- This is an assessment test.
- These tests focus on geometry and mensuration and are meant to indicate your preparation level for the subject.
- Kindly take the tests in this series with a pre-defined schedule.

## Geometry and Mensuration: Test 13

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*Geometry and Mensuration: Test 13*. You scored %%SCORE%% out of %%TOTAL%%. You correct answer percentage: %%PERCENTAGE%% . Your performance has been rated as %%RATING%% Your answers are highlighted below.

Question 1 |

If the median drawn on the base of a triangle is half its base, the triangle will be:

right-angled | |

acute-angled | |

obtuse-angled | |

equilateral |

Question 1 Explanation:

Since the median is = ½ the base therefore, the triangle has been divided into 2

Isosceles congruent triangles. So each of the triangles is 45

^{0}-45

^{0 }-90

^{0}.

Therefore the sum of the base angles of the original triangle is 45

^{0}+ 45

^{0}= 90

^{0}.

So the vertex angle must be 90 degrees.

Question 2 |

In a right-angled triangle ABC, ∠ABC= 90

^{o}, AB=5 cm and BC=12 cm. The radius of the circumcircle of the triangle ABC is7.5 cm | |

6 cm | |

6.5 cm | |

7 cm |

Question 3 |

The exterior angles obtained on producing the base BC of a triangle ABC in both ways are 120

^{o}and 105^{o}, then the vertical ∠A of the triangle is of measure36 ^{o} | |

40 ^{o} | |

45 ^{o} | |

55 ^{o} |

Question 4 |

If AD, BE and DF are medians of ΔABC, then which one of the following statements is correct?

$ \displaystyle \left( AD+BE+CF \right) | |

$ \displaystyle AD+BE+CF>AB+BC+CA$ | |

$ \displaystyle AD+BE+CF=AB+BC+CA$ | |

$ \displaystyle AD+BE+CF=\sqrt{2}\,\left( AB+BC+CA \right)$ |

Question 5 |

In a regular polygon, the exterior and interior angles are in the ratio 1: 4. The number of sides of the polygon is

10 | |

12 | |

15 | |

16 |

Question 5 Explanation:

The number of sides of the polygon is

$ \begin{array}{l}\frac{\frac{360}{n}}{\left( n-2 \right)\times \frac{180}{n}}=\frac{1}{4}\\n=10\end{array}$

$ \begin{array}{l}\frac{\frac{360}{n}}{\left( n-2 \right)\times \frac{180}{n}}=\frac{1}{4}\\n=10\end{array}$

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