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Question

In a ABC,ifa=3,b=4,c=5, then the distance between its incenter and circumcenter is


A

12

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B

32

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C

32

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D

52

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Solution

The correct option is D

52


Determine the distance between its incenter and circumcenter.

Given, a=3,b=4,c=5.

Step 1: Determine the area and semi-perimeter of the triangle.

We can observe that a2+b2=c2, therefore by the converse of Pythagoras theorem, We can say the triangle is a right-angle triangle.

We know that for right-angle triangle : Area=12×Base×Height

ar(ABC)=12×3×4ar(ABC)=6unitsq.

We know that Semi-perimeter is given by : s=(a+b+c)2.

s=3+4+52=122s=6units

Step 2: Determine value of R and r.

We know that the distance between incenter and circumcenter is given by (R22rR)...(1),

where, R is Circumradius, and r is inner circle radius.

we know that, R=a×b×c4×ar(ABC)

Substituting the values we get,

R=3×4×54×6R=52...(2)

For value of r, we know r=ar(ABC)s

Substituting the values we have,

r=66r=1...(3)

Step 3: Determine the distance between its incenter and circumcenter

We know that the distance between incenter and circumcenter is given by (R22rR),

Therefore, substituting the values of (2)and(3)inequation(1),

5222×1×52254-525-20452

Therefore, the distance between its incenter and circumcenter is 52units

Hence, option (D) is correct.


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