A circle is inscribed in a right-angled triangle ABC right-angled at A Find the area of the area not included - Maths - Areas Related to Circles

Question

A circle is inscribed in a right-angled triangle ABC
right-angled at A Find the area of the area not included in the
circle but included in the triangle if AB = 6cm and BC = 6cm and I
is the centre of the circle

Gursheen kaur. · Accepted Answer

Given: ABC is a right angled triangle at A such that ∠ A = 90°.

BC = 8 cm, AB = 6 cm.

Let I 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,

∴By Pythagoras theorem,

⇒CB² = AB² + AC²

⇒ 8 ²= 6² + AC²

⇒ AC ²= 64 - 36

⇒ AC ²= 28

⇒ AC = √28 = 2√7

Now,

Area of ∆ ABC = ½ × base × height

=½ × AC × AB = ½ ×2√7 × 6

= 6√7 cm² .

Area of ∆ABC = Area ∆IAB + Area ∆IBC + Area ∆ICA

$\Rightarrow 6 \sqrt{7} = \frac{1}{2} r (A B) + \frac{1}{2} r (B C) + \frac{1}{2} r (C A) \Rightarrow 6 \sqrt{7} = \frac{1}{2} r (A B + B C + C A) \Rightarrow 6 \sqrt{7} = \frac{1}{2} r (6 + 8 + 2 \sqrt{7}) \Rightarrow 6 \sqrt{7} = r (7 + \sqrt{7}) \Rightarrow r = \frac{7 + \sqrt{7}}{6 \sqrt{7}} .......... (I) A r e a o f t h e c i r c l e = π r^{2} = π {(\frac{7 + \sqrt{7}}{6 \sqrt{7}})}^{2} [F r o m (I)] = π \frac{{(7 + \sqrt{7})}^{2}}{252} = π (\frac{49 + 14 \sqrt{7} + 7}{252}) = π (\frac{4 + \sqrt{7}}{18}) = \frac{π}{18} (4 + \sqrt{7})$

Thus,

area of shaded region = area of ∆ ABC - area of circle

$= 6 \sqrt{7} - \frac{π}{18} (4 + \sqrt{7}) = 6 \sqrt{7} - \frac{4 π}{18} - \frac{\sqrt{7} π}{18} = \frac{108 \sqrt{7} - 4 π - \sqrt{7} π}{18} {c m}^{3}$

7

A circle is inscribed in a right-angled triangle ABC right-angled at A.Find the area of the area not included in the circle but included in the triangle if AB = 6cm and BC = 6cm and I is the centre of the circle.