triangle\(=a\). Then, the area of an equilateral triangle \( = \frac{{\sqrt 3 }}{4} \times {a^2}\). Therefore, the radius of the incircle of an equilateral triangle is given by

\(r = \frac{{2 \times {\rm{ Area}}\,{\rm{of}}\,{\rm{an}}\,{\rm{equilateral}}\,{\rm{triangle }}}}{{{\rm{ Perimeter}}\,{\rm{of}}\,{\rm{an}}\,{\rm{equilateral}}\,{\rm{triangle }}}}\)

\(r = \frac{{2 \times \frac{{\sqrt 3 }}{4}{a^2}}}{{a + a + a}}\)

\(r = \frac{{\sqrt 3 {a^2}}}{{2 \times 3a}} = \frac{a}{{2\sqrt 3 }}\)

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Solved Example

Q.1. Identify the incentre of the \(△PQR\).

Ans: Incentre is the point of intersection of angle bisectors of a triangle. 
Here, \(PM\) and \(QN\) are angle bisectors of \(∠P\) and \(∠Q,\) respectively, intersecting at \(B.\) Hence, the incentre of the \(△PQR\) is \(B.\)

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Q.2. Construct a \(△ABC\) with \(AB = 5\;{\rm{cm}},\angle B = {60^{\rm{o}}}\) and \(BC = 6.4\;{\rm{cm}}{\rm{.}}\) Draw the incircle of the \(△ABC.\)
Ans: Steps of Construction:

(i) Draw a line segment \(AB = 5\;{\rm{cm}}\).
(ii) Take \(B\) as a centre and draw an angle \(\angle B = {60^{\rm{o}}}\) with the help of a compass. Cut the line with an arc \(BC = {\rm{6}}{\rm{.4}}\,{\rm{cm}}{\rm{.}}\)
(iii) Join \(AC.\)
(iv) Now, from \(A\) and \(B\) cut the bisector of \(∠A\) and \(∠B,\) intersecting each other at point \(D.\)
(v) With \(D\) as a centre, draw an in the circle which touches all the three sides of \(△ABC.\)

Q.3. Construct a \(△PQR\) in which, \(PQ = QR = RP = 5.7\,{\rm{cm}}\). Draw the incircle of the triangle and measure its radius.
Ans: Steps of Construction:
(i) Draw an equilateral \(△RPQ\) in which \(PQ = QR = RP = 5.7\,{\rm{cm}}\) each.
(ii) From \(P\) and \(Q\) cut the bisector of \(∠P\) and \(∠Q,\) intersecting each other at point \(O.\)
(iii) With \(P\) as a centre, draw an in the circle which touches all the three sides of \(△RPQ\)

Q.4. In the given figure, a circle inscribed in a \(△ABC,\) touches the sides \(AB, BC\) and \(AC\) at points \(D, E, F\) respectively. If \(AB = 12\,{\rm{cm}},\,BC = 8\,{\rm{cm}}\) and \(AC = 10\,{\rm{cm}},\) find the length of \(AD, BE\) and \(CF.\)

Ans: We know that the lengths of the two tangent segments to a circle drawn from an external point are equal.
Now, we have
\(AD = AF,BD = BE\,{\rm{\& }}\,CE = CF\)
Now, \(AD + BD = 12\;{\rm{cm}} \ldots \ldots .{\rm{ (i)}}\)
\(AF + FC = 10\;{\rm{cm}}\)
\( \Rightarrow AD + FC = 10\;{\rm{cm}} \ldots \ldots .{\rm{ (ii)}}\)
\(BE + EC = 8\;{\rm{cm}}\)
\( \Rightarrow BD + FC = 8\;{\rm{cm}} \ldots \ldots .{\rm{ (iii)}}\)
Adding all these, we get
\(AD+BD+AD+FC+BD+FC=30\)
\(⇒2(AD+BD+FC)=30\)
\( \Rightarrow AD + BD + FC = 15\;{\rm{cm}} \ldots \ldots …….{\rm{ (iv) }}\)
Solving \({\rm{(i)}}\) and \({\rm{(iv)}}\), we get
\(FC = 3\;{\rm{cm}}\)
Solving \({\rm{(ii)}}\) and \({\rm{(iv)}}\), we get
\(BD = 5\;{\rm{cm}}\)
Solving (iii) and (iv), we get
and \(\therefore AD = AF = 7\;{\rm{cm}},BD = BE = 5\;{\rm{cm}}\) and \(CE = CF = 3\;{\rm{cm}}\)

