A circle is inscribed in an equilateral triangle ABC is side 12 cm, touching its sides (the following figure). Find the radius of the inscribed circle and the area of the shaded part.
Answers
Answer:
The radius of the inscribed circle is 2√3 cm and Area of the shaded region is 24.64 cm².
Step-by-step explanation:
Given : ABC is an equilateral triangle of side 12 cm.
Join OA, OB, and OC. O is the incentre of a circle.
OP, OR & OQ are Radius of a circle and they are equal .
Let OP = OR = OQ = r
Area of ∆AOB + Area of ∆BOC + Area of ∆AOC = Area of ∆ABC
(½ × AB × OR) + (½ × BC × OP) + (½ × AC × OQ) = √3/4 × side²
[Area of ∆ = ½ × base × height , Area of equilateral ∆ = √3/4 side²]
√3/4× 12² = (½ × 12 × r) + (½ × 12 × r) + (½ × 12 × r)
√3/4 × 144 = (½ × 12 × r) (1 + 1 +1 )
√3/4 × 144 = (½ × 12 × r) × 3
√3 × 36 = 18r
r = √3 × 36/18
r = 2√3
Radius of the inscribed circle = 2√3 cm.
Area of the shaded region = Area of ∆ABC − Area of the inscribed circle
= √3/4 side² - πr²
= [ √3/4 × 12² − π(2√3)²
= √3 × 36 - π × 12
= 36√3 − 12π
= 36 × 1.732 − 12 × 22/7 ]
= 62.35 - 264/7
= 62.35 − 37.71
= 24.64 cm²
Area of the shaded region = 24.64 cm²
Hence, the radius of the inscribed circle is 3.46 cm and Area of the shaded region is 24.64 cm².
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