Question

in fig a circle is inscribed in an equilateral triangle abc of side 12cm find the radius of inscribed circle and the area of the shaded region
Use pie 3.14 and rootunder 3 =1.73

Answer 1

Given that ABC is an equilateral triangle of side 12 cm.  Construction:Join O and A, O and B, and O and C. 

Now.

the points on BC, CA and AB respectively then, OP⊥BC OQ⊥AC OR⊥AB Assume the radius of the circle as r cm. Area of ∆AOB + Area of ∆BOC + Area of ∆AOC = Area of ∆ABC ⇒( 1/2× AB × OR) + (1/2× BC × OP) + (1/2× AC × OQ) = √3/4× (side)2. ⇒ (1/2× 12 × r) + (1/2× 12 × r) + (1/2× 12 × r) = √3/4× (12)2 ⇒ 3 × 1/2× 12 × r = √3/4× 12 × 12 ⇒ r = 2√3 = 2 × 1.73 = 3.46 Hence, the radius of the inscribed circle is 3.46 cm. Area of the shaded region = Area of ∆ABC − Area of the inscribed circle = [√3/4×(12)2 − π(2√3)2] = [36√3 − 12π]  = [36 × 1.73 − 12 × 3.14]  = [62.28 − 37.68] = 24.6 cm2 ∴ The area of the shaded region is 24.6 cm2

Answer 2

Answer:

Given that ABC is an equilateral triangle of side 12 cm. 

Construction:Join O and A, O and B, and O and C.

P, Q, R are the points on BC, CA and AB respectively then,

OP⊥BC

OQ⊥AC

OR⊥AB

Assume the radius of the circle as r cm.

Area of ∆AOB + Area of ∆BOC + Area of ∆AOC = Area of ∆ABC

⇒( 1/2× AB × OR) + (1/2× BC × OP) + (1/2× AC × OQ) = √3/4× (side)2.

⇒ (1/2× 12 × r) + (1/2× 12 × r) + (1/2× 12 × r) = √3/4× (12)2

⇒ 3 × 1/2× 12 × r = √3/4× 12 × 12

⇒ r = 2√3 = 2 × 1.73 = 3.46

Hence, the radius of the inscribed circle is 3.46 cm.

Area of the shaded region = Area of ∆ABC − Area of the inscribed circle

= [√3/4×(12)2 − π(2√3)2]

= [36√3 − 12π] 

= [36 × 1.73 − 12 × 3.14] 

= [62.28 − 37.68] 

= 24.6 cm2

∴ The area of the shaded region is 24.6 cm2