Question

Please Helpppp!!!!!

In the figure,ABC is a right angled triangle in which angle BAC = 90° AC = 8cm and BC = 10 cm. O is the centre of the incircle touching the three sides. Find the area of the circle.​

Answer 1

Step-by-step explanation:

Let ABC be the right angled triangle such that ∠B = 90° , BC = 6 cm, AB = 8 cm. Let O be the centre and r be the radius of the in circle.

AB, BC and CA are tangents to the circle at P, N and M.

∴ OP = ON = OM = r (radius of the circle)

Answer 2

Answer:
$4\pi \: {cm}^{2}$

Solution :

Given, ABC is a right angled triangle,
So,
${AB}^{2} = {BC}^{2} - {AC}^{2} \\ AB = \sqrt{{BC}^{2} - {AC}^{2}} \\ AB = \sqrt{ {10}^{2} - {8}^{2} } \\ AB = 6$
so , area of triangle (A) = 1/2 x AB x AC = 1/2 x 8 x 6 = 24 sq.cm
and semi perimeter (s) = perimeter (p) /2 = (10 + 6 + 8)/2 = 24/2 = 12 cm

and we know that,

in-radius (r)= A/s =24/12 = 2cm

So, the area of circle will be,

$a = \pi {r}^{2} \\ a = \pi {2}^{2} = 4\pi \: {cm}^{2}$


an alternate short approach can be :

for a right angled triangle
inradius (r) = ( p + b - h)/2
= (8 + 6 - 10)/3 = 2
so area = π r^2 = 4π sq. cm