Question

show that the radius of incircle of a
right angled triangle is equal to the
difference of half of the
the perimeter and
twice the
length of
of its
its hypotenuse of
that triangle.​

Answer 1

$\huge\red{\mid{\underline{\overline{\textbf{ANSWER}}}\mid}}$

$\bold{\underline{{Given}}}$

According to figure-

ABC is a right angle triangle and r is the radius of the circle inscribed in the triangle.

$\bold{\underline{{TO \;PROVE}}}$

r = Perimeter of the ΔABC/2 - Hypotenuse of the triangle

i.e. r = Perimeter of the ΔABC/2 - AC

$\bold{\underline{{SOLUTION}}}</p><p>$

From the figure,

OP = OQ = r

Again in quadrilateral OPBQ

∠B = 90, ∠P = 90 {since PB is a tangent and OP is radius}

∠Q = 90, {since BQ is a tangent and OQ is radius}

Since 3 angles are right angle,

So, ∠O is also right angle

i.e. ∠O = 90

Again, adjacent sides OP and OQ are equal.

Hence, quadrilateral OPBQ is a square.

Hence, PB = BQ = r

Now, perimeter of the ΔABC = AB + BC + AC

=> Perimeter of the ΔABC = (AP + PB) + (BQ + QC) + (AR + RC)

Since, AP = AR, PB = BQ, QC = CR {tangent drawn from external point to the circle are equal}

=> Perimeter of the ΔABC = AP + r + r + QC + AR + RC

=> Perimeter of the ΔABC = AP + QC + AR + RC + 2r

=> Perimeter of the ΔABC = AR + RC + AR + RC + 2r

=> Perimeter of the ΔABC = 2AR + 2RC + 2r

=> Perimeter of the ΔABC = 2(AR + RC) + 2r

=> Perimeter of the ΔABC = 2AC + 2r

=> Perimeter of the ΔABC = 2(AC + r)

=> Perimeter of the ΔABC/2 = AC + r

=> r = Perimeter of the ΔABC/2 - AC