- 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 20

Congratulations - you have completed

*Geometry and Mensuration: Test 20*. 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 side QR of an equilateral triangle PQR is produced to the point S in such a way that QR= RS and P is joined to S. Then the measure of ∠PSR is

30 ^{o} | |

15 ^{o} | |

60 ^{o} | |

45 ^{o} |

Question 1 Explanation:

Here ∠PRQ = 60

^{0}so ∠PRS = 120

^{0}. Now QR = PR = RS.

Hence ∠PSR = ∠RPS

Now ∠PRS + ∠PSR + ∠RPS = 180

^{0}

=>2 ∠PSR = 180

^{0}- 120

^{0}= 60

^{0}

=> ∠PSR = 30

^{0}.

Question 2 |

ABC is an isosceles triangle with AB= AC. A circle through B touching AC at the middle point intersects AB at P, Then AP: AB is:

4: 1 | |

2: 3 | |

3: 5 | |

1: 4 |

Question 2 Explanation:

$ \displaystyle \begin{array}{l}\begin{array}{*{35}{l}}
AM\text{ }is\text{ }the\text{ }tangent\text{ }to\text{ }the\text{ }circle\text{ }and \\
APB\text{ }is\text{ }a\text{ }secant\text{ }to\text{ }the\text{ }circle. \\
So~AM{}^\text{2}\text{ }=\text{ }AP\text{ }*\text{ }AB \\
\end{array}\\\begin{array}{*{35}{l}}
But\text{ }M\text{ }is\text{ }the\text{ }mid-point\text{ }of\text{ }AC\text{ }and \\
AC\text{ }=\text{ }AB \\
Thus\text{ }AM\text{ }=\text{ }\left( 1/2 \right)AB \\
\end{array}\\AM{}^\text{2}\text{ }=\text{ }\frac{1}{4}AB{}^\text{2}\\\begin{array}{*{35}{l}}
and\text{ }we\text{ }have\text{ }got\text{ }AM{}^\text{2}\text{ }=\text{ }AP\text{ }\times \text{ }AB \\
Hence,\text{ }AP:AB\text{ }=\text{ }1:4 \\
\end{array}\end{array}$

Question 3 |

$ \displaystyle \sqrt{15}\,\,a\,\,\,unit$ | |

$ \displaystyle \frac{\sqrt{15}}{2}\,\,a\,\,unit$ | |

$ \displaystyle \sqrt{17}\,\,a\,\,unit$ | |

$ \displaystyle \frac{\sqrt{17}}{2}\,\,a\,\,unit$ |

Question 4 |

30 ^{o} | |

60 ^{o} | |

45 ^{o} | |

65 ^{o} |

Question 4 Explanation:

$ \begin{array}{l}\angle DCB={{60}^{o}}\\=>\angle DCE={{30}^{o}}\\Thus,\\\angle DEC=90-30={{60}^{o}}\end{array}$

Question 5 |

The sum of interior angles of a regular polygon is 720

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

12 | |

6 | |

8 |

Question 5 Explanation:

$ \displaystyle \begin{array}{l}If\text{ }the\text{ }number\text{ }of\text{ }sides\text{ }of\text{ }regular\text{ }polygon\text{ }be\text{ }n,\text{ }then\\\left( 2n-4 \right)\times {{90}^{o}}={{720}^{o}}\\\Rightarrow 2n-4=\frac{720}{90}=8\\\Rightarrow 2n-4=8\\\Rightarrow 2n=12\\\Rightarrow n=6\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 |