- This is an assessment test.
- These tests focus on geometry and mensuration and are meant to indicate your preparation level for the subject.
- Kindly take the tests in this series with a pre-defined schedule.

## Geometry and Mensuration: Test 14

Please wait while the activity loads.

If this activity does not load, try refreshing your browser. Also, this page requires javascript. Please visit using a browser with javascript enabled.

If this activity does not load, try refreshing your browser. Also, this page requires javascript. Please visit using a browser with javascript enabled.

Congratulations - you have completed

*Geometry and Mensuration: Test 14*.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 difference between the exterior and interior angles at a vertex of a regular polygon is 150

^{o}. The number of sides of the polygon is10 | |

15 | |

24 | |

30 |

Question 1 Explanation:

Question 2 |

The length of the diagonal BD of the parallelogram ABCD is 18 cm. If P and Q are the centroid of the ΔABC and ΔADC respectively then the length of the line segment PQ is

4 cm | |

6 cm | |

9 cm | |

12 cm |

Question 3 |

ABCD is a parallelogram where AB is parallel to CD. BC is produced to Q such that BC= CQ, where P is the point on DC

area Δ( BCP)= areaΔ ( DPQ) | |

area Δ( BCP) >areaΔ ( DPQ) | |

area Δ( BCP) | |

area Δ( BCP) +areaΔ( DPQ) |

Question 3 Explanation:

We know that: Triangles ΔAPC and ΔBCP have the common base PC.

Also, the two triangles ΔAPC and ΔBCP are located between the same two parallel lines: AB and PC

Thus, we can say: AD||CQ and AD=CQ……………………….i

From the above, we can conclude ADQC is a parallelogram.

Also, triangles ΔADC and ΔDAQ have the common base: AD

Also, the two triangles ΔADC and ΔDAQ are located between the same two parallel lines: AD and CQ.

Thus, we can conclude: Area of Δ( ADC)= Area of Δ( ADQ)

Let’s subtract area of ( DADP) from both sides: area of Δ( APC)= area of Δ( DPQ)……..(ii)

From (i) and (ii), we can conclude: area of Δ( BPC) = area of Δ( DPQ)

Question 4 |

ABCD is a rhombus. A straight line through C cuts AD produced at P and AB produced at Q. if DP= ½ AB, Then the ratio of the lengths of BQ and AB is

2: 1 | |

1: 2 | |

1: 1 | |

3: 1 |

Question 5 |

In a quadrilateral ABCD, with unequal sides if the diagonals AC and BD intersect at right angles, then

AB ^{2} +BC^{2}= DC^{2} +DA^{2} | |

AB ^{2} +CD^{2} = BC^{2}+ DA^{2} | |

AB ^{2} +AD^{2}= BC^{2}+ CD^{2} | |

AB ^{2}+ BC^{2} =2(CD^{2} +DA^{2}) |

Question 5 Explanation:

$ \begin{array}{l}Let\text{ }the\text{ }diagonals\text{ }intersect\text{ }at\text{ }O.\\A{{O}^{2}}+B{{O}^{2}}=A{{B}^{2}}~,~~\\B{{O}^{2}}+C{{O}^{2}}=B{{C}^{2}}~\\C{{D}^{2}}=O{{C}^{2}}+O{{D}^{2}}\\A{{D}^{2}}=A{{O}^{2}}+A{{D}^{2}}\\From\text{ }the\text{ }equations\text{ }we\text{ }can\text{ }check\text{ }the\text{ }options\text{ }.\\A{{B}^{2}}+C{{D}^{2}}=B{{C}^{2}}+D{{A}^{2}}=B{{O}^{2}}+O{{D}^{2}}+A{{O}^{2}}+C{{O}^{2}}\\We\text{ }can\text{ }conclude\text{ }that\text{ }correct\text{ }option\text{ }is\text{ }\left( b \right)\end{array}$

Once you are finished, click the button below. Any items you have not completed will be marked incorrect.

There are 5 questions to complete.

List |