Question

in given figure abc is right triangle angle A=90° find area of shaded region if AB=6cm BC = 10 cm k is the center of incircle of ∆ABC

Answer 1

Answer. ABC is a right angled triangle at right angled at A.  BC = 10 cm, AB = 6 cm.Let ‘O’ be the centre and  r be the radius of the incircle. AB, BC and CA are tangents to the circle at P,   M and N. ∴ IP = IM = IN = r   (radius of the circle)  In right ΔBAC,BC2   = AB2 + AC2  [Pythagoras theorem,]⇒ 102 = 62 + AC2 ⇒ AC2   = 100 - 36 ⇒ AC2   = 64 ⇒ AC = √64 = 8 cm Area of ∆ ABC = 1/2 × base × height                           = 1/2 × AC × AB                          = 1/2 × 8 × 6                           = 24 cm2 Area of ∆ABC = Area ∆IAB + Area ∆IBC + Area ∆ICA  ⇒ 24 = 1/2 r (AB) + 1/2 r (BC) + 1/2 r (CA)      = 1/2 r (AB + BC + CA)      = 1/2 r (6 + 8 + 10)      = 12 r    r = 2  Area of the circle =  πr2 = 3.14 x 2 x 2 = 12.56 cm2 Area of shaded region = Area of ∆ABC - Area of circle                                      = 24 – 12.56                                     = 11.44 cm2. Hence, the area of the shaded region is 11.44 cm2.

Answer 2

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