Abc is a right angled triangle right angled at a find the area of the shaded region if ab=6cm,bc=10 cm and o is the cengre of the incircle of abc
Answers
hey mate!
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
Seems that the shaded region is the part of the triangle expect the incircle.
FInding AC,
AC² = BC² - AB²
⇒ AC² = 10² - 6²
⇒ AC² = 100 - 36
⇒ AC² = 64
⇒ AC = 8
Finding the area of triangle ABC,
1/2 × AB × AC
⇒ 1/2 × 6 × 8
⇒ 24 cm²
Finding the semiperimeter of ΔABC,
(AB + BC + AC) / 2
⇒ (8 + 6 + 10) / 2
⇒ 24 / 2
⇒ 12 cm
Find the inradius according to A = rs, (A = Area of triangle ABC ; r = inradius)
A = rs
⇒ 24 = 12r
⇒ r = 2 cm
Finding the area of the incircle,
πr²
⇒ π × 2²
⇒ 4π cm²
Finding the area of the shaded region by subtracting the area of the incircle from that of ΔABC,
24 - 4π
⇒ 4(6 - π)
⇒ 4(6 - 3.14)
⇒ 4 × 2.86
⇒ 11.44 cm²
So 11.44 cm² is the area of the shaded region, which is the part of the triangle ABC excepting the incircle.
Hope this helps. Plz ask me if you have any doubt on my answer.
Thank you. :-))