- This is an assessment test.
- These tests focus on geometry and mensuration and are meant to indicate your preparation level for the subject.
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## Geometry and Mensuration: Test 11

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Question 1 |

ABC is a right angled triangle, right angled at C and p is the length of the perpendicular from C on AB. If a, b and c are the lengths of the sides BC, CA and AB respectively, then

$ \displaystyle \frac{1}{{{p}^{2}}}=\frac{1}{{{b}^{2}}}-\frac{1}{{{a}^{2}}}$ | |

$ \displaystyle \frac{1}{{{p}^{2}}}=\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}$ | |

$ \displaystyle \frac{1}{{{p}^{2}}}+\frac{1}{{{a}^{2}}}=\frac{1}{{{b}^{2}}}$ | |

$ \displaystyle \frac{1}{{{p}^{2}}}=\frac{1}{{{a}^{2}}}-\frac{1}{{{b}^{2}}}$ |

Question 1 Explanation:

Question 2 |

$ \displaystyle \begin{array}{l}If\vartriangle ABC\text{ }is\text{ }similar\text{ }to\vartriangle DEF\text{ }such\text{ }that\text{ }BC=\text{ }3\text{ }cm,\text{ }\\EF=\text{ }4cm\text{ }and\text{ }area\text{ }of\vartriangle ABC=\text{ }54\text{ }c{{m}^{2}},\text{ }\\then\text{ }the\text{ }area\text{ }of\vartriangle DEF\text{ }is:\end{array}$

66 cm ^{2} | |

78 cm ^{2} | |

96 cm ^{2} | |

54 cm ^{2} |

Question 2 Explanation:

In case of similar triangles, the ratio of the areas of the triangles is equal to the square of the sides in proportion.

The ratio of the area of Î”ABCÂ to Î”DEF = 3

The Area of Î”DEF= (16/9)x54 = 96 cm

The ratio of the area of Î”ABCÂ to Î”DEF = 3

^{2}: 4^{2}= Â 9:16The Area of Î”DEF= (16/9)x54 = 96 cm

^{2}Thus, the correct option (c)Question 3 |

$ \displaystyle \begin{array}{l}In\,ABC,\angle A=\text{ }{{90}^{o}}and\,AD\bot BC\,where\text{ }D\text{ }lies\text{ }on\text{ }BC.\text{ }\\If\text{ }BC=\text{ }8\text{ }cm,\text{ }AC=\text{ }6cm,\text{ }\\then\,\vartriangle ABC:\,\vartriangle ACD=\text{ }?\end{array}$

4: 3 | |

25: 16 | |

16: 9 | |

25: 9 |

Question 3 Explanation:

From the figure the Î”ABC and Î”ACD are similar triangles by AAA similarity.

In case of similar triangles, the ratio of the areas of the triangles is equal to the square of the sides in proportion.

We have Î”ABC/Î”ACD =8

^{2}/6

^{2}= 16/9

Question 4 |

The areas of two similar triangles ABC and DEF are 20 cm

^{2}and 45 cm^{2}respectively. If AB= 5cm, then DE is equal to:6.5 cm | |

7.5 cm | |

8.5 cm | |

5.5 cm |

Question 4 Explanation:

In case of similar triangles, the ratio of the areas of the triangles is equal to the square of the sides in proportion.

Hence we have AB

AB/DE = 2/3

Now if AB = 5 cm, then DE = 5 x 3/2 = 7.5cm

Hence we have AB

^{2}/DE^{2}= 20/45 = 4/9AB/DE = 2/3

Now if AB = 5 cm, then DE = 5 x 3/2 = 7.5cm

Question 5 |

130 ^{o} | |

80 ^{o} | |

100 ^{o} | |

120 ^{o} |

Question 5 Explanation:

We have âˆ ABC + âˆ BAC + âˆ ACB = 180

^{0}

=> 5âˆ ACB + 3âˆ ACB + âˆ ACB = 180

^{0}

=> 9 âˆ ACB = 180

^{0}

=>âˆ ACB = 20

^{0}

So âˆ ABC = 5 âˆ ACB = 5 Ã— 20 = 100

^{0}.

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