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## Geometry and Mensuration: Test 2

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Question 1 |

The area of a rectangular field is 460 square metres. If the length is 15 per cent more than the breadth, what is breadth of the rectangular field?

15 metres | |

26 metres | |

34.5 metres | |

none |

Question 1 Explanation:

Let the breadth = b cm. Therefore, length = 1.15b cm.

Thus, $ \begin{array}{l}1.15{{b}^{2}}=460\\=>b=20\,cm\end{array}$

Thus, $ \begin{array}{l}1.15{{b}^{2}}=460\\=>b=20\,cm\end{array}$

Question 2 |

A sector of a circle of radius 15 cm has the angle 120°. It is rolled up so that two bounding radii are joined together to form a cone. The volume of the cone is

$ \displaystyle \left( 250\sqrt{2}\pi \right)\,c{{m}^{3}}$ | |

$ \displaystyle \left( 500\sqrt{2} \right)\frac{\pi }{3}\,c{{m}^{3}}$ | |

$ \displaystyle \left( 250\sqrt{2} \right)\frac{\pi }{3}\,c{{m}^{3}}$ | |

$ \displaystyle \left( 1000\sqrt{2} \right)\frac{\pi }{3}\,c{{m}^{3}}$ |

Question 2 Explanation:

The circumference of the sector

$ \begin{array}{l}\frac{1}{3}\times 2\pi 15=10\pi \\2\pi r=10,r=5\\Radius\,\,of\,cone=5\pi \,cm\\Lateral\,\,Height\,=15\,cm\\Height=\sqrt{{{15}^{2}}-{{25}^{2}}}\\=10\sqrt{2\,}\,cm\\Volume\,of\,cone=\frac{1}{3}\pi {{r}^{2}}h\\=\frac{1}{3}\pi \times 25\times 10\sqrt{2}\\=\left( 250\sqrt{2} \right)\frac{\pi }{3}c{{m}^{3}}\end{array}$

$ \begin{array}{l}\frac{1}{3}\times 2\pi 15=10\pi \\2\pi r=10,r=5\\Radius\,\,of\,cone=5\pi \,cm\\Lateral\,\,Height\,=15\,cm\\Height=\sqrt{{{15}^{2}}-{{25}^{2}}}\\=10\sqrt{2\,}\,cm\\Volume\,of\,cone=\frac{1}{3}\pi {{r}^{2}}h\\=\frac{1}{3}\pi \times 25\times 10\sqrt{2}\\=\left( 250\sqrt{2} \right)\frac{\pi }{3}c{{m}^{3}}\end{array}$

Question 3 |

A toy is in the shape of a hemisphere surmounted by a cone. If radius of base of the cone is 3 cm and its height is 4 cm, the total surface area of the toy is___Πcm

^{2}33 | |

42 | |

66 | |

56 |

Question 3 Explanation:

$ \displaystyle \begin{array}{l}Slant\,Height=l=\sqrt{{{4}^{2}}+{{3}^{2}}}=5\\Surface\,area\,of\,hemi\,sphere\\=2\pi {{\left( 3 \right)}^{2}}=18\pi \,\,c{{m}^{2}}\\Surface\,\,\,\,area\,\,\,of\,\,\,cone\,\,=\pi \,rl\\=\pi \,\times 3\times 5=15\,\pi \,\,c{{m}^{2}}\\Therefore\,required\,area=18\pi +15\pi =33\pi cm{}^{2}\end{array}$

Question 4 |

What will be the cost of gardening 1 metre broad boundary around a rectangular plot having perimeter of 340 metres at the rate of Rs.10 per square metre?

Rs.3, 400/- | |

Rs.1, 700/- | |

Rs.3, 440/- | |

Cannot be determined |

Question 4 Explanation:

Given Perimeter is = 340 m

Let the length and breadth of the plot are l and b respectively.

Now total length = l+2

Total width = b+2

So are of the boundary = [(l+2)(b+2) – lb]

2(l+b)+4

344

Total cost = Rs. 3440

Let the length and breadth of the plot are l and b respectively.

Now total length = l+2

Total width = b+2

So are of the boundary = [(l+2)(b+2) – lb]

2(l+b)+4

344

Total cost = Rs. 3440

Question 5 |

What will be the area (in square metres) of 1.5 metre wide garden developed around all the four sides of a rectangular field having area equal to 300 square metres and breadth equal to three-fourth of the length?

96 | |

105 | |

114 | |

Cannot be determined |

Question 5 Explanation:

Let Length = p cm and breadth = 3p/4 cm.

Total area = 300 = 3p

Width = 15 m.

Area of garden = {(20 + 3) × (15 + 3) - 20 × 15} = (23 × 18 - 20 × 15) = 414 - 300 = 114 m.

Correct option is (c)

Total area = 300 = 3p

^{2}/4 => p= 20 m.Width = 15 m.

Area of garden = {(20 + 3) × (15 + 3) - 20 × 15} = (23 × 18 - 20 × 15) = 414 - 300 = 114 m.

Correct option is (c)

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