- 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 17

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*Geometry and Mensuration: Test 17*.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 |

In a triangle ABC, the lengths of the sides AB, AC and BC are 3, 5 and 6 cm respectively. If a point D on BC is drawn such that the line AD bisects the ∠A internally, then what is the length of BD?

2 cm | |

2.25 cm | |

2.5 cm | |

3 cm |

Question 2 |

If P and Q are the mid-points of the sides CA and CD respectively of a triangle ABC right angled at C. Then, the value of 4 (AQ

^{2}+ BP^{2}) is equal to4 BC ^{2} | |

5 AB ^{2} | |

2AC ^{2} | |

2BC ^{2} |

Question 2 Explanation:

$ \displaystyle \begin{array}{l}A{{D}^{2}}=\text{ }A{{C}^{2}}+\text{ }C{{D}^{2}}\\In\,\,PCB,\\\begin{array}{*{35}{l}} B{{P}^{2}}=\text{ }P{{C}^{2}}+\text{ }C{{B}^{2}} \\ A{{D}^{2}}+B{{P}^{2}}=\text{ }A{{C}^{2}}+C{{D}^{2}}+P{{C}^{2}}+\text{ }C{{B}^{2}} \\ =\text{ }\left( A{{C}^{2}}+C{{B}^{2}} \right)\text{ }+\text{ }\left( C{{D}^{2}}+\text{ }P{{C}^{2}} \right) \\ =\text{ }A{{B}^{2}}=\text{ }P{{D}^{2}} \\ \end{array}\\=A{{B}^{2}}{{\left( \frac{1}{2}\,AB \right)}^{2}}\\A{{D}^{2}}+\text{ }B{{P}^{2}}=\frac{4A{{B}^{2}}+A{{B}^{2}}}{4}\\Therefore\text{ }4\text{ }\left( A{{D}^{2}}+\text{ }B{{P}^{2}} \right)\text{ }=\text{ }5A{{B}^{2}}\end{array}$

Question 3 |

If the sides of a right triangle are x,x+1 and x-1,then the hypotenuse is

5 | |

4 | |

1 | |

0 |

Question 3 Explanation:

Since the triangle is a right angled triangle, We have,
(x+1)

=> x

=> x

=> x(x-4) =0

=>x = 4 [x cannot be equal to 0 ]

Therefore Hypotenuse = 4+1 = 5

^{2}= x^{2}+(x-1)^{2}=> x

^{2}+2x +1 = x^{2}+ x^{2}-2x +1=> x

^{2}- 4x =0=> x(x-4) =0

=>x = 4 [x cannot be equal to 0 ]

Therefore Hypotenuse = 4+1 = 5

Question 4 |

Let ABC be an acute-angled triangle and CD be the altitude through C. If AB = 8 and CD= 6, then the distance between the mid-points of AD and BC is

30 | |

25 | |

27 | |

5 |

Question 4 Explanation:

Let P and Q be the mid points of AD and BC respectively.

Draw QR perpendicular on AB. Since Q is the midpoint of BC, so R is the midpoint of BD. Hence by mid point theorem QR = 3.

Also AD = 4 = DB. So DR = PD = 2 Þ PR = 4. Hence by Pythagoras theorem, we have

PQ

^{2}= PR

^{2}+ QR

^{2}= 16 + 9 = 25

=>PQ = 5.

Question 5 |

4 cm | |

√6 cm | |

3 cm | |

3.5 cm |

Question 5 Explanation:

AD = CD= 6/2 = 3 cm.

BD

. Correct option is (c)

BD

^{2}**=ADxCD = 9.**

BD= 3 cm. Correct option is (c)

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