- This is an assessment test.
- These tests focus on geometry and mensuration and are meant to indicate your preparation level for the subject.
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## Geometry and Mensuration: Test 27

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Question 1 |

Each internal angle of regular polygon is two times its external angle. Then the number of sides of the polygon is:

8 | |

6 | |

5 | |

7 |

Question 1 Explanation:

The interior angle + exterior angle =180

Thus the interior angle = 120

The sum of all the exterior angles = 360

The number of sides = 360

Correct option is (b)

^{o}Thus the interior angle = 120

^{o}and each exterior angle Â = 60^{o}The sum of all the exterior angles = 360

^{0}The number of sides = 360

^{0}/60^{0}= 6Correct option is (b)

Question 2 |

Ratio of the number of sides of two regular polygons is 5: 6 and the ratio of their each interior angle is 24: 25. Then the number of sides of these tow polygons are

20, 24 | |

15, 18 | |

10, 12 | |

5, 6 |

Question 2 Explanation:

Question 3 |

Measure of each interior angle of a regular polygon can never be;

150 ^{o} | |

105 ^{o} | |

108 ^{o} | |

144 ^{o} |

Question 3 Explanation:

Question 4 |

ABCD is a cyclic quadrilateral and O is the center of the circle. If âˆ COD= 140

^{o}and âˆ BAC= 40^{o}, then the value of âˆ BCD is equal to70 ^{o} | |

90 ^{o} | |

60 ^{o} | |

80 ^{o} |

Question 4 Explanation:

âˆ COD=140, thusâˆ CAD= Â½ of âˆ COD= 70.

âˆ BAD= 40+70=110.

âˆ BCD= 180-110=70.

Correct option is (a)

Question 5 |

ABCD is a cyclic trapezium with AB||DC and AB= diameter of the circle. If âˆ CAB= 30

^{o}, then âˆ ADC is60 ^{o} | |

120 ^{o} | |

150 ^{o} | |

30 ^{o} |

Question 5 Explanation:

Here âˆ CAB = 30

^{0}and âˆ ACB = 90

^{0}Hence âˆ ABC = 60

^{0}

Now âˆ ADC + âˆ ABC = 180

^{0}

âˆ ADC = 180

^{0}- 60

^{0}= 120

^{0}

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