Q.5. In the given figure, an isosceles \(△ABC,\) with \(AB=AC,\) circumscribes a circle. Prove that point of contact \(P\) bisects the base \(BC\).

Ans: We know that the lengths of the two tangent segments to a circle drawn from an external point are equal.
Now, we have \(AR=AQ, BR=BP\) and \(CP=CQ\)
Now, \(AB=AC\)
\(⇒AR+RB=AQ+QC\)
\(⇒AR+RB=AR+OC\)
\(⇒RB=QC\)
\(⇒BP=CP\)
Hence, \(P\) bisects \(BC\) at \(P.\)

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Summary

In this article, we learned about the definition of the incircle of a triangle. Also, we learned the meaning of incentre, the inradius of a triangle. Also, we have seen how to construct the incircle of a triangle and how to find the radius of the incircle of a triangle. With the help of this article, one can easily solve the problems related to the incircle of a triangle.

FAQs

Q.1. What is the formula of the radius of incircle of a triangle?
Ans: The radius of incircle \((r)\) of a triangle is given by
\(r = \frac{{2 \times {\rm{ Area}}\,{\rm{of}}\,{\rm{triangle }}}}{{{\rm{ Perimeter}}\,{\rm{of}}\,{\rm{triangle }}}}\)

Q.2. Explain the incircle of a triangle with an example?
Ans: A circle is drawn inside a triangle such that it touches all three sides of the triangle is called the incircle of a triangle.
The sides of the triangle which touches the circle are tangents to the circle. Hence, the centre of the circle is situated at the intersection of the internal bisectors of the angles of the triangle. This point is called the incentre of the triangle and is equidistant from the sides of the triangle. The length of the perpendicular is called the inradius.
For example: Imagine that there are three busy roads that form a triangle. And we want to open a store that is at the same distance from each road to get as many customers as possible. Finding the incenter would help you find this point because the incenter is equidistant from all sides of a triangle.

Q.3. Where is the incentre of an obtuse-angled triangle located?
Ans: The incentre of an obtuse-angled triangle is always located inside the triangle because it is the cutting point of the internal angle bisector of the triangle.

Q.4. Where do we use the incenter of a triangle in real life?
Ans: A man wants to install a new triangular countertop. And he wants to put a stove in the incenter of it so that it is easy to access from all sides. So, he uses the incenter of the counter to place the stove, so it is at the same distance from all the sides of the counter.

Q.5. How do you find the incentre of a triangle when the coordinates of the vertices of the triangle are given?
Ans: If \(A\left( {{x_1},{y_1}} \right),B\left( {{x_2},{y_2}} \right)\) and \(C\left( {{x_3},{y_3}} \right)\) are the vertices of a triangle \(ABC.\)
The steps to calculate the incentre of a triangle is:
Let \(a\) be the length of the side opposite to the vertex \(A, b\) be the length of the side opposite to the vertex \(B,\) and \(c\) be the length of the side opposite to the vertex \(C.\)
That is,
\(AB=c, BC=a\) and \(CA=b\)
Then we use the formula given below to find the incentre of the triangle
\(\frac{{a{x_1} + b{x_2} + c{x_3}}}{{a + b + c}},\frac{{a{y_1} + b{y_2} + c{y_3}}}{{a + b + c}}\)

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