- 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 10

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*Geometry and Mensuration: Test 10*.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 |

The length of a rectangle is three fifth of the side of a square. The radius of a circle is equal to side of the square. The circumference of the circle is 132 cm. What is the area of the rectangle if the breadth of the rectangle is 8 cm?

112.4 sq. cm. | |

104.2 sq. cm. | |

100.8 sq. cm. | |

Cannot be determined |

Question 1 Explanation:

$ \begin{array}{l}2\pi r=132\\\Rightarrow r=21cm\\Side\,\,of\,\,the\,\,square=21cm\\Length\,\,of\,\,the\,\,rec\tan gle=\frac{3}{5}\times 21=12.6cm\\Breadth=8cm\\Area=12.6\times 8=100.8c{{m}^{2}}\end{array}$

Question 2 |

Smallest side of a right angled triangle is 8 cm less than the side of a square of perimeter 56 cm. Second largest side of the right angled triangle is 4 cm less than the length of rectangle of area 96 sq. cm and breadth 8 cm. What is the largest side of the right angled triangle?

20 cm | |

12 cm | |

10 cm | |

15 cm |

Question 2 Explanation:

$ \displaystyle \begin{array}{l}Side\text{ }of\text{ }{{1}^{st}}square\text{ }=\text{ }56/4\text{ }=\text{ }14\text{ }cm.\\Smallest\text{ }side\text{ }of\text{ }right\text{ }angled\text{ }triangle=\text{ }14\text{ }-8~=\text{ }6\text{ }cm.\\length\text{ }of\text{ }{{2}^{nd}}rectangle\text{ }=\text{ }96/8\text{ }=\text{ }12\text{ }cm.\\Second\text{ }largest\text{ }side\text{ }of\text{ }the\text{ }{{1}^{st}}rectangle\text{ }=~12-4\text{ }=\text{ }8\text{ }cm.\\largest\text{ }side\text{ }=\text{ }hypotenuse=\sqrt{{{8}^{2}}+{{6}^{2}}}=10cm~\end{array}$

Question 3 |

In a swimming pool measuring 90 m by 40 m. 150 men take a dip. If the average displacement of water by a man is 8 cu m, what will be the rise in water level?

33.33 cm | |

30 cm | |

20 cm | |

25 cm |

Question 3 Explanation:

Total volume displaced by men = 150X 8 = 1200 cu. m.

Area of the surface of swimming pool = 90m X 40m = 3600 sq.m.

Rise in water level = 1200/3600 m = 1/3 th of a metre = 33.33 cm

Area of the surface of swimming pool = 90m X 40m = 3600 sq.m.

Rise in water level = 1200/3600 m = 1/3 th of a metre = 33.33 cm

Question 4 |

A 4 cm cube is cut into 1 cm cubes. Find the percentage increase in surface area.

200% | |

100% | |

400% | |

300% |

Question 4 Explanation:

Total surface area of 4 cm cube= 6a

=6 x 4 x 4 = 96cm

Number of new cubes = 4

Total surface area of new cube = 64x6x1

Increase = 384-96 = 288

Increase Percentage = 288/96 x 100 = 300%

^{2}=6 x 4 x 4 = 96cm

^{2}Number of new cubes = 4

^{3}/1^{3 }= 64Total surface area of new cube = 64x6x1

^{2}= 384cm^{2}Increase = 384-96 = 288

Increase Percentage = 288/96 x 100 = 300%

Question 5 |

The radius of the base of a conical tent is 5 m. If the tent is 12 m high, then the area of the canvas required in making the tent is

300 Π ^{2 } | |

60 Π ^{2} | |

90 Π ^{2 } | |

None of these |

Question 5 Explanation:

$ \displaystyle \begin{array}{l}The\text{ }lateral\text{ }length=\sqrt{{{12}^{2}}+{{5}^{2}}}=13m\\Lateral\text{ }surface\text{ }area=\pi \times r\times l\\=\pi \times 5\times 13=65\pi {{m}^{2}}\end{array}$

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