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Question: In a \[\Delta ABC\] if \[a = 3,b = 4,c = 5\] then the distance between its incentre and circumcenter...

In a ΔABC\Delta ABC if a=3,b=4,c=5a = 3,b = 4,c = 5 then the distance between its incentre and circumcenter is
A.12\dfrac{1}{2}
B.32\dfrac{{\sqrt 3 }}{2}
C.32\dfrac{3}{2}
D.52\dfrac{{\sqrt 5 }}{2}

Explanation

Solution

The incentre of a triangle is the intersection point of all the three interior angle bisectors of the triangle. The circumcenter of a triangle is defined as the point where the perpendicular bisectors of the sides of that particular triangle intersect.

Complete answer:
Incentre can be defined as the point where the internal angle bisectors of the triangle cross. This point will be equidistant from the sides of a triangle, as the central axis’s junction point is the centre point of the triangle’s inscribed circle.
The incentre of a triangle is the center of its inscribed circle which is the largest circle that will fit inside the triangle. This circle is also called an incircle of a triangle.
The point of concurrency of the bisector of the sides of a triangle is called the circumcenter. The circumcenter is also the centre of the circumcircle of that triangle and it can be either inside or outside the triangle.
Formulas used in the solution part:
The distance between the incentre and the circumcenter of a triangle =R22rR= \sqrt {{R^2} - 2rR}
Area =12absinC = \dfrac{1}{2}ab\sin C
Circumradius, R=abc4×areaR = \dfrac{{abc}}{{4 \times area}}
Inradius, r=areasr = \dfrac{{area}}{s}
Given sides are a=3,b=4,c=5a = 3,b = 4,c = 5
Here a2+b2=c2{a^2} + {b^2} = {c^2} (Pythagoras Theorem) is satisfied. So this is a right angled triangle.
Area =12absinC = \dfrac{1}{2}ab\sin C
=12×3×4×sin90= \dfrac{1}{2} \times 3 \times 4 \times \sin {90^ \circ }
Hence we get area =6 = 6
s=a+b+c2s = \dfrac{{a + b + c}}{2} =122=6 = \dfrac{{12}}{2} = 6
Circumradius, R=abc4×areaR = \dfrac{{abc}}{{4 \times area}}
=3×4×54×6=52= \dfrac{{3 \times 4 \times 5}}{{4 \times 6}} = \dfrac{5}{2}
Inradius, r=areasr = \dfrac{{area}}{s}
=66=1= \dfrac{6}{6} = 1
Therefore The distance between incentre and circumcenter =R22rR= \sqrt {{R^2} - 2rR}
=(52)22×1×52=52= \sqrt {{{\left( {\dfrac{5}{2}} \right)}^2} - 2 \times 1 \times \dfrac{5}{2}} = \dfrac{{\sqrt 5 }}{2}
Therefore option (4) is the correct answer.

Note:
The incentre of a triangle is the intersection point of all the three interior angle bisectors of the triangle. The circumcenter of a triangle is defined as the point where the perpendicular bisectors of the sides of that particular triangle intersect.