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## Geometry and Mensuration: Level 3 Test 7

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*Geometry and Mensuration: Level 3 Test 7*. 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 figure shows the rectangle ABCD with a semicircle and a circle inscribed inside in it as shown. What is the ratio of the area of the circle to that of the semicircle?

$ \displaystyle {{\left( \sqrt{2}-1 \right)}^{2}}:1$ | |

$ \displaystyle 2{{\left( \sqrt{2}-1 \right)}^{2}}:1$ | |

$ \displaystyle {{\left( \sqrt{2}-1 \right)}^{2}}:2$ | |

None of these |

Question 1 Explanation:

$ \displaystyle \begin{array}{l}Let\,\,R\,\,be\,\,the\,\,radius\,\,of\,\,the\,\,semicircle\,\\Therefore\,\,OC=R=CB\\\sin ce\,\,OCB\,\,is\,a\,\,right\,\,angle\,\,triangle\,\,\\so\,\,OB=R\sqrt{2}\\Now\,\,area\,\,of\,semicircle=\frac{\pi {{r}^{2}}}{2}=\frac{\pi {{R}^{2}}}{2}\\Now\,\,FB=OB-OF=R\sqrt{2}-R=R\left( \sqrt{2}-1 \right)\\therfore\,\,the\,\,area\,\,of\,\,circle=\frac{\pi {{R}^{2}}{{\left( \sqrt{2}-1 \right)}^{2}}}{{{2}^{2}}}\\therfore\,\,the\,\,required\,\,ratio=\frac{\frac{\pi {{R}^{2}}{{\left( \sqrt{2}-1 \right)}^{2}}}{{{2}^{2}}}}{\frac{\pi {{R}^{2}}}{2}}={{\left( \sqrt{2}-1 \right)}^{2}}:2\end{array}$

Question 2 |

A wooden box (open at the top) of thickness 0•5 cm, length 21 cm, width 11 cm and height 6 cm is painted on the inside. The expenses of painting are Rs.70. What is the rate of painting per square centimeter?

Re. 0•7 | |

Re 0•5 | |

Re 0•1 | |

Re 0•2 |

Question 2 Explanation:

The actual surface area painted = 2{(11-1)(21-1)+20X5+10X5} = 700 sq. units.

Thus cost per sq. cm =70/700 =0.10 Rs

Correct option is (c)

Thus cost per sq. cm =70/700 =0.10 Rs

Correct option is (c)

Question 3 |

10 | |

11 | |

12 | |

None of these |

Question 3 Explanation:

$ \begin{array}{l}Now\,\,AB=\sqrt{{{\left( x+3 \right)}^{2}}+{{(x-3)}^{2}}}=\sqrt{2({{x}^{2}}+9)}\\BD=\sqrt{100-{{(x-3)}^{2}}}=\sqrt{2\left( {{x}^{2}}+9 \right)}\\BD=\sqrt{100-{{(x-3)}^{2}}}\\thus\\\sqrt{100-(x-3){}^{2}}+x=\sqrt{2({{x}^{2}}+9)}\\Putting\text{ }the\text{ }values\text{ }of\text{ }x\text{ }none\text{ }of\text{ }them\text{ }satisfy\text{ }the\text{ }equation.\\Thus\text{ }the\text{ }correct\text{ }option\text{ }is\text{ }\left( d \right)\end{array}$

Question 4 |

A certain city has a circular wall around it, and this wall has four gates pointing north south, east and west. A house stands outside the city, three kms north of the north gate, and it can just be seen from a point nine kms east of the south gate. What is the diameter of the wall that surrounds the city?

6 km | |

9 km | |

12 km | |

None of these |

Question 4 Explanation:

$\displaystyle \begin{array}{l}Let\text{ }the\text{ }radius\text{ }be\text{ }r\text{ }and\text{ }the\text{ }center\text{ }be\text{ }0,0\\Now\text{ }since\text{ }between\text{ }the\text{ }point\text{ }from\text{ }where\text{ }it\text{ }can\text{ }be\text{ }just\text{ }seen\text{ }\left( -9,-r \right)\\\sqrt{6r+9}+9=\sqrt{{{(2r+3)}^{2}}+{{9}^{2}}}\\By\,\,putting\,the\,\,values\,\,only\,\,b\,\,satisfies\end{array}$

Question 5 |

Consider a cylinder of height h cm and radius r = r/2cm as shown in the figure (not drawn to scale).

A string of a certain length, when wound on its cylindrical surface, starting at point A and ending at

point B, gives a maximum of n turns (in other words, the string's length is the minimum length required

to wind n turns.

What is the vertical spacing in ems between two consecutive turns

A string of a certain length, when wound on its cylindrical surface, starting at point A and ending at

point B, gives a maximum of n turns (in other words, the string's length is the minimum length required

to wind n turns.

What is the vertical spacing in ems between two consecutive turns

h/n | |

h/√n | |

h/h ^{2} | |

Cannot be determined with given information |

Question 5 Explanation:

Opening up the cylinder into a rectangle we can see that the

string is wound to a height of h with equal vertical spacing for n times.

Thus the distance = h/n

Correct option is (a)

string is wound to a height of h with equal vertical spacing for n times.

Thus the distance = h/n

Correct option is (a)

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