∵ 1/2 x 2 √ x h

= √3/4 x (2 √3)²

∴ h = 3 cm.

Correct Option: D

∵ 1/2 x 2 √ x h

= √3/4 x (2 √3)²

∴ h = 3 cm.

Let circumference = 100 cm .
Then, ∵ 2πr = 100
⇒ r = 100/2π
=50/π

⇒ New circumference
= 105 cm

Then, 2πR = 105
⇒ R = 105 / (2π)
&rArr Original area = [ π x (50/π) x (50/π) ]
= 2500/π cm²

⇒ New Area = [π x (105/2π) x (105/2π)]
= 11025 / (4π) cm²

⇒ Increase in area = [11025/(4π)] - 2500/π cm²
= 1025 / 4π cm²

Correct Option: B

Let circumference = 100 cm .
Then, ∵ 2πr = 100
⇒ r = 100/2π
=50/π

⇒ New circumference
= 105 cm

Then, 2πR = 105
⇒ R = 105 / (2π)
&rArr Original area = [ π x (50/π) x (50/π) ]
= 2500/π cm²

⇒ New Area = [π x (105/2π) x (105/2π)]
= 11025 / (4π) cm²

⇒ Increase in area = [11025/(4π)] - 2500/π cm²
= 1025 / 4π cm²

Required increase percent [1025/(4π)] x 2500/π x 100 = 41/4%
= 10.25%

Angle swept in 30 min= 180°

Correct Option: D

Angle swept in 30 min= 180°
Area swept = [(22/7) x 7 x 7] x [180°/360°] cm²
= 77 cm²

∵ 22/7 x r² = 462
⇒ r² = (462 x 7) /22 = 147
⇒ r = 7√3 cm

∴ Height of the triangle = 3r = 21√3 cm
Now, ∵ a² = a²/4 + (3r)²
⇒ 3a²/4 = (21√3)²
⇒ a² = (1323 x 4)/3
⇒ a = 21x 2 = 42 cm

∴ Perimeter = 3a = 3 x 42
=126 cm

Correct Option: B

∵ 22/7 x r² = 462
⇒ r² = (462 x 7) /22 = 147
⇒ r = 7√3 cm

∴ Height of the triangle = 3r = 21√3 cm
Now, ∵ a² = a²/4 + (3r)²
⇒ 3a²/4 = (21√3)²
⇒ a² = (1323 x 4)/3
⇒ a = 21x 2 = 42 cm

∴ Perimeter = 3a = 3 x 42
=126 cm

Area of park = 100 x 100 = 10000 m²
Area of circular lawn = Area of park - area of park excluding circular lawn
= 10000 - 8614
= 1386

Now area of circular lawn = (22/7) x r² = 1386 m²

Correct Option: A

Area of park = 100 x 100 = 10000 m²
Area of circular lawn = Area of park - area of park excluding circular lawn
= 10000 - 8614
= 1386

Now again area of circular lawn = (22/7) x r² = 1386 m²
⇒ r² = (1386 x 7) / 22
= 63 x 7
= 3 x 3 x 7 x 7

∴ r = 21 m