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In a right angle triangle $ABC$ with vertex $B$ being the right angle, the mutually perpendicular sides $AB$ and $BC$ are $p$ cm. and $q$ cm. long respectively. If the length of hypotenuse is $\left ( p+q-6 \right )$ cm., then the radius of the largest possible circle that can be inscribe in the triangle is ____________
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r = 3 cm ? $(a-r) + (b-r) = 2R$

$\Rightarrow (p-r) + (q-r) = p+q-6$ ( given a=p, b=q, 2R = p+q-6)

$\Rightarrow p+q-2r = p+q-6$

$\Rightarrow 2r = 6$

$\Rightarrow r = 3$

$\therefore r = 3 cm$

by Boss (23.8k points)
edited by
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why p-r??
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ok, let me chk once again
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@Satbir

formula for right angle triangle

$a+b=c$

or

$a^{2}+b^{2}=c^{2}$??

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a^2 + b^2 = c^2

i.e. p^2 + q^2 = (p+q-6)^2
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then how using a+b=c-2r formula

I am not getting the proof
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If you look at the diagram above ${\color{Red} = }$ which is marked as a-r and on 2R part are actually part of a sector of another circle of radius (a-r) and ${\color{Blue} \equiv }$ marked as b-r and 2R part are part of another sector of circle of radius (b-r).

So adding the above two we are getting (a-r) +(b-r) = 2R

It is a theorem.

In a right angled triangle, △ ABC, with sides a and b adjacent to the right angle, the radius of the inscribed circle is equal to r and the radius of the circumscribed circle is equal to R.

then in △ABC,

a+b=2⋅(r+R)

In this they have prooved it by 2 methods.

See the 3rd method if you dont undertand it i will proove it.