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## Geometry and Mensuration: Level 3 Test 6

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Question 1 |

The length of the common chord of two circles of radii 15 cm and 20 cm whose centers are 25 cm apart, is (in cm):

24 | |

25 | |

15 | |

20 |

Question 1 Explanation:

$ \begin{array}{l}Let\,\,the\,\,total\,\,length\,\,be\,\,2x\,\,units\\\frac{1}{2}x\times 25=\frac{1}{2}\times 15\times 20\\x=12units\\thus\,\,2x=24\,units\\Thus\,\,the\,\,correct\,\,option\,\,is\,\,a\end{array}$

Question 2 |

In a triangle ABC, the internal bisector of the angle A meets BC at D. If AB =4, AC= 3 and âˆ A= 60

^{o}, then, the length of AD is:$ \displaystyle 2\sqrt{3}$ | |

$ \displaystyle \frac{12\sqrt{3}}{7}$ | |

$ \displaystyle \frac{15\sqrt{3}}{8}$ | |

$ \displaystyle \frac{6\sqrt{3}}{7}$ |

Question 2 Explanation:

$ \begin{array}{l}let\,\,the\,\,length\,\,of\,\,AD\,\,be\,\,x\,\,units\\\frac{1}{2}\times 4x\times Sin30=\frac{4}{7}\times 4\times 3\times Sin60\\x=\frac{12}{7}\sqrt{3}\\Correct\,\,option\,\,is\,\,b\end{array}$

Question 3 |

In the figure (not drawn to scale) given below, P is a point on AB such that AP: PB= 4: 3. PQ is parallel to AC and QD is parallel to CP. In âˆ ARC, âˆ ARC= 90

^{o}, and in Î”PSQ, âˆ PSQ= 90^{o}. The length of QS is 6 cms. What is ratio AP: PD?10: 3 | |

2: 1 | |

7: 3 | |

8: 3 |

Question 3 Explanation:

$ \displaystyle \begin{array}{l}AP:\text{ }PB\text{ }=\text{ }4:3\\Thus\text{ }AP=\text{ }4x\text{ }and\text{ }PB=3x\\Again,\text{ }PD:DB\text{ }=\text{ }CQ:AB\text{ }=\text{ }4:3\\PD=\frac{4}{7}\times BP=\frac{4}{7}\times \frac{3}{4}AP=\frac{3}{7}AP\end{array}$

Question 4 |

32 ^{o} | |

84 ^{o} | |

64 ^{o} | |

Cannot be determined |

Question 4 Explanation:

Let angle CAD= DCA = x

and thus CDB = 2x and CBD = 2x.

Thus, 180-2x+x+96= 180

=> 2x= 64

Correct option is (c)

and thus CDB = 2x and CBD = 2x.

Thus, 180-2x+x+96= 180

=> 2x= 64

Correct option is (c)

Question 5 |

In the figure given below (not drawn to scale), A, B and C are three points on a circle with centre O, The chord BA is extended to a points T such that CT becomes a tangent at the circleÂ at pointÂ C. If âˆ ATC= 30

^{o}and âˆ ACT= 50^{o}, then the angle BOA is:100 ^{o} | |

150 ^{o} | |

80 ^{o} | |

Not possible to determine |

Question 5 Explanation:

Since angle âˆ CAT = 180-50-30 = 100

By alternate segment theorem âˆ ABC = âˆ TCA = 50

Thus âˆ BOA =2{180-50-(180-100)}=100

Correct option is (a)

By alternate segment theorem âˆ ABC = âˆ TCA = 50

^{o}Thus âˆ BOA =2{180-50-(180-100)}=100

^{o}Correct option is (a)

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