- 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 15

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*Geometry and Mensuration: Test 15*.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 |

In a right-angled triangle, the product of two sides is equal to half of the square of the third side i.e,. hypotenuse. One of the acute angles must be

60 ^{o} | |

30 ^{o} | |

45 ^{o} | |

15 ^{o} |

Question 1 Explanation:

$ \displaystyle \begin{array}{*{35}{l}} Let\text{ }the\text{ }three\text{ }sides\text{ }be\text{ }a,b,c\text{ }units. \\ Where\text{ }c\text{ }is\text{ }the\text{ }hypotenuse. \\ \begin{array}{l}~ab=\frac{1}{2}{{c}^{2}}\\Thus,2ab={{a}^{2}}+{{b}^{2}}Thus\,\,we\,\,can\,\,see\,\,that\,\,a=b.\\i.e.\,\,One\,\,of\,the\,\,angles\,\,is\,\,45.\\The\text{ }correct\text{ }option\text{ }is\text{ }\left( c \right)\end{array} \\ \end{array}$

Question 2 |

A point D is taken from the side BC of a right-angled triangle ABC, where AB is hypotenuse. Then

AB ^{2} +CD^{2}= BC^{2} +AD^{2} | |

CD ^{2} +BD^{2}= 2 AD^{2} | |

AB ^{2} +AC^{2}= 2 AD^{2} | |

AB ^{2}= AD^{2} +BD^{2} |

Question 3 |

In ABC, the internal bisectors of ∠ABC and ∠ACB meet at I and ∠BAC= 50

^{o}. The measure of ∠BIC is105 ^{o} | |

115 ^{o} | |

125 ^{o} | |

130 ^{o} |

Question 4 |

Q is a point in the interior of a rectangle ABCD. If QA= 3 cm, QB=4 cm and QC= 5 cm, then the length QD in centimeter is

$ \displaystyle 3\sqrt{2}$ | |

$\displaystyle 5\sqrt{2}$ | |

$ \displaystyle \sqrt{34}$ | |

$ \displaystyle \sqrt{41}$ |

Question 5 |

Each edge of a regular tetrahedron is 3 cm, then its volume is

$ \displaystyle \frac{9\sqrt{2}}{4}\,\,c.\,c.$ | |

$ \displaystyle 27\sqrt{3}\,\,c.\,\,c.$ | |

$ \displaystyle \frac{4\sqrt{2}}{9}\,\,c.\,\,c.$ | |

$ \displaystyle 9\sqrt{3}\,\,c.\,\,c.$ |

Question 5 Explanation:

$ \begin{array}{l}The\,\,formula\,\,for\,this\\\frac{{{a}^{3}}}{6\sqrt{2}}=\frac{27}{6\sqrt{2}}=\frac{3\times 3\times 3}{3\times 2\times \sqrt{2}}\\=\frac{3\times 3\times 3}{3\times 2\times \sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}\\=\frac{9\sqrt{2}}{4}\end{array}$

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