Important for :

1

D

Given that, radius (r) = 36 cmAnd, Area of sector = 72π cm

⇒ |
πr^{2}Θ |
= 72π |

360° |

∴ Θ = | 72π × 360° | = | 72 × 360° | = 20° |

πr^{2} |
36 × 36 |

Now, length of arc = | πrΘ | = | π × 36 × 20° | = 4π cm |

180° | 180° |

Hence, optiion D is correct.

2

B

Given that, Diameter = 2a

For inscribed square,

Diameter of circle = Diagonal of inner square

For circumscribed square,

Diameter of circle = Side of outer square

∴ Area of inner square = | 1 |
(diagonal)^{2} = |
1 |
× (2a)^{2} = 2a^{2} |

2 | 2 |

And, Area of outer square = (side)

Now, Required difference = 4a

Hence, option B is correct.

3

C

In ΔABC, by Pythagoras theorem,

BC = |
AB^{2} + AC^{2} |
= |
6^{2} + 8^{2} |
= | 100 | = 10 cm |

Now, Area of that semi-circle which diameter is AB = |
π(3)^{2} |

2 |

∴ x = | 9π |
cm^{2} |

2 |

Similarly, Area of that semi-circle which diameter is AC = |
π(4)^{2} |

2 |

∴ y = | 16π |
cm^{2} |

2 |

Similarly, Area of that semi-circle which diameter is BC = |
π(5)^{2} |

2 |

∴ z = | 25π |
cm^{2} |

2 |

Now, x + y – z = | ( | 9π | + | 16π | ) | – | 25π | = 0 |

2 | 2 | 2 |

Hence, option C is correct.

4

C

Perimeter of all triangles = (3 × 1) + (3 × 0.5) + (3 × 0.25) + (3 × 0.125)

= 3 + 1.5 + 0.75 + 0.375 = 5.625 ≈ 6 units

Hence, option C is correct.

5

B

Required ratio = Area of circle : Area of ΔACD

= πr^{2} : |
1 | × 2r × r = π |

2 |

Hence, option B is correct.