For Delta ABC,a=5,b=12 and c=13. The radius o | vedclass

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(D) The sides of $\Delta ABC$ are $a=5$,$b=12$,and $c=13$.Since $5^2 + 12^2 = 25 + 144 = 169 = 13^2$,$\Delta ABC$ is a right-angled triangle where the

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$2$

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(D) The sides of $\Delta ABC$ are $a=5$,$b=12$,and $c=13$.Since $5^2 + 12^2 = 25 + 144 = 169 = 13^2$,$\Delta ABC$ is a right-angled triangle where the hypotenuse is $13$.The radius $r$ of the incircle of a right-angled triangle is given by the formula $r = \frac{a+b-c}{2}$,where $a$ and $b$ are the legs and $c$ is the hypotenuse.Substituting the values: $r = \frac{5+12-13}{2} = \frac{17-13}{2} = \frac{4}{2} = 2$.Thus,the radius of the circle touching all three sides is $2$.

Part $I$	Part $II$
$1.$ In $\Delta ABC, AB=3, BC=4, AC=5$	$a.$ In-radius $= 1$
$2.$ In $\Delta PQR, PQ=5, QR=12, PR=13$	$b.$ In-radius $= 2$
$3.$ In $\Delta XYZ, XY=8, YZ=15, XZ=17$	$c.$ In-radius $= 3$
$4.$ In $\Delta MNP, MN=20, NP=21, MP=29$	$d.$ In-radius $= 6$

For $\Delta ABC$,$a=5$,$b=12$ and $c=13$. The radius of a circle touching all the three sides of $\Delta ABC$ is ..... .

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