In $△ A B C, m ∠ B = 90^{o}, A B = 4$ and $B C = 3$ , then the radius of circle touching all three vertices of a triangle will be:

2.5

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3.5

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Solution

The correct option is A $2.5$
In $△ A B C, m ∠ B = 90^{o}, A B = 4, B C = 3$
By Pythagoras theorem, ${A C}^{2} = {A B}^{2} + {B C}^{2} = {(4)}^{2} + {(3)}^{2} = 25$
$∴ A C = 5$
Then the radius of a circle touching all the three sides of triangle $A B C = \frac{A C}{2} = \frac{5}{2} = 2.5$

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In △ABC,m∠B=90o,AB=4 and BC=3, then the radius of circle touching all three vertices of a triangle will be:

The correct option is A 2.5In △ABC,m∠B=90o,AB=4,BC=3By Pythagoras theorem, AC2=AB2+BC2=(4)2+(3)2=25∴AC=5Then the radius of a circle touching all the three sides of triangle ABC=AC2=52=2.5

In $△ A B C, m ∠ B = 90^{o}, A B = 4$ and $B C = 3$ , then the radius of circle touching all three vertices of a triangle will be:

The correct option is A $2.5$
In $△ A B C, m ∠ B = 90^{o}, A B = 4, B C = 3$
By Pythagoras theorem, ${A C}^{2} = {A B}^{2} + {B C}^{2} = {(4)}^{2} + {(3)}^{2} = 25$
$∴ A C = 5$
Then the radius of a circle touching all the three sides of triangle $A B C = \frac{A C}{2} = \frac{5}{2} = 2.5$